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✓ 23 results · all verified · 8 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 15 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Dual Spaces, Bilinear and Quadratic Forms, and Sylvester's Law of Inertia

1 · Prerequisites

2 · Summary

The page builds on finite-dimensional bases and unique coordinates (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis), linear maps and their matrices (Linear map between vector spaces over the same field, Coordinate columns $[v]_{B}$ and matrices $[T]_{B}^{C}$ of linear maps relative to ordered bases), and rank-nullity (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ). The basis-extension theorem supplies the explicitly Choice-dependent infinite-dimensional separations (Zorn's lemma gives a basis between any linearly independent set and any spanning set containing it: if $L \subseteq S \subseteq V$ with $L$ independent and $span (S) = V$ , there is a basis $B$ of $V$ with $L \subseteq B \subseteq S$ ), while the finite-dimensional extension theorem avoids that cost where applicable (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ). Matrix transpose and determinant laws provide the algebra used for congruence and Schur complements (Transpose is linear and involutive, and $(A B)^{T} = B^{T} A^{T}$ , For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ ).

Dual families lead to the finite and infinite dual-space boundary, the canonical double-dual map, annihilators, and transpose identities. Bilinear forms are identified with maps into the dual; their matrices transform by congruence, while sesquilinear and Hermitian forms receive the corresponding involutive formula. Quadratic forms are defined in every characteristic, with polarization and symmetric diagonalization restricted to characteristic not two. Alternating forms acquire symplectic normal form and even rank. Over the reals, diagonal normalization and intrinsic positive and negative dimensions prove Sylvester's law of inertia; Schur-complement elimination then yields the leading-principal-minor criterion for positive definiteness.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

Linear functionals and the algebraic dual $V^{*} = L (V, F)$

Definition

Let $V$ be a vector space over $F$ . A linear functional on $V$ is a linear map $f : V \to F$ , where $F$ is regarded as a vector space over itself. The algebraic dual space of $V$ is

$V^{*} : = L (V, F),$

with pointwise addition and scalar multiplication. This is the full algebraic dual: no topology, norm, or continuity condition is imposed.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

The dual family $(b^{})_{b \in B}$ associated to a Hamel basis $B$ , defined by $b^{} (c) = δ_{b c}$

Definition

Let $B$ be a Hamel basis of a vector space $V$ . For each $b \in B$ , the coordinate functional $b^{*} : V \to F$ is defined by

$b^{*} (c) = δ_{b c} (c \in B),$

and extended linearly: if the unique finite basis expansion of $v$ is $v = \sum_{c \in S} a_{c} c$ , then $b^{*} (v) = a_{b}$ , with $a_{b} : = 0$ when $b \notin S$ . Uniqueness of finite basis expansions makes this single-valued, and coordinatewise addition and scalar multiplication make $b^{*}$ linear. The family $(b^{*})_{b \in B}$ is the dual family associated to $B$ .

When $B$ is finite this family is the usual dual basis. When $B$ is infinite it remains a family in $V^{*}$ , but it need not span the whole algebraic dual.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

The dual family of a finite basis is a basis of the dual space, with the same dimension

Statement

If $B = (b_{1}, \dots, b_{n})$ is a basis of a finite-dimensional $F$ -vector space $V$ , then its dual family $B^{*} = (b_{1}^{*}, \dots, b_{n}^{*})$ is a basis of $V^{*}$ . Consequently $\dim V^{*} = \dim V = n$ .

Facts & Assumptions

Given: A finite basis $B = (b_{1}, \dots, b_{n})$ of $V$ and its dual family.

[L1]

The dual functionals satisfy $b_{i}^{*} (b_{j}) = δ_{i j}$ (The dual family $(b^{*})_{b \in B}$ associated to a Hamel basis $B$ , defined by $b^{*} (c) = δ_{b c}$ ).

[L2]

Every $v \in V$ has unique coordinates $v = \sum_{j = 1}^{n} a_{j} b_{j}$ (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis).

[L3]

The dimension of a finite-dimensional space is the cardinality of any finite basis (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis).

Proof

technique · direct

1.1

If $\sum_{i = 1}^{n} c_{i} b_{i}^{*} = 0$ , evaluation at $b_{j}$ and [L1] give $c_{j} = 0$ for every $j$ ; hence the dual family is linearly independent.

L1algebra

1.2

For $f \in V^{*}$ , set $g = \sum_{i = 1}^{n} f (b_{i}) b_{i}^{*}$ . If $v = \sum_{j} a_{j} b_{j}$ as in [L2], then $g (v) = \sum_{j} a_{j} f (b_{j}) = f (v)$ , so $f = g$ and the dual family spans $V^{*}$ .

L1L2algebra

2.1

Steps 1.1 and 1.2 make $B^{*}$ a basis, and [L3] gives $\dim V^{*} = n = \dim V$ . For $n = 0$ , both sums are empty and the unique functional on the zero space is zero, so the same proof applies.

step 1.1step 1.2L3∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

For an infinite Hamel basis, its dual family is linearly independent but does not span the algebraic dual

Statement

Let $B$ be an infinite Hamel basis of $V$ . Its dual family $(b^{*})_{b \in B}$ is linearly independent in $V^{*}$ but does not span $V^{*}$ .

Facts & Assumptions

Given: An infinite Hamel basis $B$ of $V$ and its dual family.

[L1]

The dual family satisfies $b^{*} (c) = δ_{b c}$ for $b, c \in B$ (The dual family $(b^{*})_{b \in B}$ associated to a Hamel basis $B$ , defined by $b^{*} (c) = δ_{b c}$ ).

[L2]

Linear independence tests only finite linear relations (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent).

[L3]

A Hamel basis is a linearly independent spanning set, with span defined through finite linear combinations (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

Proof

technique · direct

1.1

If a finite relation $\sum_{b \in S} a_{b} b^{*} = 0$ holds, evaluating it at each $c \in S$ gives $a_{c} = 0$ by [L1]. Thus the dual family is linearly independent by [L2].

L1L2

1.2

By [L3], every vector has a finite basis expansion. It is unique after zero coefficients are discarded, because subtracting two such expansions gives a finite relation in the independent set $B$ . Hence $ϕ (\sum_{b \in S} a_{b} b) = \sum_{b \in S} a_{b}$ is well defined and linear, and $ϕ (b) = 1$ for every $b \in B$ .

L3algebra

2.1

Every finite linear combination of members of the dual family vanishes at all basis vectors outside its finite support. Since $B$ is infinite while $ϕ (b) = 1$ for every $b$ , $ϕ$ is not in their span.

step 1.2L1given

3.1

The dual family is therefore independent but not spanning, so it is not a Hamel basis of $V^{*}$ .

step 1.1step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Assuming choice, if $v \notin U \leq V$ , some $f \in V^{*}$ vanishes on $U$ and satisfies $f (v) = 1$

Statement

Assume the axiom of choice. If $U$ is a linear subspace of $V$ and $v \in V ∖ U$ , then there exists $f \in V^{*}$ such that $f ∣_{U} = 0$ and $f (v) = 1$ .

Facts & Assumptions

Given: The axiom of choice, a subspace $U \leq V$ , and $v \notin U$ .

[L1]

Assuming choice, every linearly independent subset of a vector space extends to a basis (Zorn's lemma gives a basis between any linearly independent set and any spanning set containing it: if $L \subseteq S \subseteq V$ with $L$ independent and $span (S) = V$ , there is a basis $B$ of $V$ with $L \subseteq B \subseteq S$ ).

[L2]

A subspace is closed under finite linear combinations (Linear subspace of a vector space).

[L3]

Elements of $V^{*}$ are linear maps $V \to F$ (Linear functionals and the algebraic dual $V^{*} = L (V, F)$ ).

Proof

technique · basis extension

1.1

By [L1], choose a basis $C$ of $U$ . The set $C \cup {v}$ is linearly independent: a relation with nonzero coefficient of $v$ would put $v$ in the span of $C$ , which is $U$ by [L2].

L1L2givenchoose

2.1

Extend $C \cup {v}$ by [L1] to a basis $B$ of $V$ . Prescribe $f (v) = 1$ and $f (b) = 0$ for every $b \in B ∖ {v}$ , then extend by the unique finite basis expansion. This defines a linear functional $f \in V^{*}$ .

step 1.1L1L3choose

3.1

Every element of $U$ is a linear combination of elements of $C$ , so $f ∣_{U} = 0$ , while the prescription gives $f (v) = 1$ .

step 2.1L2∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

The canonical evaluation map $J_{V} : V \to V^{* *}$ given by $J_{V} (v) (f) = f (v)$

Definition

For a vector space $V$ , write $V^{* *} : = (V^{*})^{*}$ . The canonical evaluation map into the double dual is

$J_{V} : V ⟶ V^{* *}, J_{V} (v) (f) : = f (v) .$

For fixed $v$ , evaluation at $v$ is linear in $f$ , so $J_{V} (v) \in V^{* *}$ . The construction uses no basis and is therefore canonical.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Assuming choice, the canonical map $J_{V} : V \to V^{* *}$ is linear and injective

Statement

Assume the axiom of choice. For every vector space $V$ , the canonical map $J_{V} : V \to V^{* *}$ is linear and injective.

Facts & Assumptions

Given: The axiom of choice and an $F$ -vector space $V$ .

[L1]

The canonical map is defined by $J_{V} (v) (f) = f (v)$ for $v \in V$ and $f \in V^{*}$ (The canonical evaluation map $J_{V} : V \to V^{* *}$ given by $J_{V} (v) (f) = f (v)$ ).

[L2]

If $v \notin U \leq V$ , some functional vanishes on $U$ and takes value $1$ at $v$ (Assuming choice, if $v \notin U \leq V$ , some $f \in V^{*}$ vanishes on $U$ and satisfies $f (v) = 1$ ).

[L3]

A linear map is injective if and only if its kernel is trivial (The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial).

Proof

technique · direct

1.1

For $a, b \in F$ , $u, v \in V$ , and $f \in V^{*}$ , [L1] gives $J_{V} (a u + b v) (f) = f (a u + b v) = a J_{V} (u) (f) + b J_{V} (v) (f)$ . Equality at every $f$ proves that $J_{V}$ is linear.

L1algebra

1.2

If $v \neq 0$ , apply [L2] to $U = {0}$ to obtain $f$ with $f (v) = 1$ . Then $J_{V} (v) (f) = 1$ , so $J_{V} (v) \neq 0$ . Hence $\ker J_{V} = {0}$ .

L1L2given

2.1

By [L3], step 1.2 makes $J_{V}$ injective; step 1.1 supplies linearity. The zero space is included, since its unique map has trivial kernel.

step 1.1step 1.2L3∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Assuming choice, $J_{V} : V \to V^{* *}$ is surjective if and only if $V$ is finite-dimensional

Statement

Assume the axiom of choice. The canonical map $J_{V} : V \to V^{* *}$ is surjective if and only if $V$ is finite-dimensional.

Facts & Assumptions

Given: The axiom of choice and an $F$ -vector space $V$ .

[L1]

The canonical map is linear and injective (Assuming choice, the canonical map $J_{V} : V \to V^{* *}$ is linear and injective).

[L2]

For a finite basis $(b_{i})$ , its dual family $(b_{i}^{*})$ is a basis of $V^{*}$ (The dual family of a finite basis is a basis of the dual space, with the same dimension).

[L3]

For an infinite Hamel basis $B$ , its coordinate functionals span a proper subspace of $V^{*}$ (For an infinite Hamel basis, its dual family is linearly independent but does not span the algebraic dual).

[L4]

Assuming choice, a vector outside a subspace is separated from it by a linear functional (Assuming choice, if $v \notin U \leq V$ , some $f \in V^{*}$ vanishes on $U$ and satisfies $f (v) = 1$ ).

[L5]

Assuming choice, every vector space has a basis (Every vector space has a basis); for an infinite-dimensional $V$ such a basis is infinite.

Proof

technique · direct, proving both implications

1.1

Suppose $V$ has finite basis $(b_{1}, \dots, b_{n})$ and let $L \in V^{* *}$ . Put $v = \sum_{i} L (b_{i}^{*}) b_{i}$ . By [L2], every $f \in V^{*}$ is $\sum_{i} f (b_{i}) b_{i}^{*}$ , so $L (f) = \sum_{i} f (b_{i}) L (b_{i}^{*}) = f (v) = J_{V} (v) (f)$ . Thus $L = J_{V} (v)$ and $J_{V}$ is surjective.

L2algebra

1.2

Conversely, suppose $V$ is infinite-dimensional. By [L5], choice supplies a basis $B$ of $V$ , necessarily infinite. Let $Φ = span {b^{*} : b \in B}$ ; [L3] makes $Φ$ proper, so choose $ϕ \in V^{*} ∖ Φ$ . By [L4] applied inside $V^{*}$ , choose $L \in V^{* *}$ with $L ∣_{Φ} = 0$ and $L (ϕ) = 1$ .

L3L4L5givenchoose

2.1

If $L = J_{V} (v)$ , then $b^{*} (v) = J_{V} (v) (b^{*}) = L (b^{*}) = 0$ for every $b \in B$ . All basis coordinates of $v$ vanish, so $v = 0$ ; then $L = J_{V} (0) = 0$ , contradicting $L (ϕ) = 1$ . Hence $L$ is not in the image and $J_{V}$ is not surjective.

step 1.2algebra

3.1

Step 1.1 proves the forward finite-dimensional case and steps 1.2–2.1 prove its contrapositive. Therefore surjectivity is equivalent to finite dimensionality; [L1] additionally shows the finite-dimensional map is an isomorphism.

step 1.1step 1.2step 2.1L1∎

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

The annihilator $U^{\circ} \leq V^{}$ of $U \leq V$ and the preannihilator $^{\circ} S \leq V$ of $S \leq V^{}$

Definition

Let $U \leq V$ be a linear subspace. Its annihilator in the algebraic dual is

$U^{\circ} : = {f \in V^{*} : f (u) = 0 for every u \in U} \leq V^{*} .$

For a linear subspace $S \leq V^{*}$ , its preannihilator in $V$ is

$^{\circ} S : = {v \in V : f (v) = 0 for every f \in S} \leq V .$

Both are linear subspaces because all their defining equations are homogeneous and linear. The superscript position records which side of the evaluation pairing $V^{*} \times V \to F$ is being annihilated.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

Assuming choice, $^{\circ} (U^{\circ}) = U$ ; in finite dimension, $\dim U^{\circ} = \dim V - \dim U$

Statement

Assume the axiom of choice. For every subspace $U \leq V$ ,

$^{\circ} (U^{\circ}) = U .$

If $V$ is finite-dimensional, then

$\dim U^{\circ} = \dim V - \dim U .$

Facts & Assumptions

Given: The axiom of choice, an $F$ -vector space $V$ , and a subspace $U \leq V$ .

[L1]

The annihilator $U^{\circ}$ consists of the functionals vanishing on $U$ , and $^{\circ} S$ consists of vectors annihilated by every member of $S$ (The annihilator $U^{\circ} \leq V^{*}$ of $U \leq V$ and the preannihilator $^{\circ} S \leq V$ of $S \leq V^{*}$ ).

[L2]

If $v \notin U$ , some $f \in V^{*}$ vanishes on $U$ and has $f (v) = 1$ (Assuming choice, if $v \notin U \leq V$ , some $f \in V^{*}$ vanishes on $U$ and satisfies $f (v) = 1$ ).

[L3]

The dual of a finite-dimensional space has the same dimension (The dual family of a finite basis is a basis of the dual space, with the same dimension).

[L4]

Rank-nullity gives $\dim X = \dim \ker R + \dim im R$ for a linear map with finite-dimensional domain $X$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

[L5]

In finite dimension, a basis of a subspace extends without Choice to a basis of the ambient space (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ , clause 3).

Proof

technique · direct

1.1

Every $u \in U$ is killed by every $f \in U^{\circ}$ , so [L1] gives $U \subseteq^{\circ} (U^{\circ})$ . If $v \notin U$ , [L2] gives $f \in U^{\circ}$ with $f (v) = 1$ , so $v \notin^{\circ} (U^{\circ})$ . Hence $^{\circ} (U^{\circ}) = U$ .

L1L2

1.2

Now suppose $V$ is finite-dimensional and consider restriction $R : V^{*} \to U^{*}$ , $R (f) = f ∣_{U}$ . Its kernel is $U^{\circ}$ by [L1]. It is surjective: extend a basis of $U$ to one of $V$ by [L5], prescribe any given functional's values on the former basis, and prescribe zero on the added vectors.

L1L5choose

2.1

Rank-nullity and surjectivity give $\dim V^{*} = \dim U^{\circ} + \dim U^{*}$ . Applying [L3] to $V$ and $U$ , then rearranging, gives $\dim U^{\circ} = \dim V - \dim U$ .

step 1.2L3L4algebra

3.1

Step 1.1 proves the double-annihilator identity in arbitrary dimension under Choice, and step 2.1 proves the finite-dimensional formula, including $U = 0$ and $U = V$ .

step 1.1step 2.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

The transpose or algebraic adjoint $T^{} : W^{} \to V^{}$ , $T^{} (g) = g \circ T$ , of a linear map $T : V \to W$

Definition

Let $T : V \to W$ be linear. Its transpose, or algebraic adjoint, is the map

$T^{*} : W^{*} ⟶ V^{*}, T^{*} (g) : = g \circ T .$

The composite $g \circ T$ is linear, so it belongs to $V^{*}$ . This definition is algebraic and does not use an inner product.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Transpose is linear, sends identities to identities, and reverses composition: $(S \circ T)^{} = T^{} \circ S^{*}$

Statement

For linear maps $T, T^{'} : V \to W$ , scalars $a, b$ , and $S : W \to X$ ,

$(a T + b T^{'})^{*} = a T^{*} + b T^{' *}, ({id}_{V})^{*} = {id}_{V^{*}}, (S \circ T)^{*} = T^{*} \circ S^{*} .$

Facts & Assumptions

Given: The displayed compatible linear maps and scalars.

[L1]

The transpose is $T^{*} (g) = g \circ T$ (The transpose or algebraic adjoint $T^{*} : W^{*} \to V^{*}$ , $T^{*} (g) = g \circ T$ , of a linear map $T : V \to W$ ).

[L2]

Composition of linear maps is associative, and identity maps are its identities (Identity maps and composites of linear maps are linear).

Proof

technique · pointwise evaluation

1.1

For $g \in W^{*}$ and $v \in V$ , $((a T + b T^{'})^{*} g) (v) = g (a T (v) + b T^{'} (v)) = a (T^{*} g) (v) + b (T^{' *} g) (v)$ , proving linearity in the map.

L1algebra

1.2

For $f \in V^{*}$ and $v \in V$ , $(({id}_{V})^{*} f) (v) = f (v)$ , so $({id}_{V})^{*} = {id}_{V^{*}}$ .

L1L2

1.3

For $h \in X^{*}$ , [L1] and associativity give $(S \circ T)^{*} (h) = h \circ S \circ T = T^{*} (S^{*} (h))$ , hence $(S \circ T)^{*} = T^{*} \circ S^{*}$ .

L1L2

2.1

Equality at every functional and vector proves all three asserted identities.

step 1.1step 1.2step 1.3∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

Assuming choice, $\ker T^{} = (im T)^{\circ}$ and $im T^{} = (\ker T)^{\circ}$ ; in finite dimensions $rank T^{*} = rank T$

Statement

Assume the axiom of choice. For a linear map $T : V \to W$ ,

$\ker T^{*} = (im T)^{\circ}, im T^{*} = (\ker T)^{\circ} .$

If $V$ and $W$ are finite-dimensional, then $rank T^{*} = rank T$ .

Facts & Assumptions

Given: The axiom of choice and a linear map $T : V \to W$ .

[L1]

The transpose satisfies $T^{*} (g) = g \circ T$ (The transpose or algebraic adjoint $T^{*} : W^{*} \to V^{*}$ , $T^{*} (g) = g \circ T$ , of a linear map $T : V \to W$ ).

[L2]

An annihilator consists exactly of the functionals vanishing on the named subspace (The annihilator $U^{\circ} \leq V^{*}$ of $U \leq V$ and the preannihilator $^{\circ} S \leq V$ of $S \leq V^{*}$ ).

[L3]

In finite dimension, $\dim U^{\circ} = \dim X - \dim U$ for $U \leq X$ (Assuming choice, $^{\circ} (U^{\circ}) = U$ ; in finite dimension, $\dim U^{\circ} = \dim V - \dim U$ ).

[L4]

Assuming choice, every independent set extends to a basis (Zorn's lemma gives a basis between any linearly independent set and any spanning set containing it: if $L \subseteq S \subseteq V$ with $L$ independent and $span (S) = V$ , there is a basis $B$ of $V$ with $L \subseteq B \subseteq S$ ).

[L5]

Rank-nullity gives $\dim V = \dim \ker T + rank T$ when $V$ is finite-dimensional (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

Proof

technique · direct

1.1

A functional $g \in W^{*}$ lies in $\ker T^{*}$ exactly when $g (T v) = 0$ for every $v \in V$ , which by [L1] and [L2] is exactly $g \in (im T)^{\circ}$ .

L1L2

1.2

Every $T^{*} (g) = g \circ T$ vanishes on $\ker T$ , so $im T^{*} \subseteq (\ker T)^{\circ}$ .

L1L2

1.3

Conversely let $f \in (\ker T)^{\circ}$ . Define $h_{0} : im T \to F$ by $h_{0} (T v) = f (v)$ . If $T v = T v^{'}$ , then $v - v^{'} \in \ker T$ and $f (v) = f (v^{'})$ , so $h_{0}$ is well defined and linear. Extend a basis of $im T$ to a basis of $W$ using [L4], and extend $h_{0}$ by value $0$ on the added basis vectors to obtain $h \in W^{*}$ . Then $T^{*} (h) = f$ .

L1L2L4choosealgebra

2.1

Steps 1.2 and 1.3 prove $im T^{*} = (\ker T)^{\circ}$ .

step 1.2step 1.3

3.1

If $V, W$ are finite-dimensional, [L3] and step 2.1 give $rank T^{*} = \dim (\ker T)^{\circ} = \dim V - \dim \ker T$ , which equals $rank T$ by [L5].

step 2.1L3L5

4.1

Steps 1.1, 2.1, and 3.1 give the two identities and the finite-dimensional rank equality, including zero source or target.

step 1.1step 2.1step 3.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13Open item page →

In dual bases, the matrix of $T^{*}$ is the transpose of the matrix of $T$

Statement

Let $T : V \to W$ be linear between finite-dimensional spaces, with ordered bases $V = (v_{1}, \dots, v_{n})$ and $W = (w_{1}, \dots, w_{m})$ . In the dual bases,

$[T^{*}]_{W^{*}}^{V^{*}} = ([T]_{V}^{W})^{T} .$

Facts & Assumptions

Given: The displayed map, bases, and their dual bases.

[L1]

The transpose satisfies $T^{*} (g) = g \circ T$ (The transpose or algebraic adjoint $T^{*} : W^{*} \to V^{*}$ , $T^{*} (g) = g \circ T$ , of a linear map $T : V \to W$ ).

[L2]

The dual families of the finite bases are bases of the dual spaces (The dual family of a finite basis is a basis of the dual space, with the same dimension).

[L3]

The $j$ th column of a representing matrix is the coordinate column of the image of the $j$ th basis vector (Coordinate columns $[v]_{B}$ and matrices $[T]_{B}^{C}$ of linear maps relative to ordered bases).

[L4]

The transpose of an $m \times n$ matrix has $(j, i)$ entry equal to the original $(i, j)$ entry (The transpose $A^{T}$ of a matrix).

Proof

technique · entrywise comparison

1.1

Write $[T]_{V}^{W} = (a_{i j})$ , so [L3] gives $T (v_{j}) = \sum_{i = 1}^{m} a_{i j} w_{i}$ and therefore $w_{i}^{*} (T (v_{j})) = a_{i j}$ .

L2L3algebra

2.1

The $(j, i)$ entry of $[T^{*}]_{W^{*}}^{V^{*}}$ is the coefficient of $v_{j}^{*}$ in the expansion of $T^{*} (w_{i}^{*}) \in V^{*}$ along the dual basis $V^{*}$ . Since every $ϕ \in V^{*}$ satisfies $ϕ = \sum_{j} ϕ (v_{j}) v_{j}^{*}$ , that coefficient is $T^{*} (w_{i}^{*}) (v_{j})$ , which by [L1] equals $w_{i}^{*} (T (v_{j})) = a_{i j}$ .

step 1.1L1L2L3

3.1

By [L4], step 2.1 says exactly that the $n \times m$ matrix of $T^{*}$ is the transpose of the $m \times n$ matrix of $T$ . The calculation also covers $m = 0$ or $n = 0$ , where the matrices are empty rectangles.

step 2.1L4∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms

Definition

Let $V$ be a vector space over $F$ . A bilinear form on $V$ is a function $B : V \times V \to F$ that is linear in each variable separately.

The form is

symmetric when $B (u, v) = B (v, u)$ for all $u, v \in V$ ;
skew-symmetric when $B (u, v) = - B (v, u)$ for all $u, v \in V$ ;
alternating when $B (v, v) = 0$ for every $v \in V$ .

These conditions are kept distinct because their relations depend on the characteristic of $F$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

Alternating forms are skew-symmetric; the converse holds when $char F \neq 2$ , while in characteristic $2$ alternating forms are symmetric

Statement

Every alternating bilinear form is skew-symmetric. If $char F \neq 2$ , every skew-symmetric bilinear form is alternating. If $char F = 2$ , every alternating bilinear form is symmetric.

Facts & Assumptions

Given: A bilinear form $B : V \times V \to F$ .

[L1]

Alternating means $B (v, v) = 0$ for all $v$ , skew-symmetric means $B (u, v) = - B (v, u)$ , and symmetric means $B (u, v) = B (v, u)$ (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms).

[L2]

The characteristic is the least positive natural multiple of $1_{F}$ that is zero, or $0$ if none exists (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise); a field has $0_{F} \neq 1_{F}$ and every nonzero element is invertible (Field).

Proof

technique · direct

1.1

If $B$ is alternating, bilinearity gives $0 = B (u + v, u + v) = B (u, v) + B (v, u)$ , so $B (u, v) = - B (v, u)$ and $B$ is skew-symmetric.

L1algebra

1.2

If $B$ is skew-symmetric, then $B (v, v) = - B (v, v)$ , so $2 B (v, v) = 0$ . When $char F \neq 2$ , [L2] makes $2 = 1_{F} + 1_{F}$ nonzero and invertible, giving $B (v, v) = 0$ for every $v$ and hence alternation.

L1L2algebra

2.1

If $char F = 2$ and $B$ is alternating, [L2] gives $1_{F} + 1_{F} = 0_{F}$ and hence $- a = a$ for every $a \in F$ . Step 1.1 therefore gives $B (u, v) = - B (v, u) = B (v, u)$ , so $B$ is symmetric.

step 1.1L1L2algebra

3.1

These arguments prove each asserted implication without claiming that every symmetric form in characteristic $2$ is alternating.

step 1.1step 1.2step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Bilinear forms on $V$ correspond linearly and bijectively to linear maps $V \to V^{*}$

Statement

The assignment

$B ⟼ B^{♭}, B^{♭} (v) (w) = B (v, w),$

is a linear bijection from the vector space of bilinear forms on $V$ to $L (V, V^{*})$ .

Facts & Assumptions

Given: An $F$ -vector space $V$ .

[L1]

A bilinear form is linear separately in both variables (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms).

[L2]

The algebraic dual $V^{*}$ consists of all linear maps $V \to F$ (Linear functionals and the algebraic dual $V^{*} = L (V, F)$ ).

[L3]

Linear maps form a vector space under pointwise addition and scalar multiplication (The space $L (V, W)$ of linear maps with pointwise addition and scalar multiplication).

Proof

technique · explicit inverse

1.1

For a bilinear $B$ , fixing $v$ makes $w \mapsto B (v, w)$ a member of $V^{*}$ by [L1] and [L2], and linearity in $v$ makes $B^{♭} : V \to V^{*}$ linear.

L1L2

1.2

Conversely, for $L \in L (V, V^{*})$ , define $B_{L} (v, w) = L (v) (w)$ . Linearity of $L$ gives linearity in $v$ , and $L (v) \in V^{*}$ gives linearity in $w$ , so $B_{L}$ is bilinear.

L1L2L3

2.1

Pointwise, $B_{B^{♭}} (v, w) = B (v, w)$ and $(B_{L})^{♭} (v) (w) = L (v) (w)$ , so the constructions are inverse. They also preserve addition and scalar multiplication pointwise.

step 1.1step 1.2L3algebra

3.1

Therefore $B \mapsto B^{♭}$ is a linear bijection. No finite-dimensionality assumption is used.

step 2.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space

Definition

Let $B$ be a bilinear form on an $n$ -dimensional vector space $V$ , and let $B = (v_{0}, \dots, v_{n - 1})$ be an ordered basis. The matrix of $B$ in $B$ is

$[B]_{B} : = (B (v_{i}, v_{j}))_{i, j < n} .$

For coordinate columns $x = [u]_{B}$ and $y = [v]_{B}$ , one has $B (u, v) = x^{T} [B]_{B} y$ .

The left radical and right radical are

${rad}_{L} (B) : = {u : B (u, v) = 0 for every v}, {rad}_{R} (B) : = {v : B (u, v) = 0 for every u} .$

The rank of $B$ is the rank of the associated map $B^{♭} : V \to V^{*}$ , $B^{♭} (u) (v) = B (u, v)$ . The form is nondegenerate when both radicals are zero. In finite dimension this is equivalent to $B^{♭}$ being an isomorphism, and also to $[B]_{B}$ being invertible.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13Open item page →

A basis change by $P$ changes the matrix of a bilinear form from $A$ to $P^{T} A P$

Statement

Let $B$ be a bilinear form on a finite-dimensional space. If $A$ is its matrix in an old basis and the columns of an invertible matrix $P$ are the new basis vectors in old coordinates, then its matrix in the new basis is

$P^{T} A P .$

Matrices related by $A^{'} = P^{T} A P$ with $P$ invertible are called congruent.

Facts & Assumptions

Given: The form, the two bases, and the change-of-basis matrix $P$ described above.

[L1]

If $x, y$ are coordinate columns, the matrix $A$ of $B$ satisfies $B (u, v) = x^{T} A y$ (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space).

[L2]

A representing matrix acts on coordinate columns by multiplication ( $[T (v)]_{C} = [T]_{B}^{C} [v]_{B}$ ).

[L3]

Transpose reverses products: $(X Y)^{T} = Y^{T} X^{T}$ (Transpose is linear and involutive, and $(A B)^{T} = B^{T} A^{T}$ ).

[L4]

Matrix multiplication is associative and represents the relevant finite sums (Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication).

Proof

technique · coordinate calculation

1.1

If $x, y$ are the new coordinate columns of $u, v$ , respectively, [L2] makes their old coordinate columns $P x, P y$ .

L2given

2.1

By [L1], $B (u, v) = (P x)^{T} A (P y) = x^{T} P^{T} A P y$ , using [L3] and [L4].

step 1.1L1L3L4

3.1

Since step 2.1 holds for every $x, y$ , the new matrix is $P^{T} A P$ . The displayed relation is therefore exactly the equivalence generated by basis changes and is called congruence, not similarity.

step 2.1∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Congruent matrices have the same rank; hence rank and nondegeneracy of a bilinear form are basis-independent

Statement

If $A^{'} = P^{T} A P$ with $P$ invertible, then $rank A^{'} = rank A$ . Consequently the rank and nondegeneracy of a bilinear form do not depend on the basis used to represent it.

Facts & Assumptions

Given: Congruent matrices $A^{'} = P^{T} A P$ with $P$ invertible.

[L1]

Congruence is precisely the matrix relation arising from a basis change for a bilinear form (A basis change by $P$ changes the matrix of a bilinear form from $A$ to $P^{T} A P$ ).

[L2]

Matrix multiplication represents composition of the associated linear maps ( $[S \circ T]_{B}^{D} = [S]_{C}^{D} [T]_{B}^{C}$ ).

[L3]

Transpose reverses products and sends an inverse to the inverse transpose (Transpose is linear and involutive, and $(A B)^{T} = B^{T} A^{T}$ ).

[L4]

An invertible linear map is a linear isomorphism and has a two-sided linear inverse (Invertible linear maps, linear isomorphisms, and inverse linear maps). Two finite-dimensional spaces are linearly isomorphic if and only if they have the same dimension (Two finite-dimensional vector spaces over $F$ are linearly isomorphic if and only if they have the same dimension).

Proof

technique · direct

1.1

Right multiplication by the invertible $P$ is precomposition with an isomorphism, so $im (A P) = im (A)$ . Left multiplication by $P^{T}$ maps this image isomorphically to $im (P^{T} A P)$ , because [L3] makes $P^{T}$ invertible.

L2L3L4given

2.1

Therefore the two image spaces have the same dimension, so $rank A^{'} = rank A$ .

step 1.1L4

3.1

By [L1], matrices of one bilinear form in two bases are congruent. Rank is thus basis-independent, and nondegeneracy is also basis-independent because it is equivalent to invertibility of a representing matrix, which the congruence relation preserves in both directions.

step 2.1L1L3L4∎

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-13Open item page →

Sesquilinear and Hermitian forms over a field with an involution, using the convention linear in the first variable

Definition

Let $F$ be a field with an involution, an automorphism $σ : F \to F$ satisfying $σ^{2} = {id}_{F}$ . A function $H : V \times V \to F$ is sesquilinear, with the convention used here, when it is linear in the first variable and $σ$ -linear in the second:

$H (a u + b v, w) = a H (u, w) + b H (v, w),$

$H (u, a v + b w) = σ (a) H (u, v) + σ (b) H (u, w) .$

It is Hermitian when

$H (u, v) = σ (H (v, u))$

for all $u, v \in V$ . When $σ$ is the identity, sesquilinear forms are bilinear and Hermitian forms are symmetric bilinear forms.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13Open item page →

For the linear-first convention, a basis change by $P$ sends a sesquilinear matrix $A$ to $P^{T} A σ (P)$ ; Hermitian forms satisfy $A = σ (A)^{T}$

Statement

Let $H$ be sesquilinear over a field with involution $σ$ , using the convention linear in the first variable. If its old matrix is $A$ and a basis change has matrix $P$ , then its new matrix is

$P^{T} A σ (P),$

where $σ (P)$ is obtained entrywise. Moreover, $H$ is Hermitian if and only if $A = σ (A)^{T}$ .

Facts & Assumptions

Given: A field involution $σ$ , a sesquilinear form $H$ , and the displayed basis data.

[L1]

Sesquilinearity is linear in the first variable and $σ$ -linear in the second; Hermitian symmetry is $H (u, v) = σ (H (v, u))$ (Sesquilinear and Hermitian forms over a field with an involution, using the convention linear in the first variable).

[L2]

Matrix multiplication expands as the corresponding finite row-column sums (Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication).

[L3]

Transpose reverses products (Transpose is linear and involutive, and $(A B)^{T} = B^{T} A^{T}$ ).

Proof

technique · coordinate calculation

1.1

If $x, y$ are new coordinate columns, their old columns are $P x, P y$ . Expanding [L1] in an old basis gives $H (u, v) = (P x)^{T} A σ (P y) = x^{T} P^{T} A σ (P) σ (y)$ .

L1L2L3algebra

1.2

In a basis $(e_{i})$ , Hermitian symmetry is $A_{i j} = H (e_{i}, e_{j}) = σ (H (e_{j}, e_{i})) = σ (A_{j i})$ for every $i, j$ , which is exactly $A = σ (A)^{T}$ .

L1algebra

2.1

Since step 1.1 holds for every $x, y$ , the new matrix is $P^{T} A σ (P)$ . This includes the identity involution, where it reduces to ordinary congruence.

step 1.1

2.2

Conversely, if $A = σ (A)^{T}$ , the coordinate formula and $σ^{2} = id$ give $H (u, v) = σ (H (v, u))$ for arbitrary coordinate columns, so $H$ is Hermitian.

step 1.2L1L2L3algebra

3.1

Steps 2.1, 1.2, and 2.2 prove the basis-change formula and both directions of the Hermitian criterion.

step 2.1step 1.2step 2.2∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

A quadratic form $q$ in arbitrary characteristic and its polar form $b_{q} (u, v) = q (u + v) - q (u) - q (v)$

Definition

Let $V$ be a vector space over a field $F$ . A quadratic form on $V$ is a function $q : V \to F$ such that

$q (a v) = a^{2} q (v)$

for all $a \in F$ and $v \in V$ , and such that its polar form

$b_{q} (u, v) : = q (u + v) - q (u) - q (v)$

is bilinear. This definition is valid in every characteristic. In characteristic $2$ , the polar form need not determine $q$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

If $char F \neq 2$ , quadratic forms and symmetric bilinear forms correspond by $q (v) = B (v, v)$ and $B (u, v) = \frac{1}{2} b_{q} (u, v)$

Statement

Let $char F \neq 2$ . The assignments

$B ⟼ q_{B}, q_{B} (v) = B (v, v), q ⟼ B_{q}, B_{q} (u, v) = \frac{1}{2} b_{q} (u, v)$

are inverse bijections between symmetric bilinear forms and quadratic forms.

Facts & Assumptions

Given: A field $F$ with $char F \neq 2$ and an $F$ -vector space $V$ .

[L1]

A quadratic form satisfies $q (a v) = a^{2} q (v)$ and has bilinear polar form $b_{q} (u, v) = q (u + v) - q (u) - q (v)$ (A quadratic form $q$ in arbitrary characteristic and its polar form $b_{q} (u, v) = q (u + v) - q (u) - q (v)$ ).

[L2]

A symmetric bilinear form is bilinear and satisfies $B (u, v) = B (v, u)$ (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms).

[L3]

The characteristic is the least positive natural multiple of $1_{F}$ that is zero, or $0$ if none exists (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise); in a field $0_{F} \neq 1_{F}$ and every nonzero scalar is invertible (Field).

Proof

technique · explicit inverse

1.1

Because $char F \neq 2$ , [L3] makes the scalar $2 = 1_{F} + 1_{F}$ nonzero and hence invertible. If $B$ is symmetric bilinear, then $q_{B} (a v) = a^{2} q_{B} (v)$ and $b_{q_{B}} (u, v) = B (u + v, u + v) - B (u, u) - B (v, v) = 2 B (u, v)$ . Thus $q_{B}$ is a quadratic form.

L1L2L3algebra

1.2

If $q$ is quadratic, [L1] makes $b_{q}$ bilinear, and its defining formula is symmetric. Hence $B_{q} = \frac{1}{2} b_{q}$ is symmetric bilinear by [L2] and [L3].

L1L2L3

2.1

Step 1.1 gives $B_{q_{B}} = B$ . Conversely, $b_{q} (v, v) = q (2 v) - 2 q (v) = 4 q (v) - 2 q (v) = 2 q (v)$ , so $q_{B_{q}} (v) = \frac{1}{2} b_{q} (v, v) = q (v)$ .

step 1.1step 1.2L1L3algebra

3.1

The two assignments are therefore mutually inverse bijections. The use of $2^{- 1}$ identifies precisely where the characteristic hypothesis enters.

step 2.1L3∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Every symmetric bilinear form on a finite-dimensional space over a field of characteristic not $2$ has an orthogonal basis

Statement

Let $V$ be finite-dimensional over a field of characteristic not $2$ . Every symmetric bilinear form $B$ on $V$ admits a basis whose distinct vectors are pairwise orthogonal for $B$ .

Facts & Assumptions

Given: A finite-dimensional $F$ -vector space $V$ , $char F \neq 2$ , and a symmetric bilinear form $B$ .

[L1]

In characteristic not $2$ , a symmetric bilinear form is recovered from $q_{B} (v) = B (v, v)$ by $B (u, v) = \frac{1}{2} b_{q_{B}} (u, v)$ (If $char F \neq 2$ , quadratic forms and symmetric bilinear forms correspond by $q (v) = B (v, v)$ and $B (u, v) = \frac{1}{2} b_{q} (u, v)$ ).

[L2]

A subspace of a finite-dimensional space is finite-dimensional, and an independent subset extends without Choice to a basis (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ).

[L3]

A basis is a linearly independent spanning set; the zero space has the empty basis (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

Proof

technique · induction on $n=\dim V$

1.1

If $n = 0$ , the empty basis is orthogonal. If $B = 0$ , any basis is orthogonal.

baseL3given

1.2

Assume $n > 0$ , $B \neq 0$ , and the theorem below dimension $n$ . By [L1], choose $v \in V$ with $B (v, v) \neq 0$ . Put $v^{⊥} = {w : B (w, v) = 0}$ .

ihL1choose

2.1

Every $w \in V$ has the decomposition $w = a v + z$ , where $a = B (w, v) B (v, v)^{- 1}$ and $z = w - a v \in v^{⊥}$ . If $a v \in v^{⊥}$ , then $a B (v, v) = 0$ , so $a = 0$ . Hence $V = F v \oplus v^{⊥}$ .

step 1.2algebra

3.1

Since $v \notin v^{⊥}$ , this subspace is proper and [L2] gives $\dim v^{⊥} < n$ . The restricted form is symmetric, so the induction hypothesis gives it an orthogonal basis; adjoining $v$ gives an orthogonal basis of $V$ .

step 1.2step 2.1ihL2L3

4.1

The base cases and induction step prove the theorem, including degenerate forms and the zero space.

step 1.1step 3.1discharge-induction∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Over a field of characteristic not $2$ , every symmetric matrix is congruent to a diagonal matrix

Statement

If $A \in M_{n} (F)$ is symmetric and $char F \neq 2$ , then there is an invertible $P$ such that $P^{T} A P$ is diagonal.

Facts & Assumptions

Given: A symmetric matrix $A \in M_{n} (F)$ over a field of characteristic not $2$ .

[L1]

The standard coordinate vectors form a basis of $F^{n}$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[L2]

Every symmetric bilinear form in the stated characteristic has an orthogonal basis (Every symmetric bilinear form on a finite-dimensional space over a field of characteristic not $2$ has an orthogonal basis).

[L3]

A basis change by $P$ sends a bilinear-form matrix $A$ to $P^{T} A P$ (A basis change by $P$ changes the matrix of a bilinear form from $A$ to $P^{T} A P$ ).

Proof

technique · direct

1.1

In the standard basis [L1], let $B (u, v) = u^{T} A v$ . Symmetry of $A$ makes $B$ symmetric.

L1givenalgebra

1.2

Choose an orthogonal basis for $B$ by [L2], and let $P$ have those basis vectors as its columns in standard coordinates. Then $P$ is invertible and the matrix of $B$ in that basis is diagonal.

L2choose

2.1

By [L3], this diagonal matrix is $P^{T} A P$ . When $n = 0$ , the empty matrix is already diagonal and the same conclusion holds.

step 1.2L3∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

Over a field of characteristic not $2$ , every quadratic form has diagonal coordinates $q (x) = a_{1} x_{1}^{2} + \dots + a_{n} x_{n}^{2}$

Statement

Let $q$ be a quadratic form on an $n$ -dimensional vector space over a field of characteristic not $2$ . Some basis gives

$q (x_{1} e_{1} + \dots + x_{n} e_{n}) = a_{1} x_{1}^{2} + \dots + a_{n} x_{n}^{2}$

for scalars $a_{1}, \dots, a_{n}$ , which may include zeros.

Facts & Assumptions

Given: The quadratic form $q$ in the stated characteristic.

[L1]

Polarization gives a symmetric bilinear form $B_{q} = \frac{1}{2} b_{q}$ satisfying $q (v) = B_{q} (v, v)$ (If $char F \neq 2$ , quadratic forms and symmetric bilinear forms correspond by $q (v) = B (v, v)$ and $B (u, v) = \frac{1}{2} b_{q} (u, v)$ ).

[L2]

Every finite-dimensional symmetric bilinear form in characteristic not $2$ has an orthogonal basis (Every symmetric bilinear form on a finite-dimensional space over a field of characteristic not $2$ has an orthogonal basis).

Proof

technique · direct

1.1

Form $B_{q}$ by [L1] and choose an orthogonal basis $(e_{1}, \dots, e_{n})$ by [L2]. Set $a_{i} = B_{q} (e_{i}, e_{i})$ .

L1L2choose

2.1

For $v = \sum_{i} x_{i} e_{i}$ , bilinearity and orthogonality give $q (v) = B_{q} (v, v) = \sum_{i} a_{i} x_{i}^{2}$ ; every cross term vanishes.

step 1.1L1algebra

3.1

This is the required diagonal expression. Zero coefficients are allowed, so degenerate forms and the zero form are included; for $n = 0$ the sum is empty.

step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Every alternating form on a finite-dimensional space has a basis of symplectic pairs followed by a basis of its radical; in particular its rank is even

Statement

Let $B$ be an alternating bilinear form on a finite-dimensional vector space $V$ . There is a basis

$e_{1}, f_{1}, \dots, e_{r}, f_{r}, z_{1}, \dots, z_{s}$

such that $B (e_{i}, f_{i}) = 1$ , $B (f_{i}, e_{i}) = - 1$ , every other pairing of distinct listed blocks is zero, and $z_{1}, \dots, z_{s}$ is a basis of $rad B$ . Thus the matrix is a direct sum of $r$ blocks $(\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ and an $s \times s$ zero block, so $rank B = 2 r$ is even.

Facts & Assumptions

Given: A finite-dimensional $F$ -vector space $V$ and an alternating bilinear form $B$ .

[L1]

Alternation means $B (v, v) = 0$ for every $v$ (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms), and every alternating form is skew-symmetric (Alternating forms are skew-symmetric; the converse holds when $char F \neq 2$ , while in characteristic $2$ alternating forms are symmetric).

[L2]

The radical consists of vectors pairing to zero with every vector, and the rank is the rank of the associated map into the dual (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space).

[L3]

Subspaces of finite-dimensional spaces are finite-dimensional and admit bases (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ ); a basis is an independent spanning set, with the empty basis for zero space (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

Proof

technique · induction on $n=\dim V$

1.1

If $n = 0$ , the empty basis has the asserted form. If $B = 0$ , any basis of $V = rad B$ works with $r = 0$ .

baseL2L3

1.2

Assume $B \neq 0$ and the result below dimension $n$ . Choose $e, f$ with $B (e, f) \neq 0$ and rescale $f$ so $B (e, f) = 1$ ; [L1] gives $B (f, e) = - 1$ .

ihL1givenchoose

2.1

Put $W = span {e, f}$ and $W^{⊥} = {v : B (v, e) = B (v, f) = 0}$ . For every $v \in V$ , the vector $z = v - B (v, f) e + B (v, e) f$ lies in $W^{⊥}$ , so $V = W + W^{⊥}$ . If $a e + b f \in W^{⊥}$ , pairing with $e$ and $f$ gives $b = 0 = a$ , hence $V = W \oplus W^{⊥}$ .

step 1.2L1algebra

3.1

Since $e \notin W^{⊥}$ , this subspace is proper and [L3] gives $\dim W^{⊥} < n$ . By induction its restricted alternating form has symplectic pairs followed by a basis of its radical. Adjoining $e, f$ gives the displayed basis of $V$ ; because $W$ is nondegenerate and orthogonal to $W^{⊥}$ , the remaining radical is exactly the radical of the whole form.

step 1.2step 2.1ihL2L3

4.1

In that basis the associated map has one invertible rank-two block per pair and is zero on the radical block, so its rank is $2 r$ . This remains valid in characteristic $2$ , where $- 1 = 1$ .

step 3.1L1L2algebra

5.1

The base cases and induction step establish the normal form and even-rank conclusion in every finite dimension, including odd-dimensional and degenerate forms.

step 1.1step 4.1discharge-induction∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

Positive and negative definiteness, the inertia $(p, q, r)$ , rank $p + q$ , and signature $p - q$ of a real symmetric bilinear or quadratic form

Definition

Let $B$ be a symmetric bilinear form on a finite-dimensional real vector space, and write $q (v) : = B (v, v)$ . The form is positive definite when $q (v) > 0$ for every $v \neq 0$ , and negative definite when $q (v) < 0$ for every $v \neq 0$ . The same terminology applies to the associated quadratic form.

Suppose a basis gives a diagonal matrix with $p$ positive diagonal entries, $q$ negative diagonal entries, and $r$ zero entries. Its inertia is the triple $(p, q, r)$ , its rank is $p + q$ , and its signature is $p - q$ . Sylvester's law of inertia proves that this triple is independent of the diagonalizing basis, and therefore justifies the notation.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Sylvester's law of inertia: every real symmetric form is congruent to $diag (I_{p}, - I_{q}, 0_{r})$ , and $(p, q, r)$ is unique

Statement

Every symmetric bilinear form on a finite-dimensional real vector space is congruent to exactly one normal form

$diag (I_{p}, - I_{q}, 0_{r}), p + q + r = \dim V .$

Equivalently, the numbers of positive, negative, and zero diagonal entries are independent of the diagonalizing basis.

Facts & Assumptions

Given: A symmetric bilinear form $B$ on a finite-dimensional real vector space $V$ .

[L1]

Every real symmetric matrix is congruent to a diagonal matrix (Over a field of characteristic not $2$ , every symmetric matrix is congruent to a diagonal matrix).

[L2]

Positive and negative definiteness and the inertia data have the meanings stated for real symmetric forms (Positive and negative definiteness, the inertia $(p, q, r)$ , rank $p + q$ , and signature $p - q$ of a real symmetric bilinear or quadratic form).

[L3]

The constructed real field has the least-upper-bound property and hence is complete ordered (The Cauchy-sequence reals have the least-upper-bound property), so every positive real has a nonzero positive square root (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[L4]

For finite-dimensional subspaces $U, W$ , $\dim (U + W) + \dim (U \cap W) = \dim U + \dim W$ (The dimension formula: for finite-dimensional linear subspaces $U$ and $W$ of $V$ , the subspaces $U + W$ and $U \cap W$ are finite-dimensional and $\dim_{F} (U + W) + \dim_{F} (U \cap W) = \dim_{F} U + \dim_{F} W$ ).

[L5]

A subspace of a finite-dimensional $n$ -dimensional space has dimension at most $n$ (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ , clause 1).

[L6]

Rank-nullity gives the dimension of a kernel as ambient dimension minus rank (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

Proof

technique · diagonal normalization and intrinsic dimension bounds

1.1

By [L1], choose a basis in which the matrix is diagonal, say with positive entries $d_{1}, \dots, d_{p}$ , negative entries $d_{p + 1}, \dots, d_{p + q}$ , and $r$ zero entries. For each nonzero $d_{i}$ , [L3] supplies $s_{i} = \sqrt{∣ d_{i} ∣} > 0$ ; replacing the corresponding basis vector by $s_{i}^{- 1}$ times it changes $d_{i}$ to $1$ or $- 1$ . This proves existence of the displayed normal form, including $p = q = 0$ and the empty basis when $V = 0$ .

L1L3choosealgebra

2.1

In this normal form, the positive coordinate subspace $P$ has dimension $p$ and the form is positive definite on it. Let $N_{0}$ be the span of the negative and zero coordinates, of dimension $q + r$ . If a positive-definite subspace $U$ had $\dim U > p$ , [L4] and [L5] applied to $U + N_{0} \leq V$ would give $\dim (U \cap N_{0}) \geq \dim U + q + r - \dim V > 0$ . A nonzero vector there has form value at most $0$ , a contradiction. Thus $p$ is the intrinsic maximum dimension of a positive-definite subspace.

step 1.1L2L4L5algebra

3.1

Applying step 2.1 to $- B$ shows that $q$ is the intrinsic maximum dimension of a negative-definite subspace. The radical is the kernel of the associated map; in normal form its rank is $p + q$ , so [L6] gives its dimension $r = \dim V - p - q$ .

step 1.1step 2.1L2L6algebra

4.1

Any congruent normal form represents the same bilinear form and therefore has the same two intrinsic maxima and radical dimension. Hence its triple is the same $(p, q, r)$ , proving uniqueness.

step 2.1step 3.1L2

5.1

Steps 1.1 and 4.1 prove existence and uniqueness for all finite dimensions and for degenerate as well as nondegenerate forms.

step 1.1step 4.1∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Two real symmetric bilinear forms are congruent if and only if they have the same inertia

Statement

Two real symmetric bilinear forms on vector spaces of the same finite dimension are congruent if and only if they have the same inertia $(p, q, r)$ .

Facts & Assumptions

Given: Real symmetric bilinear forms $B$ and $C$ in the same finite dimension.

[L1]

Sylvester's law gives each form a unique normal form $diag (I_{p}, - I_{q}, 0_{r})$ (Sylvester's law of inertia: every real symmetric form is congruent to $diag (I_{p}, - I_{q}, 0_{r})$ , and $(p, q, r)$ is unique).

[L2]

A basis change acts on a bilinear-form matrix by congruence $A \mapsto P^{T} A P$ (A basis change by $P$ changes the matrix of a bilinear form from $A$ to $P^{T} A P$ ).

Proof

technique · direct, proving both implications

1.1

If $B$ and $C$ are congruent, [L2] says they are two matrix representations of the same form after an invertible coordinate identification. The uniqueness clause of [L1] therefore gives them the same inertia.

L1L2given

1.2

Conversely, suppose both have inertia $(p, q, r)$ . By [L1], choose bases in which both matrices equal $D = diag (I_{p}, - I_{q}, 0_{r})$ . The linear map sending the first chosen basis to the second is an isomorphism and carries one form to the other, so the forms are congruent; equivalently, compose the two invertible change-of-basis matrices in [L2].

L1L2choose

2.1

Steps 1.1 and 1.2 prove both directions, including the zero-dimensional and degenerate cases.

step 1.1step 1.2∎

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

For symmetric $M = (\begin{array}{cc} A & B \\ B^{T} & C \end{array})$ with $A$ invertible, a block-unitriangular congruence gives $A \oplus (C - B^{T} A^{- 1} B)$ and factors $\det M$

Statement

Let

$M = (\begin{array}{cc} A & B \\ B^{T} & C \end{array})$

be symmetric over a field, with square diagonal blocks and $A$ invertible. Put $S = C - B^{T} A^{- 1} B$ . Then for

$P = (\begin{array}{cc} I & - A^{- 1} B \\ 0 & I \end{array})$

one has $P^{T} M P = A \oplus S$ , and

$\det M = \det A \det S,$

where the determinant of a $0 \times 0$ block is interpreted as $1$ .

Facts & Assumptions

Given: The displayed symmetric block matrix with $A$ invertible; an empty square block has determinant $1$ .

[L1]

A basis change by $P$ changes a bilinear-form matrix to $P^{T} M P$ (A basis change by $P$ changes the matrix of a bilinear form from $A$ to $P^{T} A P$ ).

[L2]

For a positive-sized square matrix over a commutative ring, the determinant is the signed sum over permutations of products selecting one entry in each row and column (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L3]

Determinants of positive-sized square matrices over a commutative ring are multiplicative (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ ).

[L4]

A triangular matrix has determinant equal to the product of its diagonal entries (The determinant of a triangular matrix is the product of its diagonal entries).

[L5]

Block products obey associative and distributive matrix arithmetic (Matrix arithmetic over a commutative ring is associative, unital and distributive, and transpose reverses products).

Proof

technique · explicit block elimination

1.1

If the total size is zero, every block and $P$ is empty, both displayed matrix identities are the empty identity, and the determinant formula reads $1 = 1 \cdot 1$ . Otherwise $P$ is block upper triangular with identity diagonal blocks, and $P^{- 1} = (\begin{array}{cc} I & A^{- 1} B \\ 0 & I \end{array})$ ; hence it is invertible and [L4] gives $\det P = 1 = \det P^{T}$ .

L4L5givenalgebra

2.1

Since symmetry of $M$ makes $A^{T} = A$ and therefore $(A^{- 1})^{T} = A^{- 1}$ , direct multiplication gives $P^{T} M P = (\begin{array}{cc} A & 0 \\ 0 & C - B^{T} A^{- 1} B \end{array}) = A \oplus S$ .

step 1.1L1L5algebra

3.1

In positive total size, [L3] and step 1.1 give $\det (P^{T} M P) = \det M$ . In the Leibniz sum [L2] for the block diagonal matrix of step 2.1, every nonzero term preserves both index blocks, so its determinant is $\det A \det S$ , with the stated empty-block convention when one block has size zero. Thus $\det M = \det A \det S$ .

step 1.1step 2.1L2L3algebra

4.1

The calculation proves both assertions. Zero-sized blocks, zero entries inside $A, B, C$ , and singular $S$ require no cancellation and are all included.

step 1.1step 2.1step 3.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Sylvester's criterion: a real symmetric $n \times n$ matrix with $n \geq 1$ is positive definite if and only if all leading principal minors are positive

Statement

Let $A \in M_{n} (R)$ be symmetric, with $n \geq 1$ , and let $Δ_{k}$ be the determinant of its leading $k \times k$ principal submatrix. Then $A$ is positive definite if and only if

$Δ_{k} > 0 (1 \leq k \leq n) .$

Facts & Assumptions

Given: A real symmetric $n \times n$ matrix $A$ with $n \geq 1$ .

[L1]

Schur block elimination gives $P^{T} M P = A_{0} \oplus S$ and $\det M = \det A_{0} \det S$ when the leading block $A_{0}$ is invertible (For symmetric $M = (\begin{array}{cc} A & B \\ B^{T} & C \end{array})$ with $A$ invertible, a block-unitriangular congruence gives $A \oplus (C - B^{T} A^{- 1} B)$ and factors $\det M$ ).

[L2]

Every real symmetric form is congruent to exactly one matrix $diag (I_{p}, - I_{q}, 0_{r})$ (Sylvester's law of inertia: every real symmetric form is congruent to $diag (I_{p}, - I_{q}, 0_{r})$ , and $(p, q, r)$ is unique).

[L3]

A product of two positive or two negative elements is positive, and multiplication by a positive scalar preserves and reflects strict inequalities (Sign rules for products and monotonicity of multiplication, clauses 1, 3, and 4).

[L4]

A triangular matrix has determinant equal to the product of its diagonal entries, so $\det I_{k} = 1$ (The determinant of a triangular matrix is the product of its diagonal entries).

[L5]

Determinants are multiplicative (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ ).

[L6]

The determinant of a positive-sized square matrix equals the determinant of its transpose (For every square matrix over a commutative ring, $\det (A^{T}) = \det (A)$ ).

Proof

technique · forward determinant comparison and reverse induction on $n$

1.1

Suppose $A$ is positive definite. Its restriction $A_{k}$ to the first $k$ coordinate subspace is positive definite for every $1 \leq k \leq n$ . In the unique normal form of [L2], a negative or zero diagonal entry would give a nonzero vector of nonpositive value, so the normal form is $I_{k}$ ; choose invertible $P_{k}$ with $P_{k}^{T} A_{k} P_{k} = I_{k}$ . Taking determinants using [L4]–[L6] gives $1 = (\det P_{k})^{2} Δ_{k}$ . Since $\det P_{k} \neq 0$ , its square is positive by [L3], so $Δ_{k} > 0$ .

L2L3L4L5L6givenchoose

1.2

For the converse, the case $n = 1$ reads $A = [Δ_{1}]$ with $Δ_{1} > 0$ , which is positive definite.

basegiven

1.3

Assume $n > 1$ , all $Δ_{k} > 0$ , and the converse in size $n - 1$ . Write $A = (\begin{array}{cc} a & b^{T} \\ b & C \end{array})$ , so $a = Δ_{1} > 0$ , and set $S = C - a^{- 1} b b^{T}$ . For $1 \leq j < n$ , apply [L1] to the leading $(j + 1) \times (j + 1)$ block: $Δ_{j + 1} = a \det S_{j}$ , where $S_{j}$ is the leading $j \times j$ block of $S$ . Thus [L3] gives $\det S_{j} = Δ_{j + 1} / a > 0$ .

ihL1L3given

2.1

The induction hypothesis makes $S$ positive definite. The congruence in [L1] gives $P^{T} A P = [a] \oplus S$ , which is positive definite because $a > 0$ and $S$ is. Since $P$ is invertible, every nonzero vector is $P y$ for a unique nonzero $y$ , so $A$ is positive definite as well.

step 1.3ihL1L3algebra

3.1

Step 1.1 proves the forward implication. Steps 1.2–2.1 prove the reverse implication for every $n \geq 1$ by induction.

step 1.1step 1.2step 2.1discharge-induction∎

5 · Examples, counterexamples and false statements

None yet.

Sources

Standard references

Recommended treatments; not extraction sources.

Dual Spaces, Bilinear and Quadratic Forms, and Sylvester's Law of Inertia

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Linear functionals and the algebraic dual V∗=L(V,F)V∗=L(V,F)

Definition

The dual family (b∗)b∈B(b∗)b∈B​ associated to a Hamel basis BB, defined by b∗(c)=δbcb∗(c)=δbc​

Definition

The dual family of a finite basis is a basis of the dual space, with the same dimension

Statement

Facts & Assumptions

Proof

For an infinite Hamel basis, its dual family is linearly independent but does not span the algebraic dual

Statement

Facts & Assumptions

Proof

Assuming choice, if v∉U≤Vv∈/U≤V, some f∈V∗f∈V∗ vanishes on UU and satisfies f(v)=1f(v)=1

Statement

Facts & Assumptions

Proof

The canonical evaluation map JV:V→V∗∗JV​:V→V∗∗ given by JV(v)(f)=f(v)JV​(v)(f)=f(v)

Definition

Assuming choice, the canonical map JV:V→V∗∗JV​:V→V∗∗ is linear and injective

Statement

Facts & Assumptions

Proof

Assuming choice, JV:V→V∗∗JV​:V→V∗∗ is surjective if and only if VV is finite-dimensional

Statement

Facts & Assumptions

Proof

The annihilator U∘≤V∗U∘≤V∗ of U≤VU≤V and the preannihilator ∘S≤V∘S≤V of S≤V∗S≤V∗

Definition

Assuming choice, ∘(U∘)=U∘(U∘)=U; in finite dimension, dim⁡U∘=dim⁡V−dim⁡UdimU∘=dimV−dimU

Statement

Facts & Assumptions

Proof

The transpose or algebraic adjoint T∗:W∗→V∗T∗:W∗→V∗, T∗(g)=g∘TT∗(g)=g∘T, of a linear map T:V→WT:V→W

Definition

Transpose is linear, sends identities to identities, and reverses composition: (S∘T)∗=T∗∘S∗(S∘T)∗=T∗∘S∗

Statement

Facts & Assumptions

Proof

Assuming choice, ker⁡T∗=(im⁡T)∘kerT∗=(imT)∘ and im⁡T∗=(ker⁡T)∘imT∗=(kerT)∘; in finite dimensions rank⁡T∗=rank⁡TrankT∗=rankT

Statement

Facts & Assumptions

Proof

In dual bases, the matrix of T∗T∗ is the transpose of the matrix of TT

Statement

Facts & Assumptions

Proof

Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms

Definition

Alternating forms are skew-symmetric; the converse holds when char⁡F≠2charF=2, while in characteristic 22 alternating forms are symmetric

Statement

Facts & Assumptions

Proof

Bilinear forms on VV correspond linearly and bijectively to linear maps V→V∗V→V∗

Statement

Facts & Assumptions

Proof

The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space

Definition

A basis change by PP changes the matrix of a bilinear form from AA to PTAPPTAP

Statement

Facts & Assumptions

Proof

Congruent matrices have the same rank; hence rank and nondegeneracy of a bilinear form are basis-independent

Statement

Facts & Assumptions

Proof

Sesquilinear and Hermitian forms over a field with an involution, using the convention linear in the first variable

Definition

For the linear-first convention, a basis change by PP sends a sesquilinear matrix AA to PTA σ(P)PTAσ(P); Hermitian forms satisfy A=σ(A)TA=σ(A)T

Statement

Facts & Assumptions

Proof

A quadratic form qq in arbitrary characteristic and its polar form bq(u,v)=q(u+v)−q(u)−q(v)bq​(u,v)=q(u+v)−q(u)−q(v)

Definition

If char⁡F≠2charF=2, quadratic forms and symmetric bilinear forms correspond by q(v)=B(v,v)q(v)=B(v,v) and B(u,v)=12bq(u,v)B(u,v)=21​bq​(u,v)

Linear functionals and the algebraic dual $V^{*} = L (V, F)$

The dual family $(b^{})_{b \in B}$ associated to a Hamel basis $B$ , defined by $b^{} (c) = δ_{b c}$

Assuming choice, if $v \notin U \leq V$ , some $f \in V^{*}$ vanishes on $U$ and satisfies $f (v) = 1$

The canonical evaluation map $J_{V} : V \to V^{* *}$ given by $J_{V} (v) (f) = f (v)$

Assuming choice, the canonical map $J_{V} : V \to V^{* *}$ is linear and injective

Assuming choice, $J_{V} : V \to V^{* *}$ is surjective if and only if $V$ is finite-dimensional

The annihilator $U^{\circ} \leq V^{}$ of $U \leq V$ and the preannihilator $^{\circ} S \leq V$ of $S \leq V^{}$

Assuming choice, $^{\circ} (U^{\circ}) = U$ ; in finite dimension, $\dim U^{\circ} = \dim V - \dim U$

The transpose or algebraic adjoint $T^{} : W^{} \to V^{}$ , $T^{} (g) = g \circ T$ , of a linear map $T : V \to W$

Transpose is linear, sends identities to identities, and reverses composition: $(S \circ T)^{} = T^{} \circ S^{*}$

Assuming choice, $\ker T^{} = (im T)^{\circ}$ and $im T^{} = (\ker T)^{\circ}$ ; in finite dimensions $rank T^{*} = rank T$

In dual bases, the matrix of $T^{*}$ is the transpose of the matrix of $T$

Alternating forms are skew-symmetric; the converse holds when $char F \neq 2$ , while in characteristic $2$ alternating forms are symmetric

Bilinear forms on $V$ correspond linearly and bijectively to linear maps $V \to V^{*}$

A basis change by $P$ changes the matrix of a bilinear form from $A$ to $P^{T} A P$

For the linear-first convention, a basis change by $P$ sends a sesquilinear matrix $A$ to $P^{T} A σ (P)$ ; Hermitian forms satisfy $A = σ (A)^{T}$

A quadratic form $q$ in arbitrary characteristic and its polar form $b_{q} (u, v) = q (u + v) - q (u) - q (v)$

If $char F \neq 2$ , quadratic forms and symmetric bilinear forms correspond by $q (v) = B (v, v)$ and $B (u, v) = \frac{1}{2} b_{q} (u, v)$

Every symmetric bilinear form on a finite-dimensional space over a field of characteristic not $2$ has an orthogonal basis

Over a field of characteristic not $2$ , every symmetric matrix is congruent to a diagonal matrix

Over a field of characteristic not $2$ , every quadratic form has diagonal coordinates $q (x) = a_{1} x_{1}^{2} + \dots + a_{n} x_{n}^{2}$

Positive and negative definiteness, the inertia $(p, q, r)$ , rank $p + q$ , and signature $p - q$ of a real symmetric bilinear or quadratic form

Sylvester's law of inertia: every real symmetric form is congruent to $diag (I_{p}, - I_{q}, 0_{r})$ , and $(p, q, r)$ is unique

For symmetric $M = (\begin{array}{cc} A & B \\ B^{T} & C \end{array})$ with $A$ invertible, a block-unitriangular congruence gives $A \oplus (C - B^{T} A^{- 1} B)$ and factors $\det M$

Sylvester's criterion: a real symmetric $n \times n$ matrix with $n \geq 1$ is positive definite if and only if all leading principal minors are positive