TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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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

1.1L1L2

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.

1.2L1L2L3

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.

2.1step 1.1step 1.2L3algebra

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.

3.1step 2.1∎

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

Depends on

Dependency tree · 2 levels

Sources

H. Pinkham, Linear Algebra, Chapters 6–7

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