The Determinant of a Linear Operator, Cofactors and Cramer's Rule

For $A\in M_n(F)$, let $w_j$ be the unique vector whose $\mathcal{B}$-coordinate column is column $j$ of $A$, and define $G(A):=\omega (w_0,\dots ,w_{n-1})$. This is well defined, alternating, and column-multilinear.

The rigidity lemma gives $G(A)=G(I_n)\det (A)$.

For $A=M_{\mathcal{B}}(v_0,\dots ,v_{n-1})$, one has $G(A)=\omega (v_0,\dots ,v_{n-1})$ and $G(I_n)=\omega (b_0,\dots ,b_{n-1})$. Substitution in step 2.1 proves the formula and the final determination claim.

Fix an ordered basis $\mathcal{B}$. Define ${\omega }_T(v_0,\dots ,v_{n-1}):=\omega (T{v}_0,\dots ,T{v}_{n-1})$; this is an alternating $n$-linear form.

Define ${\delta }_{\mathcal{B}}(v_0,\dots ,v_{n-1}):=\det M_{\mathcal{B}}(v_0,\dots ,v_{n-1})$. It is alternating and $n$-linear, and ${\delta }_{\mathcal{B}}(\mathcal{B})=1$.

Applying [L1] to ${\omega }_T$ and then to $\omega$ gives ${\omega }_T(v_0,\dots ,v_{n-1})={\omega }_T(\mathcal{B})\det M_{\mathcal{B}}(v_0,\dots ,v_{n-1})$ and ${\omega }_T(\mathcal{B})=\omega (\mathcal{B})\det ([T{]}_{\mathcal{B}})$.

Combining the two formulas in step 2.1 and using [F1] and [L1] once more gives ${\omega }_T=\det (T)\omega$.

If a scalar $c$ has the stated scaling property, evaluate it for ${\delta }_{\mathcal{B}}$ at $\mathcal{B}$. The left side is $\det ([T{]}_{\mathcal{B}})$, so $c=\det ([T{]}_{\mathcal{B}})=\det (T)$.

If $\dim V=0$, [F1] defines the operator determinant directly as $1$; no determinant of a $0\times 0$ representing matrix is invoked.

Suppose $\dim V\ge 1$. Applying [L1] first with $\mathcal{B}$ and then with $\mathcal{C}$ characterises both $\det ([T{]}_{\mathcal{B}})$ and $\det ([T{]}_{\mathcal{C}})$ as the same unique basis-free scaling scalar.

The two matrix determinants are therefore equal in positive dimension; together with the separate zero-dimensional definition in step 1.1, this proves well-definedness in every finite dimension.

If $\dim V=0$, all three determinants in the formula are $1$.

Suppose $\dim V=n\ge 1$, and let $\omega$ be any alternating $n$-linear form. Applying [L1] to $S$ and then to $T$ gives $\omega (ST{v}_0,\dots ,ST{v}_{n-1})=\det (S)\det (T)\omega (v_0,\dots ,v_{n-1})\mathrm{.}$

The uniqueness clause of [L1], applied to $S\circ T$, identifies the scaling scalar in step 1.2 as $\det (S\circ T)$. Together with step 1.1, this proves the formula in every finite dimension.

By [L1], $\det (T)=\det (A)$, independently of the chosen basis.

Apply Gaussian elimination to $A$ and record, in order, every row operation used to reach an echelon matrix $U$.

Reversing the recorded effects in [L3] expresses $\det (A)$ in terms of $\det (U)$: each swap contributes a factor $-1$, each forward row scaling by $c\mathrm{\ne }0$ contributes a reverse factor $c^{-1}$, and a row addition contributes no factor.

Since the square echelon matrix $U$ is upper triangular, [L4] gives $\det (U)$ as its diagonal product. With $s$ swaps and scalings $c_1,\dots ,c_m$, step 2.1 gives $\det (U)=(-1{)}^s({\prod }_j{c}_j)\det (A)$, and step 1.1 identifies $\det (A)$ with $\det (T)$; solving for $\det (T)$ gives the displayed formula.

If $n=1$, both displayed sums have one term, $a_{00}(-1{)}^0\cdot 1=a_{00}=\det (A)$.

Suppose $n>1$. In the Leibniz sum for $\det (A)$, group the terms according to the unique column $k$ whose chosen entry lies in row $i$.

After the factor $a_{ik}$ is removed, the remaining choices are exactly the permutations contributing to $\det (A^{(i,k)})$. Moving row $i$ and column $k$ to the last positions takes $(n-1-i)+(n-1-k)$ transpositions, whose parity is the parity of $i+k$; hence the sign contribution is $(-1{)}^{i+k}$.

Summing the groups in step 2.1 gives $\det (A)={\sum }_k{a}_{ik}(-1{)}^{i+k}\det (A^{(i,k)})={\sum }_k{a}_{ik}{C}_{ik}(A)$.

Apply the row formula from step 3.1 to $A^{\mathsf{T}}$. Its row-$j$ cofactors are the column-$j$ cofactors of $A$, and [L2] identifies the two determinants, giving the column formula.

Steps 1.1, 3.1, and 4.1 prove both formulas for every $n\ge 1$.

The $(i,j)$ entry of $A\mathrm{adj}(A)$ is ${\sum }_k{a}_{ik}{C}_{jk}(A)$. If $i=j$, [L1] makes this $\det (A)$.

If $i\mathrm{\ne }j$, replace row $j$ of $A$ by row $i$. Expanding the resulting determinant along row $j$ gives ${\sum }_k{a}_{ik}{C}_{jk}(A)$, because the minors used in that row do not involve row $j$. The matrix has two equal rows, so its determinant is $0$ by alternation.

The $(i,j)$ entry of $\mathrm{adj}(A)A$ is ${\sum }_k{C}_{ki}(A)a_{kj}$. It equals $\det (A)$ when $i=j$ by column expansion. When $i\mathrm{\ne }j$, it is the column-$i$ expansion of the matrix obtained by replacing column $i$ by column $j$, whose determinant is $0$ because it has two equal columns.

Thus $A\mathrm{adj}(A)=\det (A)I_n$.

Hence $\mathrm{adj}(A)A=\det (A)I_n$.

Combining steps 2.1 and 2.2 proves both identities.

If $A$ is invertible, [L1] says directly that $\det (A)$ is a unit.

Conversely, suppose $u:=\det (A)$ is a unit, and let $u^{-1}$ be its inverse.

Multiplying both identities in [L2] by the scalar $u^{-1}$ gives $A(u^{-1}\mathrm{adj}(A))=(u^{-1}\mathrm{adj}(A))A=I_n$.

Thus $u^{-1}\mathrm{adj}(A)$ is a two-sided inverse, so $A$ is invertible. Together with step 1.1, this proves both directions.

Put $u=\det (A)$. Scaling [L1] by $u^{-1}$ gives $A(u^{-1}\mathrm{adj}(A))=(u^{-1}\mathrm{adj}(A))A=I_n$.

By [L2], $A^{-1}$ exists. Both it and $u^{-1}\mathrm{adj}(A)$ are two-sided inverses of $A$, so [L3] makes them equal.

Column $j$ of $A+u{v}^{\mathsf{T}}$ is $A_j+v_{j}u$. Expanding the determinant by column multilinearity gives one term for each subset of columns chosen from $u{v}^{\mathsf{T}}$.

The empty subset contributes $\det (A)$. Every term choosing at least two columns from $u{v}^{\mathsf{T}}$ vanishes, since those chosen columns are scalar multiples of the same column $u$ and alternation makes the determinant zero.

For the singleton subset $\{j\}$, pull out $v_j$ and expand the determinant of $A$ with column $j$ replaced by $u$ along that column. Deleting that replaced column leaves exactly the same minors as deleting column $j$ from $A$, so its contribution is $v_j{\sum }_i{u}_i{C}_{ij}(A)$.

Summing step 2.2 over $j$ and using [F1] gives ${\sum }_{i,j}{v}_j{C}_{ij}(A)u_i=v^{\mathsf{T}}\mathrm{adj}(A)u$. Together with step 2.1, this is the claimed identity.

For arbitrary columns $u,v$, $C+u{v}^{\mathsf{T}}=P^{-1}(A+(Pu)(v^{\mathsf{T}}{P}^{-1}))P\mathrm{.}$

Apply [L1] to step 1.1, then [L2] to both rank-one updates. Since $\det (C)=\det (A)$, cancellation in the additive group of $R$ gives $v^{\mathsf{T}}\mathrm{adj}(C)u=v^{\mathsf{T}}{P}^{-1}\mathrm{adj}(A)Pu\mathrm{.}$

For each $i,j$, let $v$ be the column with entry $1$ at $i$ and $0$ elsewhere, and let $u$ be the analogous column at $j$. The product formula [F2] makes $v^{\mathsf{T}}Bu=B_{ij}$ for every $B\in M_n(R)$, so step 2.1 says that the $(i,j)$ entries of $\mathrm{adj}(C)$ and $P^{-1}\mathrm{adj}(A)P$ are equal.

Equality of all entries proves $\mathrm{adj}(C)=P^{-1}\mathrm{adj}(A)P$, and substituting the definition of $C$ proves the statement.

If $Ax=b$, then $b={\sum }_k{x}_k{A}_k$, where $A_k$ is column $k$ of $A$.

Now suppose $u:=\det (A)$ is a unit. By [L3], $A$ is invertible, so $x=A^{-1}b=u^{-1}\mathrm{adj}(A)b$ is the unique solution.

Replace column $j$ by the sum from step 1.1 and use column multilinearity. Every term with $k\mathrm{\ne }j$ has two equal columns and vanishes; the term $k=j$ is $x_j\det (A)$. Thus $\det (A_j(b))=x_j\det (A)$.

Its $j$th coordinate is $u^{-1}{\sum }_i{C}_{ij}(A)b_i$. Deleting column $j$ removes the replaced column, so $C_{ij}(A_j(b))=C_{ij}(A)$; expanding $A_j(b)$ along column $j$ therefore identifies the sum as $\det (A_j(b))$.

Step 2.1 proves the unconditional cross-multiplied identity, and step 2.2 proves the quotient formula and uniqueness under the unit hypothesis.

If $\det (A)\mathrm{\ne }0$, it is a unit by [F1], so [L1] gives the unique solution and the displayed quotient formula.

Conversely, suppose $Ax=b$ has the unique solution $x$. If $Az=0$, then $A(x+z)=b$ by distributivity, so uniqueness gives $z=0$. Thus the kernel of $A$ is zero.

By [L2], step 1.2 makes $A$ invertible. By [L3] and [F1], $\det (A)$ is a unit and hence is nonzero.

Steps 1.1 and 2.1 prove both directions.

Suppose $\dim V\ge 1$ and choose an ordered basis, writing $A=[T]$. If $T$ is invertible, then $A$ is invertible by [L1]. Conversely, if $A$ is invertible, [L4] gives an operator $S$ with $[S]=A^{-1}$; the composite formula gives $[S\circ T]=I=[T\circ S]$, and injectivity of matrix representation gives $S\circ T=I_V=T\circ S$. Thus $T$ is invertible exactly when $A$ is invertible.

If $\dim V=0$, the unique endomorphism is the identity and is invertible, while [F2] gives $\det (T)=1\mathrm{\ne }0$.

By [L2] and [F1], $A$ is invertible exactly when $\det (A)\mathrm{\ne }0$; by [L3], this is exactly $\det (T)\mathrm{\ne }0$.

Steps 2.1 and 1.2 establish the equivalence in every finite dimension.

If $T$ is invertible, [L2] gives $\det (T)\mathrm{\ne }0$, so $\det (T)\in F^{\times }$ and [L3] supplies its inverse.

Put $d=\det (I_V)$. By [L1], $d=d^2$, and [L2] gives $d\mathrm{\ne }0$ because $I_V$ is invertible. Field cancellation yields $d=1$.

Multiplicativity [L1] and step 1.2 show that determinant preserves the group product and identity.

For invertible $T$, [F1] gives $T\circ T^{-1}=I_V$. Applying [L1] and step 1.2 yields $\det (T)\det (T^{-1})=1$, so uniqueness of inverses in [L3] gives $\det (T^{-1})=\det (T{)}^{-1}$.

Steps 1.1, 2.1, and 2.2 prove the homomorphism and inverse claims.