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Prove that if $A$ is ANY $n\times n$ matrix, then $\det(\operatorname{adj}(A)) = (\det(A))^{n-1}$.

first of all, since I did it about 3 or 4 times before, I started off by proving the case where $A$ is an INVERTIBLE $n\times n$ matrix from $A^{-1} = \operatorname{adj}(A)/\det(A)$.

but how do I prove it when $A$ isn't invertible? I wanna do something like plug $0$ in for $\det(A)$ but how can I use $A^{-1}$ in my proof if $A$ is not invertible? do I have to just substitute $A$ for $B$ and continue using $B^{-1}$ where needed? If so why is that even allowed?

darij grinberg
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J L
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  • http://math.stackexchange.com/q/185333/184154 – curious Oct 13 '14 at 23:33
  • One trick is to consider the matrix $X\cdot I_n + A$ over the polynomial ring $K\left[X\right]$ (where $K$ is the ring over which your matrix lives). The matrix $X\cdot I_n + A$ is not usually invertible, but its determinant is a monic polynomial in $X$ and thus can be cancelled from any equation. And once you have shown that $\det\left(\operatorname{adj}\left(X\cdot I_n + A\right)\right) = \left(\det\left(X \cdot I_n + A\right)\right)^{n-1}$, you can evaluate at $X = 0$ and get $\det\left(\operatorname{adj}A\right) = \left(\det A\right)^{n-1}$. Note that you need $n \geq 1$. – darij grinberg Oct 13 '14 at 23:34

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Use the Laplace expansion to prove the formula $$ A\times\text{adj}A=\det(A)I. $$ If $\det(A)\neq 0$, then your conclusion follows. Now, consider $\det(A)=0$. Then, you have $A\times\text{adj}(A)=0$. If $A$ is the $0$ matrix, then $\text{adj}(A)$ is also the $0$ matrix and thus has determinant $0$. If $A$ has a nonzero row, then this row together with $A\times\text{adj}(A)=0$ provides a linear dependence between the rows of $\text{adj}(A)$, which implies $\det(\text{adj}(A))=0$.

Kim Jong Un
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  • Rephrasing the last part more simply (perhaps), if $\det A = 0$, $\operatorname{im} \operatorname{adj} A \subset \operatorname{ker} A$, so if the latter isn't the whole space, then $\operatorname{adj} A$ isn't surjective and so it's singular. – Najib Idrissi Oct 15 '14 at 15:17
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If $A$ is singular and $A\neq0$, then $\text{det}(A)=0$ and $\text{rank}(A)>0$, $adj(A)A=0$ and $\text{rank}(adj(A))+\text{rank}(A)\leq\text{rank}(adj(A)A)+n$, $\text{rank}(adj(A))<n$, $adj(A)$ is singular and $\text{det}(adj(A))=0=(\text{det}(A))^{n-1}$.

If $A=0$, then $adj(A)=0$, then the result is obvious.

Alfred Chern
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$$\det (\text{adj} A) = \det(A)^{n-1}$$

is a polynomial equality with integral coefficients in variables $(a_{ij})$ and is valid on a nonvoid open subset of $M_n(\mathbb{C})$, hence it is an identity.

orangeskid
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