How to show that $D \det_A (H)$ exists and equals $\det( adj(A)H)$?

Question

Consider the function $\det : M_n(\mathbb R) \to \mathbb R$ ; how to show that for any $A , H \in M_n(\mathbb R)$ , the derivative operator of determinat of $A$ evaluated at $H$ i.e. $D \det_A (H)$ exists and equals $\det( adj(A)H)$ ? Please help . Any link or reference will also be very helpful. Thanks in advance .

Answer 1

It should be $\text{tr}(\text{adj}(A)H)$.

The derivative of $\det A$ is easiest to obtain from Laplace's expansion. Denote $A_{ij}$ the cofactor of the element $a_{ij}$. Then $$ \det A=a_{ij}A_{ij}+\text{independent on $a_{ij}$ terms}\quad\Rightarrow\quad \frac{\partial\det A}{\partial a_{ij}}=A_{ij}. $$ Then in your notations $D\det_A(H)$ is the linear form $$ D\det{}_A(H)=\sum_{ij}\frac{\partial\det A}{\partial a_{ij}}h_{ij}=\sum_{ij}A_{ij}h_{ij}=\text{tr}(\text{adj}(A)H). $$