Proof for the derivative of the determinant $d det(A)B = tr((cof(X))^{T}A)$

Question

I want to proof that the following equations is true: $$d \space det(X)A = tr((cof(X))^{T}A)$$
I tried using Jacobi's formula, but I couldn't get anywhere.
Has anyone an Idea?

Kind Regards

$cof(X)_{i, j} = (-1)^{i+j}*det(X)_{i, j}$
$(X)_{i, j}$ is equal to the Matrix $X$ where the $i^{th}$ row and $j^{th}$ column are cancelled

Answer 1

Let $X = (x_{i,j})_{i,j = 1}^n$ and let $X_{i,j}$ be the matrix obtained from $X$ by deleting the $i$-th row and $j$-th column. Then the expansion by minors formula is

$$ \det X = \sum_{j = 1}^n (-1)^{i + j} x_{i,j} \det X_{i,j}. $$

It follows that the partial derivative of $X \mapsto \det X$ with respect to $x_{i,j}$ is $(-1)^{i + j}\det X_{i,j}$ or more accurately, it is the map $A \mapsto \left( (-1)^{i + j}\det X_{i,j} \right) a_{i,j}$.

Hence the derivative at $X$ is the linear map

$$ A = (a_{i,j}) \mapsto \sum_{i,j = 1}^n \left. \frac{\partial \det}{\partial x_{i,j}} \right|_{X}(A) = \sum_{i,j = 1}^n \left( (-1)^{i + j} \det X_{i,j} \right)a_{i,j} = \operatorname{tr}\left( \operatorname{cof}(X)^T A \right). $$

So $(d\det)_XA = \operatorname{tr}( \operatorname{cof}(X)^T A )$.

Answer 2

To answer some questions I saw in the comments:

Note that cofactor expansion of determinants tells us that det is actually a polynomial in $n^2$ variables $x_{i,j}$. The derivative they are discussing is the derivative between tangent bundles from differential geometry, which in the simple case of mappings $\mathbb{R}^N \to \mathbb{R}$ is just the gradient of the function, viewed as a linear functional on $\mathbb{R}^N$. The $A$ in the equation is a matrix, since the tangent space to a euclidean space is itself and we are viewing $\mathbb{R}^{n^2}$ as $\mathbb{R}^{n \times n}$.

My suggestion is you start playing around with the polynomial giving det on small matrices and see if you can figure out the pattern for the gradient in general.