Derivative of determinant at some point

Question

Let $c:\mathbb{R} \rightarrow \mathbb{M}_n(\mathbb{R})$ defined by $$c(t)=A e^{tB}$$ where $A\in GL(n,\mathbb{R})$ and $B \in \mathbb{M}_n(\mathbb{R})$.

The question ask me to find $c'(0)$ and $(\text{det}\circ c)'(0)$.

Then find the derivative of $\text{det}:\mathbb{M}_n(\mathbb{R})\rightarrow \mathbb{R}$ at $A$

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I did the first part, where: $$c'(0)=AB$$ $$(\text{det}\circ c)'(0)=\text{det}(A) \text{Tr}(B)$$

But I don't know how to deduce $D\text{det}_A$, the derivative of $\text{det}$ at $A$.

Here is what I was thinking, since $D\text{det}_A$ takes initial tangent vector (which is $c'(0)$) to initial tangent vector of $\text{det}\circ c$.

Hence,

$$D\text{det}_A(c'(0)) =(\text{det}\circ c)'(0) $$

and I don't know what to do next.

Answer 1

The idea was to exploit the fact that $$\det A\, tr B=(\det\circ c)'(0) = \nabla_X(\det X)'\big|_{X=c(0)}:c'(0)$$ with $\nabla_X(\det X)'$ being the gradient of the determinant with respect to coefficients of the matrix.

Hence, we get $$\det A\,tr B = \nabla (\det A):(AB)$$ As $\det A\ne0$, we can chose $AB$ to be whatever matrix we want: For example, if $AB=e_{ij}$, then $(\nabla (\det A):(AB)_{ij}=\nabla (\det A)_{ij}$, and on the other hand we have $B=A^{-1}e_{ij}$, therefore $tr B = (A^{-1})_{ji}$.

Finally, by enumerating all possible indices $i,j$ we conclude that $$\nabla(\det A) = \det A(A^{-1})^T.$$

As a side note, I think that Laplace development for determinant together with the notion of comatrix allow to obtain the above formula in a much easier way.

An addendum: where the double contraction comes from: let $X $ with components $x_{ij}$ be a square matrix and we study the gradient of $f(X)$ - certain numerical function depending on $X$. This gradient $\nabla_X f(X)$ is a matrix whose components are $$(\nabla_X f(X))_{ij}=\frac{\partial f(X)}{\partial x_{ij}}.$$ Now let us suppose that $X=X(t)$ and we want the derivative of the function $t\to f(X(t))$. By the rule of composite derivative we get $$\frac{d}{dt}f(X(t)) = \sum_{ij}\frac{\partial f(X)}{\partial x_{ij}}\big|_{X=X(t)}\cdot \frac{d}{dt}x_{ij}(t). $$

Now, take into account that $\frac{d}{dt}x_{ij}(t)$ form a matrix $X'(t)$ and the definition of double contraction: $A:B = \sum_{i,j}A_{ij}B_{ij}$ to obtain $$\frac{d}{dt}f(X(t)) = \nabla_X f(X(t)):X'(t).$$

The derivative of the matrix is taken component-wise (i.e. "the derivative of a matrix is a matrix of derivatives").

Answer 2

For finding $d(\det)_A$, let's use what you've already computed. Given $B\in M_n(\mathbb{R})$, we have the curve $c(t)=A\exp{tA^{-1}B}$ with $c(0)=A$ and $c'(0)=B$. Thus \begin{align*} d(\det)_A(B)&=d(\det)_A(c'(0))\\ &=(\det\circ c)'(0)\\ &=\det{A}\operatorname{tr}(A^{-1}B). \end{align*}