Partial derivatives of determinant

Question

I have recently took and exam and one of the exercises involved finding the partial derivatives of the function $F: M_n(\mathbb{R}) \rightarrow \mathbb{R}, F(A) = \det (A)$, where $M_n(\mathbb{R})$ is the set of matrices with real entries. My approach of this problem consisted of considering $M_n(\mathbb{R})$ as equivalent to $\mathbb{R}^{n^2}$, what means that $F$ would actually a function with $n^2$ variables, each one being real. Therefore, it makes sense talking about partial derivatives of the "function determinant".
In the case $n = 2$, I was able to find those partial derivatives and "the gradient" of the function. It is easy since if we consider a square matrix of order $2, A = (a_{ij}), 1 \leq i,j \leq 2$. Obviously, $F(A) = a_{11}a_{22}-a_{12}a_{21}$ and the partial derivatives are easy to find. I was wondering which is the result for a general $n$, since I was asked to find the differential application in the "point" $I_2$, being $I_2$ the identity matrix for this size, and finding the partial derivatives for a general $n$ is the only step I am missing to find the differential application.

Answer 1

Let $A = \left[ a_{ij} \right]_{n \times n}$. Then we can calculate the determinant of $A$ as follows:

1. Expanding along row $i_0$, where $i_0 \in \{ 1, \ldots, n\}$, we have $$ |A| = a_{i_0 \, 1} A_{i_0 \, 1} + \cdots + a_{i_0 \, n} A_{i_0\, n}, $$ where, for each $j = 1, \ldots, n$, $$ A_{i_0 \, j} = (-1)^{i_0 + j} \left| M_{i_0 \, j} \right|, $$ and $M_{i_0 \, j}$ is the matrix obtained from $A$ by deleting the $i_0$th row and $j$th column of $A$.

2. Expanding along column $j_0$, where $j_0 \in \{ 1, \ldots, n\}$, we have $$ |A| = a_{1 \, j_0} A_{1 \, j_0} + \cdots + a_{n \, j_0} A_{n \, j_0}, $$ where, for each $i = 1, \ldots, n$, $$ A_{i \, j_0} = (-1)^{i + j_0} \left| M_{i \, j_0} \right|, $$ and $M_{i \, j_0}$ is the matrix obtained from $A$ by deleting the $i$th row and $j_0$th column of $A$.

Note that, for each $i, j \in \{1, \ldots, n\}$, the cofactors $A_{ij}$ do not involve $a_{ij}$ at all.

Now let $f \colon M_n(\mathbb{R}) \longrightarrow \mathbb{R}$ be the function defined by $$ f(A) \colon= |A|. $$ Then we have $$ \frac{\partial f}{\partial a_{ij} } = A_{ij} $$ for each $i = 1, \ldots, n$ and for each $j = 1, \ldots, n$.