Is the determinant differentiable? I was wondering, given an $n \times n$ square matrix, let function $\det : \left(a_1,a_2,\ldots,a_{n^2}\right) \to \textbf{R}$ give the determinant, where $a_{k}$'s are the entries of the $n \times n$ matrix.

*

*Is this function (determinant) a differentiable kind?


*If so, is the derivative continuous? That is, is $d\left(\det\right)$ a continuous function?


*Furthermore, if so, to what differentiability class does this $\det$ function belong?
Thanks in advance.
 A: The determinant of a square matrix is a polynomial of its entries so it is infinitely differentiable.
A: If you would not know that the determinant is a polynomial in the entries of the matrix you may know that it is, if considered as a function of the columns (or rows) of the matrix, mulitilinear, hence $C^{\infty}$ as a function of the columns. Since the matrices depend smoothly on their entries they also depend smoothly on the columns.
A: As others have noted, since $A=(a_{ij})_{i,j=1\dots n}$ has determinant
$$
\det A = \sum_{\sigma\in S_n} \epsilon_\sigma\prod_{k=1}^n a_{k,\sigma(k)}
$$
which is a polynomial expression in the $a_{ij}$, the map $\det: \mathbb R^{n\times n}\to\mathbb R$ is infinitely differentiable. The first derivative with respect to $a_{ij}$ is calculated as
$$
\frac{\partial}{\partial a_{ij}} \det A = (\operatorname{adj} A)_{ji},
$$
where $\operatorname{adj} A$ is the adjugate matrix of $A$.
We can also look at the total derivative (or Fréchet derivative) $\mathrm D\det: \mathbb R^{n\times n}\to L(\mathbb R^{n\times n},\mathbb R)$ which assigns to every $A\in\mathbb R^{n\times n}$ the linear map $\mathrm D \det(A) : \mathbb R^{n\times n}\to \mathbb R$ given by
$$
(\mathrm D\det(A))(B) = \sum_{i,j} \left( \frac{\partial}{\partial a_{ij}} \det A\right) b_{ij}= \sum_{i,j} (\operatorname{adj}A)_{ji}b_{ij} = \operatorname{tr}((\operatorname{adj} A)B).$$
For invertible $A$ we can use $A^{-1}=\frac{1}{\det a}\operatorname{adj}A$ to get the expression
$$
(\mathrm D\det(A))(B) = \det(A)\operatorname{tr}(A^{-1} B).
$$
This allows us to use the chain rule to calculate the derivative of functions like $f(t)=\det(A(t))$ where $A:\mathbb R\to\mathbb R^{n\times n}$ is a differentiable matrix-valued function. By the chain rule, we have
\begin{align}
f'(t) &= \left(\mathrm Df(t)\right)(1) = \left(\mathrm D(\det\circ A)(t)\right)(1) =
\left(\mathrm D \det(A(t)) \circ \mathrm D A(t)\right)(1) \\&=
\left(\mathrm D \det(A(t))\right)\left(\mathrm D A(t)(1)\right) = 
\left(\mathrm D \det(A(t))\right)\left(\frac{\mathrm d A(t)}{\mathrm dt}\right) \\&= 
\operatorname{tr}\left(\left(\operatorname{adj} A(t)\right)\frac{\mathrm d A(t)}{\mathrm dt}\right).
\end{align}
A: The determinant is a polynomial on the entries of the matrix. Hence it's differentiable infinitely many times.
