Is the determinant differentiable?

Solution 1:

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}

Solution 2:

The determinant is a polynomial on the entries of the matrix. Hence it's differentiable infinitely many times.

Solution 3:

The determinant of a square matrix is a polynomial of its entries so it is infinitely differentiable.