16
$\begingroup$

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.

  1. Is this function (determinant) a differentiable kind?

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

  3. Furthermore, if so, to what differentiability class does this $\det$ function belong?

Thanks in advance.

$\endgroup$
2
  • 1
    $\begingroup$ All of the answers so far treat the determinant as a polynomial in the matrix entries $(a_{ij})$. However, just as we can take the derivative of $\vec{y}=f(\vec{x})$ as the vector $\vec{x}$ varies, it would be nice to see someone answer the question from the perspective of $$\frac{d}{dA} det(A).$$ $\endgroup$ Apr 24, 2014 at 1:59
  • $\begingroup$ @TravisBemrose I calculated the total (or Fréchet) derivative of the determinant in my updated answer, maybe that's what you're looking for. $\endgroup$
    – Christoph
    Apr 24, 2014 at 7:42

4 Answers 4

49
$\begingroup$

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}

$\endgroup$
2
  • $\begingroup$ What exactly is D, is it another matrix ? $\endgroup$
    – Mo711
    Jun 19, 2023 at 15:16
  • $\begingroup$ It is notation for derivative. So $\mathrm{D}f$ represents the (Fréchet) derivative of $f$. $\endgroup$ Jul 7, 2023 at 11:37
26
$\begingroup$

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

$\endgroup$
9
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .