Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was wondering, given an $n\times n$ square matrix with $n^2$ many entries, the function $\det:\left(a_1,a_2,\ldots,a_{n^2}\right)\to \textbf{R}$ which gives 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.

share|improve this question
    
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 $$\frac{d}{dA} det(A).$$ –  Travis Bemrose Apr 24 at 1:59
    
@TravisBemrose I calculated the total (or Fréchet) derivative of the determinant in my updated answer, maybe that's what you're looking for. –  Christoph Apr 24 at 7:42

4 Answers 4

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

share|improve this answer

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}

share|improve this answer

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.