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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a real $n\times n$ matrix with $n\geq 3$ and consider the function $f\colon (\mathbb{R}^{n})^n\to \mathbb{R}$ defined by $f(X)=\det X$ where $\det X$ means the determinant of $X$. If $X$ has rank $n-1$, is $X$ a critical point of $f$? How can the set of critical points of $f$ be defined in terms of the rank of $X$?

I thought this was fairly straightforward (and it still might be) and that I had a solution, but then I realized I wasn't thinking about the derivative of the determinant. A point $X$ is a critical point if $f'(X)=0$ (or is undefined). Since $\text{rk} X=n-1, \det X=0\Rightarrow X$ is not invertible. However, I'm not really sure how to go about considering the derivative of the determinant function. Is there a way to define the determinant so that it's clear what the derivative is? I feel like I'm close to a solution, but I'm not really sure how to handle this part.

share|cite|improve this question
up vote 2 down vote accepted

$\def\Mat{\operatorname{Mat}}$We will compute the partial derivatives of $\det$. So let $1 \le i,j\le n$ be given, we have \[ \det X = \sum_{k=1}^n (-1)^{k+j} x_{kj}\det X_{kj} \] where $X_{kj}$ denotes the $k,j$-cofactor of $X$. So \[ \frac{\partial \det}{\partial x_{ij}}(X) = (-1)^{i+j}\det X_{ij} \] That is, $X$ is a critical point of $\det$ exactly iff all $\det X_{ij} = 0$. As the cofactors are exactly the $(n-1)$-minors, $X$ is a critical points of $X$ iff $\operatorname{rk} X < n-1$.

share|cite|improve this answer

Your Answer


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.