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

I'm looking for the dimension of the space of $n\times n$ real matrices $A$ such that $\det(A)=c$.

I apply 2 different approaches and I get different answers. which one is correct?

1) So we know by a variation of the constant rank theorem that a equation $f(x)=0$ specifies a closed submanifold of dimension $n-r$ when $r$ is the constant rank (if constant) of the partial derivatives of $f$.

Expanding $\det(A)$ using the Laplace expansion we get a $1 \times(n^2)$ matrix for $df$, the elements of which are the $(n-1)\times(n-1)$ minors of $A$. Clearly $df$ can either have rank $1$ or $0$. We consider the cases separately and consider their relation to each other.

If zero, all its elements should be zero.
$\implies$ Which means all $(n-1)\times(n-1)$ minors are zero.
$\implies$ that means that determinant of $A$ is zero.
That can only be the case for the matrices of rank less than $n$. ($\det(A)=0$)
$\hskip0.5in$ $\implies$@ if $c\neq0$ ($c= \det(A)$)
$\hskip0.5in$ this concludes that the $df$ always has rank $1$ for our solution set)
$\hskip0.5in$ $\implies$ $\dim(\text{space of such }A)= n^2 -1$
$\hskip0.5in$ $\implies$@ if $c=0$ We don't really need the steps above, as by Gaussian elimination, all such matrices can be put in the form of a matrix with a column or row all being zero.
$\hskip0.5in$ $\implies$thus we would have $n^2 -n$ free elements to choose.
$\hskip0.5in$ $\implies$$\dim(\text{space of such }A)=n^2-n$

That was approach 1.

2) By transforming the matrix to its normal form, i.e. choosing bases in both source and the target, not just like the Jordan canonical form where you use the same bases in both source and target (it is explained in p.5 of this, one can write any matrix in the diagonal form i.e. it's isomorphic to the space of diagonal matrices) and as such, we will have only $n$ elements to worry about.

$\hskip0.5in$ @if $c\neq0$
$\hskip0.5in$ $\implies$ $n-1$ free to choose, one fixed.
$\hskip0.5in$ $\implies$ $\dim(\text{space of such }A)=n-1$

$\hskip0.5in$ @if $c=0$
$\hskip0.5in$ $\implies$ at least one of the elements is zero.
$\hskip0.5in$ $\implies$ $n-1$ elements are free to choose.
$\hskip0.5in$ $\implies$ $\dim(\text{space of such }A)=n-1$

Now, I understand that the main trick in the 2nd approach is having those choices of bases and they are different for each point. But we do a similar thing when considering the special orthogonal group for instance. The isomorphism is different for each point but the points in the two spaces can be mapped bijectively.

Any idea which one is correct?
Thanks in advance for your help ;)

share|cite|improve this question
I'm confused here. The set of matrices with constant determinant is not closed under addition, so is not a vector space under it. Are you using some other operation in place of matrix addition? – Alex Becker May 22 '12 at 23:09
I assume the OP wants the dimension as a smooth manifold. The cleanest approach I can think of is to compute the tangent space to $A = I$ (for $c = 1$). – Qiaochu Yuan May 22 '12 at 23:23
@AlexBecker Yes, you are right. its not closed under addition, nor it is under taking inverse or product. I dropped the "vector" space. Qiaochu filled me in. – shahab29 May 22 '12 at 23:30
@QiaochuYuan Tangent space to Identity would be related to that of the special orthogonal group. And what would be the next step? – shahab29 May 22 '12 at 23:33
@shahab: why do you say that? – Qiaochu Yuan May 22 '12 at 23:37

Your approach 1 looks better to me.

If $f$ is a real valued function and $\nabla f(p)\ne 0$, the equation $f=c$ defines a smooth manifold of codimension $1$ near $p$, by the Implicit Function Theorem. Here we have $f(A)=\det A$. If $c\ne 0$, then at every point $A$ with $f(A)=c$ the gradient $\nabla f$ has a nonzero entry (at least one $(n-1)\times (n-1)$ minor must be nonzero), so IFT applies. The set is a smooth manifold of dimension $n^2-1$.

If $c=0$, things are more complicated because the set $\det A=0$ is not a manifold. For example, in the simplest case $n=2$ it's $\{ad-bc=0\}$. The IFT applies only when the rank of $A$ is $n-1$. Since such points are "generic" on $\{\det A=0\}$, one could still say that the dimension is $n^2-1$. However, the proper consideration of this variety belongs to the realm of algebraic geometry rather than differential geometry.

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.