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.

What is the dimension of the space of $\{A\ {^t\!A}: A\in M(n\times n,\mathbb{R})\}$? I think it should be $n(n+1)/2$ if one knows already the dimension of the special orthogonal group, but I would love to derive the latter from the former.

share|improve this question
It's not clear to me that your set is a vector space. It's contained in the $n(n+1)/2$ dimensional space of symmetric matrices, but, for example, $\pmatrix{0&1\cr1&0\cr}$ isn't in it. –  Gerry Myerson May 3 '12 at 23:40
Ah, you are right. But you have made me realize what I was missing before, I think. It is sufficient for my purposes to just note that I can define a map from the $\mathbb{R}^{2n}$ to my set considered as a subset of the symmetric matrices. Thanks! –  user21725 May 3 '12 at 23:45
The space is the same as the set of (symmetric) positive semi-definite matrices. The dimension could be defined as the dimension of the affine hull of the set, which in this case (since $0$ is in the set) is the span of the set. The span of the set is just the set of symmetric matrices, which has the given dimension. –  copper.hat May 4 '12 at 0:52
Talking properly, your set is a cone. –  Martin Argerami May 4 '12 at 1:15
Would you explain this intriguing comment, Martin? If you expanded and wrote an answer I would accept it. –  user21725 May 4 '12 at 1:18

1 Answer 1

up vote 0 down vote accepted

A cone is a subset of a vector space such that it is closed under addition and multiplication by non-negative scalars. Here we are in a finite-dimensional setting, which simplifies things a little.

Typically, a cone can be seen as the "positive part" of your space. Canonical examples are $[0,\infty)\subset\mathbb{R}$, $[0,\infty)\subset\mathbb{C}$, and the positive-semidefinite matrices (either real or complex); this last one is your example.

The main property of the cone is that it is convex (i.e. $x,y$ in the cone, $t\in[0,1]$ imply that $tx+(1-t)y$ is in the cone), and so it can be seen as generated (as a convex set) by its extreme points (that is, every point is a convex combination of extreme points). In your case, the extreme points are the scalar multiplies of the projections, i.e. $\lambda A$ with $\lambda\geq0$, $A=A^t=A^2$.

All this said, I have no idea if this helps with your problem!

share|improve this answer
It doesn't, but it's interesting and as good an answer as this question deserves. –  user21725 May 4 '12 at 2:53
Many definitions of cone that I have seen just require $\lambda x$ be in the code when $x$ is in the cone and $\lambda>0$. If $0$ is in the cone, it is sometimes called a pointed cone. That is, convexity is not necessarily part of the definition. Not that it matters here. –  copper.hat May 4 '12 at 3:09

Your Answer


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