Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Supposing $A$ is an $m \times n$ matrix where $m > n$ and $A$ has full column rank. I want to find a $C$ (an $m \times m$ matrix) such that $A^TCA$ is a diagonal matrix and also that the maximum singular value of $C$ is the smallest possible.

EDIT: $C$ also has to satisfy: $C= VDV^T$ where $V$ is an orthogonal matrix and $D$ is diagonal with positive entries on the diagonal.


share|cite|improve this question
If $A^TCA$ is diagonal, so is $A^T(\frac C2)A$. So how can you have a $C$ with the smallest possible maximum singular value? – Rahul Apr 13 '12 at 11:33
Depends on the ring that contains $C$'s entries. In other words: We are looking for $C \in GL(m,R)$. What is $R$? – m_l Apr 13 '12 at 11:34
@RahulNarain: Thanks for the comment. I have overlooked that. This just says that the maximum singular value can get as close to $1$ as we want. – jpv Apr 13 '12 at 11:38
@m_l: please look at the updated question. – jpv Apr 13 '12 at 11:41
It still depends on the ring. If $C$ has entries in $\mathbb{Q}$, for example, Rahul's comment applies and there is no $C$ with smallest maximum singular value if I am not mistaken. If $C$ has entries in $\mathbb{Z}$, that's a whole different story. Is $D$ a fixed matrix? – m_l Apr 13 '12 at 11:52
up vote 1 down vote accepted

There is no $C \in \mathbb{R}^{m \times m}$ with minimal singular values.

By the restriction $C = VDV^{tr}$, $C$ must be symmetric and positive definite. Now suppose $C \in \mathbb{R}^{m \times m}$ symmetric and positive definite such that $A^{tr}CA$ is diagonal. As Rahul pointed out in his comment, consider $\widetilde{C} := \frac{1}{2}C$. Then $\widetilde{C}$ is symmetric and positive definite and $A^{tr}\widetilde{C}A = \frac{1}{2} A^{tr}CA$ is diagonal, but the singular values of $\widetilde{C}$ are strictly smaller than those of $C$.

share|cite|improve this answer
@m_1: Thanks for the answer. It is now clear that there exists no such $C$. But for other purposes, I want to know if there is an easy way to find at least one $C$ which is symmetric and positive definite such that $A^TCA$ is diagonal? – jpv Apr 13 '12 at 13:07
I think so. Find $B \in \mathbb{R}^{m \times m}$ such that $BA$ has the identity matrix as its first $n \times n$ submatrix and all rows below contain only 0. Then $C := B^{tr}B$ should be what you are looking for. – m_l Apr 13 '12 at 13:29

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.