# Matrix with Orthonormal columns

1. Jul 25, 2009

### hotvette

If Q is an m x n (m > n) matrix with orthonormal columns, we know that $Q^TQ = I$ of dimension n x n. I have a question about $QQ^T$. It is a symmetric m x m matrix but also appears to be singular. Why would it be singular? Probably something basic I've long since forgotten.

2. Jul 25, 2009

### HallsofIvy

Staff Emeritus
Why do you say it "appears to be singular"? You can take Q= I as a matrix with orthonormal matrix such that $QQ^T$ is not singular.

3. Jul 25, 2009

### hotvette

Thanks for your reply. I'm dealing specifically with non-square matrices with more rows than columns where I'm trying to factor $QQ^T$ using QR or SVD. The result is always nonsense so I tried an experiment by generating random non-square matrices with orthonormal columns and in all cases $QQ^T$ had condition numbers ~1E15 and the factorization routines broke down. I figured there must be some rule or axiom I'd forgotten.

This came up when trying to simplify the matrix equation $A^TDAx = A^Tb$ where A is non-square with more rows than columns and D is diagonal. If A is factored into QR the simplification gets stalled because I end up with $QQ^T$ on both sides and can't go further.

The broader question is whether $A^TDAx = A^Tb$ can be solved for x without having to carry out the matrix multiplications $A^TDA$, where A is non-square with more rows than columns and D is diagonal.

Last edited: Jul 25, 2009
4. Jul 26, 2009

### hotvette

I managed to answer the broader question after realizing I made a goof in the equation. It really is ATDAx = ATDb and can be readily solved by letting B = D1/2A. The problem is then BTBx = BTD1/2b which is easy to solve.

On the other question, I'm still curious whether a generalization can be made about the nature of QQT in the case where Q is rectangular (m > n) and has orthonormal columns. I've scoured my linear algebra texts and can't find any statements about this.

5. Jul 26, 2009

### g_edgar

$QQ^T$ has rank n, so if n < m then it is singular.

6. Jul 29, 2009

### hotvette

Thanks. Actually, I realized it doesn't matter that the columns are orthonormal. I found the relation I was looking for:

rank(A) = rank(ATA) = rank(AAT)

It makes intuitive sense since you aren't adding any information by forming the matrix products.

http://en.wikipedia.org/wiki/Rank_(linear_algebra [Broken])

Last edited by a moderator: May 4, 2017