Matrix with Orthonormal columns

[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.

 

[HallsofIvy]

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.

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.

 

[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.

 

[hotvette]

I managed to answer the broader question after realizing I made a goof in the equation. It really is A^TDAx = A^TDb and can be readily solved by letting B = D^1/2A. The problem is then B^TBx = B^TD^1/2b which is easy to solve.

On the other question, I'm still curious whether a generalization can be made about the nature of QQ^T 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.

 

[g_edgar]

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.

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

 

[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(A^TA) = rank(AA^T)

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

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