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.

As we know that the Ky Fan k Norm is the sum of k-th largest singular values.

On the other hand, the trace of a matrix is the sum of its eigenvalues.

For a N by N symmetric matrix $M$, its Ky Fan N-Norm is equal to the trace of $M$.

Yet how about the matrix $M$ is square but not symmetric?

Is there any relation between the trace and the Ky Fan $N$ norm?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

The relation is the following:

Let $\Vert A \Vert$ be the Ky-Fan $N$ norm of $A$. Then, $\Vert A \Vert = trace((A^*A)^{\frac{1}{2}})$

btw, The Ky-Fan $N$ norm is equal to the Schatten norm with $p=1$. The Schatten norm is defined as the regular $l_p$ norm of the singular values:

$\Vert A \Vert_p=(\sum_{i=1}^N \sigma_i^p)^{\frac{1}{p}}$

And it is related to the trace by:

$\Vert A \Vert_p = trace((A^*A)^{\frac{p}{2}})$

If you are looking for a good reference about advanced material like that, I would highly recommend "Matrix Analysis" by R. Bhatia

share|improve this answer

Your Answer

 
discard

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.