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?
Best Answer
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