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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?

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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

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