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# Inequality of norm of matrix

Let denote . Consider and let be its eigenvalues. Prove that:

Solution

Noting that , it is clear that this norm is invariant under conjugation by a unitary matrix. Since Schur tells us that every matrix is unitarily equivalent to an upper triangular matrix (where of course the diagonal entries are just the eigenvalues) the claim follows immediately.

## 1 Comment

1. The above result has a nice interpretation; It says that for a square complex matrix the sum of squares of the singular values is greater of equal than the sum of squares of absolute values of the eigenvalues, with equality iff the matrix is normal. Of course if it’s normal then the singular values are exactly the absolute values of the eigenvalues, and the above inequality implies that the reverse implication holds as well.

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