Show that $A\in\mathbb{C}_n$ is normal $\iff$ $tr(A*A) = \sum_{i = 1}^n|\lambda_i|^2$, where $\lambda_1,...,\lambda_n$ are the eigenvalues of $A$.

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Title restated: Show that $A\in\mathbb{C}_n$ is normal $\iff$ $tr(A^*A) = \sum_{i = 1}^n|\lambda_i|^2$, where $\lambda_1,...,\lambda_n$ are the eigenvalues of $A$.

This question comes from "Matrices and Linear Transformations" by Charles Cullen. I'm studying for an exam and am trying to do some problems but I'm having trouble with this one. Any help would be appreciated.

Thank you.

Daniel · Accepted Answer · 2014-11-20T00:39:34.380

4

Exists an unitary matrix $U$ such that $A=UTU^*$, where $T$ is upper triangular matrix with diagonal $\lambda_1,\ldots,\lambda_n$.

Now $tr(A^*A)=tr(AA^*)=tr(UTU^*UT^*U^*)=tr(UTT^*U^*)=tr(TT^*)$.

If $T$ is not diagonal then $tr(TT^*)>\sum_{i=1}^n|\lambda_i|^2$. If $T$ is diagonal then $A$ is normal.

edited Nov 20 '14 at 00:39

answered Nov 20 '14 at 00:18

Daniel

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Show that $A\in\mathbb{C}_n$ is normal $\iff$ $tr(A*A) = \sum_{i = 1}^n|\lambda_i|^2$, where $\lambda_1,...,\lambda_n$ are the eigenvalues of $A$.

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