Let $A$ be $n\times n$ complex squared matrix. I want to find a lower bound for $\mathrm{tr}(A^\dagger A).$ What I could find so far is that if $A$ is Hermitian then $$\mathrm{tr}(A^\dagger A) \geq \frac{|\mathrm{tr}(A)|^2}{\mathrm{rank}(A)}.$$ I wonder whether this is correct for non-Hermitian case. Thanks.
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Please clarify in your question the meaning of the symbol $A^\dagger$. In older literature on linear algebra, $A^\dagger$ usually means Moore-Penrose pseudoinverse, whereas in physics literature, $A^\dagger$ usually means Hermitian transpose. – user1551 Jul 01 '16 at 16:39
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I meant Hermitian transpose. – Hovher Jul 01 '16 at 17:09
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Your bound is indeed correct in general. This question (with say the top answer) shows that the sum of the singular values is at least the sum of absolute values of the eigenvalues.
On the other hand, a rank-$r$ matrix has exactly $r$ nonzero singular values $\sigma_1,\ldots,\sigma_r$. Since the trace of $A^TA$ is the sum of the squares of the singular values, the quadratic mean-arithmetic mean and triangle inequalities give us $$\text{tr}(A^TA) = \sigma_1^2 + \ldots + \sigma_r^2 \geq \frac{1}{r}(\sigma_1+\ldots+\sigma_r)^2 \geq \frac{1}{r}|\text{tr}A|^2$$ as wanted.
