Comparing Rank and Trace of a Matrix

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Let $M$ be a nonzero complex $n\times n$-matrix. Prove $$\operatorname{rank}M \ge |\operatorname{trace} M|^2/\operatorname{trace}(M^\dagger M)$$ What is a necessary and sufficient condition for equality?

 

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Spoiler

Consider the inner product $\langle A,B\rangle=\text{tr}(B^*A)$ on the space of $n\times n$ complex matrices.

Let $P$ be the projection matrix onto the column space of $M.$ Note that $P^*=P$ and $PM=M.$

Then, from Cauchy-Schwarz,

$|\text{tr}(M)|^2=|\text{tr}(PM)|^2=|\langle M,P\rangle|^2 \leq \langle M,M\rangle \langle P,P\rangle=\text{tr}(M^*M) \text{rank}(M).$

Dividing both sides by $\text{tr}(M^*M)$ proves the inequality.

Equality in Cauchy Schwarz occurs when $M$ and $P$ are dependent, i.e. $M$ is a multiple of a projection matrix (which I think should be equivalent to saying that it is diagonalizable, all its nonzero eigenvalues are equal, and its nullspace is orthogonal to its columnspace).