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).