# Comparing Rank and Trace of a Matrix

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• Euge
In summary, the rank of a matrix represents the number of essential components or variables in the data, while the trace represents the sum of the eigenvalues or variances. The rank is always less than or equal to the trace and both have significant roles in solving linear systems and analyzing properties. The rank and trace can be equal in a full rank square matrix, but cannot be directly compared as they depend on the size of the matrix. However, the normalized trace can give a sense of the average importance of each variable and can be compared between different matrices.
Euge
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MHB
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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?

topsquark
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).

Last edited by a moderator:
Euge, julian, anuttarasammyak and 1 other person

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