# Prove $\text{rank}(A) \geq \frac{(\text{tr}(A))^2}{\text{tr}(A^2)}$ when $A$ is Hermitian

If $A \in \mathbb{C}^{n \times n}$ is a non-zero Hermitian matrix, I need to show that $$\text{rank}(A) \geq \frac{(\text{tr}(A))^2}{\text{tr}(A^2)}$$ and reason out when the equality is attained?

Can anyone help me in showing this result?

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Your matrix is diagonalizable to a diagonal matriz with real entries, when you diagonalize it, $\mathrm{rank}A$, $\mathrm{tr} A$ and $\mathrm{tr} A^2$ do not change. This means that you can suppose that $A$ is real and diagonal to begin with.

So suppose $A=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$ with the $\lambda_i\in\mathbb R$. Then $\operatorname{rank}A$ is the number of non-zero $\lambda_i$s, $\operatorname{tr}A=\sum_i\lambda_i$ and $\operatorname{tr}A^2=\sum_i\lambda_i^2$. It follows that we are reduced to showing that

if $\lambda_1$, $\dots$, $\lambda_n$ are real numbers, then $\big(\sum_{i=1}^n\lambda_i\big)^2\leq n\sum_{i=1}^n\lambda_i^2$.

One way to do prove this is to fix $r\geq0$, compute the maximum $M$ of the function $f(\lambda_1,\dots,\lambda_n)=(\lambda_1+\cdots+\lambda_n)^2$ in the sphere $\lambda_1^2+\cdots+\lambda_n^2=r$, and finally check that $M$ is equal to $nr$. This can be done easily using the method of Lagrange multipliers---it will, moreover, tell you where the maximum is achieved, and show when your inequality becomes an equality.

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Rank of a matrix is linked to number of independent rows/ columns in the matrix, whereas trace is linked with sum of eigen values of the matrix. How do we link them? Also number of non-zero eigen value not necessary equals rank of a matrix! – Learner Oct 14 '11 at 5:11
@Learner But number of zero eigs equal to the rank drop no? – user13838 Oct 14 '11 at 5:13
Mariano: (1) The blockquoted inequality is a case of Cauchy-Schwarz inequality. (2) Both the inequality itself and the equality case are direct once you note that RHS$-$LHS is $\sum_{i,j}(x_i-x_j)^2$. – Did Oct 14 '11 at 5:32
@Didier: my favourite proof of C-S uses Lagrange multipliers, so there is no loss :) – Mariano Suárez-Alvarez Oct 14 '11 at 5:37
Seems to me like a typical case of cracking nuts with a sledgehammer since, for example, no differential structure is involved, but as they say, gustibus non est disputandum... – Did Oct 14 '11 at 5:42