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Inequality on Hermitian matrices https://www.mathimatikoi.org/forum/viewtopic.php?f=2&t=1185 
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Author:  Riemann [ Sun Jun 11, 2017 9:18 pm ] 
Post subject:  Inequality on Hermitian matrices 
Let $\mathbb{O} \neq A \in \mathcal{M}_n(\mathbb{C})$ be a Hermitian matrix. Prove that \[{\rm rank}(A) \geq \frac{\left({\rm tr}(A) \right)^2}{{\rm tr}\left(A^2 \right)}\] 
Author:  dr.tasos [ Sat Jul 29, 2017 12:08 am ] 
Post subject:  Re: Inequality on Hermitian matrices 
I think it's direct from cauchy schwartz . Let me explain myself : because A is Hermitian from spectral theorem it is diagonizable so one may deduce that $ ker(A)=ker(T) $ with $T$ some diagonal matrix . If all eigenvalues are different from zero then $ dimker(A)=0 \Rightarrow rank(A)=n $ and the inequality becomes $ (1^2+1....+1^2)(x_{1}^2+...+x_{n}^2) \geq (x_1+...+x_n)^2 $ which follows directly from cauchy scwartz . If there is an eigenvalue which is zero with geometric multiplicity exactly $ k $ It will appear exactly $ k $ times in the diagonal of $ T $ hence $ rank(A)=nk $ and deduce the inequality ( since $ k $ terms vanish ) in a similar manner . 
Author:  Riemann [ Sat Jul 29, 2017 8:08 am ] 
Post subject:  Re: Inequality on Hermitian matrices 
Yes! Indeed, It is known that if $A$ is Hermitian then $A$ is diagonisable and of course each of its eigenvalues is real. It is also known that if $a_1, \dots, a_m \in \mathbb{R}$ then \[\left ( a_1+\cdots+a_m \right )^2 \leq m \left ( a_1^2 +\cdots +a_m^2 \right )\] Having said the above we have that if $\lambda_1, \dots, \lambda_m$ are the distinct eigenvalues of the matrix $A$ then ${\rm rank}(A)=m$ as well as \[{\rm tr}(A) = \sum_{i=1}^{m} \lambda_i \quad \text{and} \quad {\rm tr} \left ( A^2 \right ) = \sum_{i=1}^{m} \lambda_i^2\] Making use of the inequality above we get the result. My solution is not that different than yours. Thank you! 
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