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 Post subject: Inequality on Hermitian matricesPosted: Sun Jun 11, 2017 9:18 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 136
Location: Melbourne, Australia
Let $\mathbb{O} \neq A \in \mathcal{M}_n(\mathbb{C})$ be a Hermitian matrix. Prove that

${\rm rank}(A) \geq \frac{\left({\rm tr}(A) \right)^2}{{\rm tr}\left(A^2 \right)}$

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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 Post subject: Re: Inequality on Hermitian matricesPosted: Sat Jul 29, 2017 12:08 am

Joined: Tue Nov 24, 2015 7:47 pm
Posts: 13
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 $rank(A)=n-k$ and deduce the inequality ( since $k$ terms vanish ) in a similar manner .

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 Post subject: Re: Inequality on Hermitian matricesPosted: Sat Jul 29, 2017 8:08 am

Joined: Sat Nov 14, 2015 6:32 am
Posts: 136
Location: Melbourne, Australia
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!

_________________
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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