When I read a paper, I met these implications involving inequalities :

$$R-a^Ha\ge0 \ \implies \ I-R^{-1/2}aa^HR^{-1/2}\ge0 \ \implies \ 1-a^HR^{-1}a\ge0$$

$R$ is an invertible Hermitian matrix with size $(M,M)$, and $a$ is a $(M,1)$ vector.

I don't know how I can get the third inequality from the second.

Here is my solution, but it seems there is something wrong with it:

The matrix on the left of inequality sign is positive semidefinite, so its trace must be non-negative.

So $tr(I-R^{-1/2}aa^HR^{-1/2})=M-tr(R^{-1/2}aa^HR^{-1/2})=M-tr(a^HR^{-1}a)$.

Because $a^HR^{-1}a$ is a number,


Finally I can get $M-a^HR^{-1}a\ge0$, which is different from the third inequality above.

Can anyone give me some help? Thanks!


Your result is ok, just not optimal! The idea is to multiply the "inequality" from the right by $R^{-1/2}a$ and from the left by $a^HR^{-1/2}$ to obtain $a^HR^{-1}a-(a^HR^{-1}a)(a^HR^{-1}a)\geq 0$. Since $a^HR^{-1}a>0$ -- otherwise there is nothing to prove -- we find the result.


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