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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,

$$M-tr(a^HR^{-1}a)=M-a^HR^{-1}a.$$

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!

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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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