# How to use trace operator for inequalities dealing with an hermitian matrix and its inverse?

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!

## 1 Answer

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.