How to prove positive semi-definiteness using eigenvalues I'm reading a very short paper by Peter G. Casazza but I cannot understand the proof in page 2. 

Suppose $\mathbf A$ is a positive semi-definite matrix with rank $k$; that is, some of its eigenvalues $\lambda_i$ may be zero, but all the positive eigenvalues are larger than $b$ and $b'$. This is what we want to prove:
$$(\mathbf A - b\mathbf I)^{-1} - (\mathbf A - b'\mathbf I)^{-1} \succeq
\frac{\delta}{2}(\mathbf A - b'\mathbf I)^{-2}$$
where $b' = b-\delta > \delta > 0$ and $\mathbf X \succeq \mathbf Y$ means $\mathbf X - \mathbf Y$ is positive semi-definite.

He finished his proof by showing the following two things:
$$\frac{-1}{b} - \frac{-1}{b'} = \frac{b-b'}{bb'} = 
\frac{\delta}{b'(b'+\delta)} \geq \frac{\delta}{2(b')^2}$$
and 
$$\frac{1}{\lambda_i - b} - \frac{1}{\lambda_i-b'}
= \frac{b-b'}{(\lambda_i - b)(\lambda_i - b')}
\geq \frac{\delta}{(\lambda_i - b')^2}.$$

My question is:


*

*To prove $\mathbf X - \mathbf Y \succeq \mathbf Z$, is it enough to check their eigenvalues and find some relation between them?

*Can anyone explain why those two things above are enough to show the desired positive semi-definiteness?

 A: Step I
If a matrix $A\in \text{ Mat}_n(\mathbb R)$ has eigen-values $\lambda_i$, then $A-aI$ has eigen-values $\lambda_i-a$, $\forall a\in \mathbb R$, and $A^2$ has eigen-values $\lambda_i^2$.
Step II
If all the eigen-values of a matrix are $\ge0$, then the matrix is positive semi-definite.    

Now let $A$ have characteristic equation $f(x)=\Pi_{i=0}^k(x-\lambda_i)^{m_i}$, where $m_i$ are the multiplicities of $\lambda_i$, and $k$ is the number of eigen-values of $A$. Then $A^2$ satisfies the equation $g(A)=0$, where $g(x)=\Pi_{i=0}^{k}(x-\lambda_i^2)^{m_i}$. So the minimal polynomial of $A^2$ must divide $g(x)$; in particular,  all the eigen-values of $A^2$ lie in the set $\{\lambda_i^2\mid i=1,\cdots,k\}$. So the above-found eigen-values are all the ones of $A-bI$ and of $A^2$ respectively.  

This concludes our proof.  
A: This is merely a simple application of unitary diagonalisation. As $A$ is positive semidefinite, $A=U\Lambda U^\ast$ for some unitary matrix $U$ and some nonnegative diagonal matrix $\Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. Therefore
\begin{align*}
M&:=(A - bI)^{-1} - (A - b'I)^{-1} - \frac{\delta}{2}(A - b'I)^{-2}\\
&=U\ \underbrace{\left[(\Lambda - bI)^{-1} - (\Lambda - b'I)^{-1} - \frac{\delta}{2}(\Lambda - b'I)^{-2}\right]}_D\ U^\ast\\
&=UDU^\ast
\end{align*}
and $M\succeq0$ if and only if $D\succeq0$, i.e. if and only if all diagonal entries of $D$ are nonnegative. Clearly the diagonal entries of $D$ are $d_i=\dfrac{1}{\lambda_i - b} - \dfrac{1}{\lambda_i - b'} - \dfrac{\delta}{2(\lambda_i - b')^2}$. Hence the result.
