Inverse of sum of two matrix when one is PSD and the other is diagnal?

Question

See I wanna calculate the inverse of following equation:
$(A + \lambda I)^{-1}$
where $A$ is possitive semi definate (PSD).
I wanna decompose $A$ and $\lambda I$ so to treat their (if have) inversion separately. Is there any possibility to simplify and extract something from it?

Answer 1

I can't really think of any way to treat the inversion of these matrices completely separately, but I do have this:

For any symnetric matrix $A$ of $\text{size} \; n$ there exists an orthogonal matrix $O$,

$OO^T = O^TO = I, \tag 1$

such that

$O^TAO = \text{diag}(\mu_1, \mu_2, \ldots, \mu_n), \tag 2$

where the $\mu_i$, $1 \le i \le n$, are the eigenvalues of $A$; also, note that

$O^T(\lambda I)O = \lambda O^TIO = \lambda O^TO = \lambda I; \tag 3$

thus

$O^T(A + \lambda I)O = O^TAO + O^T(\lambda I)O = \text{diag}(\mu_1 + \lambda, \mu_2 + \lambda, \ldots, \mu_n + \lambda), \tag 4$

which is invertible provided

$\lambda \ne -\mu_i, \; 1 \le i \le n; \tag 5$

indeed, we have

$[O^T(A + \lambda I)O]^{-1} = \text{diag}((\mu_1 + \lambda)^{-1}, (\mu_2 + \lambda)^{-1}, \ldots, (\mu_n + \lambda)^{-1}); \tag 6$

combining this equation with

$[O^T(A + \lambda I)O]^{-1} = O^T(A + \lambda I)^{-1}O \tag 7$

yields

$O^T(A + \lambda I)^{-1}O = \text{diag}((\mu_1 + \lambda)^{-1}, (\mu_2 + \lambda)^{-1}, \ldots, (\mu_n + \lambda)^{-1}), \tag 8$

from which

$(A + \lambda I)^{-1} = O\text{diag}((\mu_1 + \lambda)^{-1}, (\mu_2 + \lambda)^{-1}, \ldots, (\mu_n + \lambda)^{-1})O^T. \tag 9$ .