Invertibility of $A-\Delta Id$. Assume that $A\in\mathbb{R}^{n\times n}$ is positive semidefinite and let
\begin{equation}
\Delta=diag(\lambda_1,\ldots,\lambda_n),
\end{equation}
for $\lambda_1,\ldots,\lambda_n\in\mathbb{R}$. What can we say about the invertibility of $A+\Delta$? I mean, if $A$ is diagonal and none of the $\lambda_j$'s is equal to any of the eigenvalues of $A$, then $A+\Delta$ is invertible, because it is a diagonal matrix having non-zero eigenvalues. But what can we say in the general situation?
 A: Aassuming $\Delta$ is invertible, let's consider
$$A-(-\Delta)$$
There is a theorem found in Rudin's Principles of Mathematical Analysis book which provides conditions guaranteeing $A$ will be invertible.
For a given invertible linear map $E$ and linear map $F$ if
$$\|F-E\|\cdot\|E^{-1}\|<1$$
then $F$ is invertible.
Note: The intuition may be that for a given linear map $$ and invertible linear map $$, as long as the distance $\|F-E\|$does not exceed (is strictly less than) $1/\|E^{-1}\|$ then  is invertible.
We first consider $F=A$ and $E=-\Delta$ then the condition reads:
$$\|A-(-\Delta)\|\cdot\|-\Delta^{-1}\|<1$$
Here, the norm $\|\cdot\|$ can be the largest absolute value of all matrix entries.
Since $\Delta$ is assumed to be diagonal we have $\Delta^{-1}=\text{diag}\{\lambda_1^{-1},...,\lambda_n^{-1}\}$ and so the norm $$\|-\Delta^{-1}\|=\text{max}_{j=1,...,n}\{|\lambda_j^{-1}|\}=|\lambda_k^{-1}|$$
for some $k\in \{1,...,n\}$.
So you will have the condition: if $\|A-(-\Delta)\|<|\lambda_k^{-1}|^{-1}=|\lambda_k|$ then $A$ will be invertible.
For $A+\Delta$ to be invertible, you can use the same idea; if
$$\|A+\Delta-\Delta\|\cdot\|\Delta^{-1}\|<1$$ then the theorem implies $A+\Delta $ is invertible.
A: In general, this matrix is almost always invertible. Most matrices are. However, there is a simple condition which guarantees invertibility, which is that all the $\lambda_j$ are positive. In this case, $A+\Delta$ is positive definite, and therefore invertible. To see this, compute
$$ x^\intercal(A+\Delta)x = x^\intercal A x + x^\intercal \Delta x. $$
Both terms are at least zero, and the second term is positive if $x \neq 0$.
