If $A$ is symetric and positive definite, when will $A + B$ be invertible?

Question

$A$ is a symmetric and positive definite matrix with size of $n \times n$, and $B$ is a matrix of the same size. Under which circumstance will $A+B$ be invertible?

Answer 1

Here I post this topic as a record.

We just need to ensure $\lVert Bx\rVert<\lVert Ax\rVert\,,\;\forall x\in\Bbb R^n. $

Note the eigenvalues of $A$ as { $\lambda_i$} , eigenvalues of $B$ as $\{\sigma_i\}.$

Use $l_2$ norm, let $\lVert x\rVert=1$ , then $\lVert Bx\rVert\leqslant\max\{|\sigma _{i}|\}$.

$\lVert Ax\rVert=\lVert\lambda_1\xi_1+\ldots+\lambda_n\xi_n\rVert$ , where $\{\xi_i\}$ is the eigenvector of $A$ , $\sum\limits_{i=1}^{n}\xi_i=x\,.$
Now, for $A$ is symetric and positive definite, so $\{\xi _i\} $ are orthogonal. As a result, $\lVert Ax\rVert=\lVert\lambda_1\xi_1+\ldots+\lambda_n\xi_n\rVert=\lVert\lambda_1\xi_1\rVert+\ldots+\lVert\lambda_n\xi_n\rVert\geqslant\min\{\lambda_i\}$

So, if $\min\{\lambda_i\}>\max\{|\sigma_{i}|\}$, we will have $A+B$ is invertible.