Can absolute max eigenvalue of A and B tell whether A+B is invertible?

Question

The teaching assistant explained to me that if

1. $A,B\in\mathbb R^{nxn}$ are symmetric,
2. $|\,\text{max eigenvalue of }B\,|\le1\le2\le|\,\text{max eigenvalue of }A\,|$, and
3. $|\,\text{min eigenvalue of }A\,|\ge2$,

then $A+B$ is invertible. Does this make any sense? I know that

• A square matrix is invertible if and only if it does not have a zero eigenvalue.
• Eigenvalues of $A$ + eigenvalues of $B$ = eigenvalues of $A+B$.

But to me, these don't gather up to prove $A+B$ is invertible..

Answer 1

Eigenvalues of A + Eigenvalues of B = Eigenvalues of A+B.

This is not true.

The true reason is that $$\forall V\in R^n, |V^T(A+B)V|\ge|V|^2.$$ So there is no zero eigenvalue.

Answer 2

If you haven't misquoted your TA, then he/she is wrong. E.g. the given conditions are satisfied by $A=\operatorname{diag}(2,1,-1,-2)$ and $B=\operatorname{diag}(0,-1,1,0)$, but $A+B=\operatorname{diag}(2,0,0,-2)$ is singular.

It is true, however, that if the maximum modulus of all eigenvalues of $B$ is smaller than the minimum modulus of all eigenvalues of $A$, i.e. if $$ |\lambda|_\max(B)<|\lambda|_\min(A)\tag{$\ast$} $$ (as opposed to $|\lambda_\max(B)|<|\lambda_\min(A)|$ in your question --- don't confuse the max/min modulus of the eigenvalues with the modulus of the max/min eigenvalue), then $A+B$ is invertible. For, if $(A+B)x=0$, then $$ |\lambda|_\min(A)\,\|x\|\le\|Ax\|=\|-Bx\|\le|\lambda|_\max(B)\|x\| $$ and hence $x$ must be zero.