Proof of $\operatorname{sep}(\lambda, A)\leq \operatorname{dist}(\lambda,\sigma(A))$ for general matrices $A$

Question

I want to prove that $\operatorname{sep}(\lambda, A)\leq \operatorname{dist}(\lambda,\sigma(A))$ for general matrices $A$.

For normal matrices it is $\operatorname{sep}(\lambda, A)=\operatorname{dist}(\lambda,\sigma(A))$, and this is provable via unitary (norm-invariant) diagonalization of $(\lambda\operatorname{I} - A)^{-1}$ as $\operatorname{sep}(\lambda, A)=1/|| (\lambda\operatorname{I} - A)^{-1}||_2$.

How do I prove the general statement of the title? Via unitary triangularization?

Thanks in advance

Answer 1

By unitary triangularization, we can assume without loss of generality that $A$ is upper triangular. That is, $$ A = \pmatrix{\lambda_1 & &*\\&\ddots\\0&&\lambda_n} $$ Fix $\lambda$. For any $\lambda_k$, we may define $E_k$ = $(\lambda - \lambda_k)e_ke_k^T$, where $e_k$ denotes the $k$th standard basis vector (so $E_k$ is diagonal with one non-zero entry). Verify that $\lambda I - (A + E_k)$ is singular. Note that $\|E_k\| = |\lambda - \lambda_k|$ It follows that $$ \operatorname{sep}(\lambda, A) \leq |\lambda - \lambda_k| \qquad k = 1,\dots,n $$ It follows that $$ \operatorname{sep}(\lambda, A) \leq \min_{k = 1,\dots,n} |\lambda - \lambda_k| = d(\lambda,\sigma(A)) $$ as desired.