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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

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    What's the definition of $\operatorname{sep}(\lambda, A)$? – 2017-02-13
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    $\operatorname{sep}(\lambda, A)=\min\{||E||_2\colon \lambda\operatorname{I}-(A+E)\text{ singular}\}=1/|| (\lambda\operatorname{I} - A)^{-1}||_2$. The context is Numerical Linear Algebra. – 2017-02-13

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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.

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    I don't know if a comment is the right place for this: In this very context, what is an example for a matrix with eigenvalue 0 and $dist(\lambda,\sigma(A))/\operatorname{sep}(\lambda,A)\to\infty,\,\lambda\to\infty$? That is, why does equality hold only in the normal case? What is an example for the separation being much smaller than the distance to the spectrum? – 2017-02-13
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    I don't know if that ratio can be made arbitrarily large. Here's a useful example, though: $$ A = \pmatrix{1&n\\0&1}, \quad E = \pmatrix{0&0\\1/n&0} $$ shows that $\operatorname{sep}(0,A)$ can be made arbitrarily small (or that $\operatorname{sep}(\lambda,A)$ can be made small for any fixed $\lambda$). – 2017-02-13
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    If you're wondering more about that though, it seems like you should make a new question – 2017-02-13