Expression for a lower bound of $||(A-z)^{-1}||^{-1}$ that depends only on the spectrum of $A$ and/or its singular values. If we have a square complex matrix $A$ and the spectral norm$||\cdot||$, we known that $||A||= max  \left(\sigma(A)\right)$ the greatest singular value. Lets denote $\Lambda(A)$ the spectrum of $A$.
For Hermitian matrices (actually for normal matrices) $||A|| = \rho(A)$, where $\rho(A)$ is the spectral radius of $A$. Thus,for $A$ normal, if $z\in \mathbb{C}$, then $||(A-z)|| = max_{\lambda \in \Lambda(A)}||\lambda-z||$. In fact, if we considere inverses, the norm has a beautiful geometric meaning, I mean, if $z\in \mathbb{C}-\Lambda(A)$, then $||(A-z)^{-1}||^{-1} = min_{\lambda \in \Lambda(A)} ||\lambda-z|| = \mbox{distance}(\Lambda(A),z)$. Moreover, for example, if we know that $||(A-z)^{-1}|| > 1/r$, then we can conclude that $\mbox{distance}(\Lambda(A),z) < r$ (This is useful for a generalized version of Gershgorin circle theorem considering block matrices).
Since in the general case (incluiding non normal matrices) $||A||$ could be much greater than $\rho(A)$, then not neccesarily $||(A-z)^{-1}||^{-1} = \mbox{distance}(\Lambda(A),z)$.
So, I have been wondering if it is possible to express $ ||A-z||$ just in terms of eigenvalues of $A$ and/or singular values of $A$ (like in the hermitian or normal case), I mean, ¿does it exist a simple expression like a function $f$ such that for all $A$, $||A-z|| = f(z,\sigma_1(A),...,\sigma_r(A),\lambda_1(A),...,\lambda_t(A))$ ? , or is it possible to find a function $g(z,\sigma_1(A),...,\sigma_r(A),\lambda_1(A),...,\lambda_t(A))$ depending just of the spectrum and/or singular values of A, such that for all $A$,  $g(z,\sigma_1(A),...,\sigma_r(A),\lambda_1(A),...,\lambda_t(A)) < ||(A-z)^{-1}||^{-1}$?.
 A: You are asking for a lower bound of $\|(A-z)^{-1}\|^{-1}=\sigma_\min(A-z)$ in terms of $z$ and the singular values of $A$. Since
$$
\|(A-z)u\|\ge
\begin{cases}
\|Au\|-\|zu\|\ge\sigma_\min(A)-|z|\\
\|zu\|-\|Au\|\ge|z|-\sigma_\max(A)
\end{cases}
$$
for every unit vector $u$, we have
$$
\sigma_\min(A-z)\ge
\begin{cases}
\sigma_\min(A)-|z|&\text{when}\quad\sigma_\min(A)\ge|z|,\\
|z|-\sigma_\max(A)&\text{when}\quad|z|\ge\sigma_\max(A).\\
\end{cases}
$$
When $\sigma_\min(A)<|z|<\sigma_\max(A)$, the best lower bound of $\sigma_\min(A-z)$ in terms of $z$ and the singular values of $A$ is zero. In fact, given any $\sigma_1\ge\sigma_2\ge\cdots\ge\sigma_n\ge0$ and any complex number $z$ such that $\sigma_n<|z|<\sigma_1$, there always exists an $n\times n$ matrix $A$ with singular values $\sigma_1,\sigma_2,\ldots,\sigma_n$ such that $A-z$ is singular. More specifically, let $z=r\omega$ where $r\ge0$ and $|\omega|=1$. Then $\sigma_n<r<\sigma_1$ and $(r-\sigma_1)(r-\sigma_n)<0$. Therefore $c:=\dfrac{r^2+\sigma_1\sigma_n}{(\sigma_1+\sigma_n)r}\in(0,1)$ and $c=\cos\theta$ for some real number $\theta$. Now let $s=\sin\theta$ and
$$
A=\omega\pmatrix{\sigma_1c&-\sigma_ns\\ \sigma_1s&\sigma_nc\\ &&\sigma_3\\ &&&\ddots\\ &&&&\sigma_{n-1}}.
$$
Clearly, the singular values of $A$ are $\sigma_1,\sigma_2,\ldots,\sigma_n$. Now $A-z$ is singular because
$$
\det\pmatrix{\sigma_1c-r&-\sigma_ns\\ \sigma_1s&\sigma_nc-r}
=\sigma_1\sigma_n-(\sigma_1+\sigma_n)rc+r^2=0.
$$
Remark. If lower bounds in terms of other quantities are allowed, there are several easily computable bounds of Gershgorin type. E.g. (c.f. C. R. Johnson, A Gershgorin-type lower bound for the smallest singular value, Linear Algebra and Its Applications, 112:1-7 (1989))
$$
\sigma_\min(B)\ge\min_k\left\{|b_{kk}|-\frac12\left(r_k'(B)+c_k'(B)\right)\right\}
$$
or (cf. Horn and Johnson, Topics in Matrix Analysis, p.231)
$$
\sigma_\min(B)\ge\frac12\min_k\left\{\left[4|b_{kk}|^2+\left(r_k'(B)-c_k'(B)\right)^2\right]^{1/2}-\left(r_k'(B)+c_k'(B)\right)\right\},
$$
where
$$
r_k'(B)=\sum_{j\ne k}|b_{kj}|
\quad\text{and}\quad
c_k'(B)=\sum_{i\ne k}|b_{ik}|.
$$
