relation between singular values and eigenvalue How is this inequality proved
$\sigma_{min}(A) \leq \min_{i}|\lambda_i|\leq\max_{i}|\lambda_i| \leq \sigma_{max}(A) $  
where $\sigma$ are the singular values and $\lambda $ are the eigen values of a matrix A
in the book i am reading (Matrix computations, Golub), it says that it can be seen using schur form

The point is that if $T=(t_{ij})\in\mathbb{C}^{n\times n}$ is triangular, then
$$\tag{$❀$}
\sigma_{\min}(T)\leq\min_i|t_{ii}|\leq\max_i|t_{ii}|\leq\sigma_{\max}(T).
$$
In other words,
$$
\sigma_{\min}(T)\leq|t_{ii}|\leq\sigma_{\max}(T), \quad i=1,\ldots,n.
$$
To see this, consider the vector $e_i$ (the $i$th column of the identity matrix). Then
$$
|t_{ii}|\leq\sqrt{\sum_{k=1}^i|t_{ki}|^2}=\|Te_i\|_2\leq\max_{\|x\|_2=1}\|Tx\|_2=\sigma_{\max}(T).
$$
The other direction can be show similarly using the inverse of $T$, the fact that $\sigma_{\min}(T)=1/\sigma_{\max}(T^{-1})$, and that the diagonal of $T^{-1}$ is equal to the inverse of the diagonal of $T$. Note that if the matrix $T$ is not invertible, then the lower bound on $|t_{ii}|$ is trivial as $\sigma_{\min}(T)=0$.
Now if $T=Q^*AQ$ is the Schur form of $A$, $T$ has the same singular values as $A$, and $t_{ii}$ are the eigenvalues of $T$ (and $A$). Then just use ($❀$).
