Prove $\min_{i}|\lambda_i| \leq |r_{jj}| \leq \max_{i}|\lambda_i|$ Let A be a normal $n \times n$ matrix with the eigenvalues $\lambda_1,...,\lambda_n$ |A| = |QR|, $|Q^HQ| = I$, $|R| = [r_{ik}]$ upper triangular matrix. Prove: $$\min_{i}|\lambda_i| \leq |r_{jj}| \leq \max_{i}|\lambda_i|, \; j = 1,...,n$$
In this problem, I know that somehow we have to use Cholesky decomposition, which is $A = LL^H$, but I don't know how can I do it to prove. 
 A: If $A$ is normal, then the absolute values of its eigenvalues are equal to its singular values. In particular, if $\sigma_{\min}(A)$ and $\sigma_{\max}(A)$ are the minimal and maximal singular values of $A$, respectively, then
$$
\sigma_{\min}(A)=\min_{\|x\|=1}\|Ax\|_2=\min_{1\leq i\leq n}|\lambda_i|
\quad\text{and}\quad
\sigma_{\max}(A)=\max_{\|x\|=1}\|Ax\|_2=\max_{1\leq i\leq n}|\lambda_i|.
$$
The easiest way to see this is to consider the spectral decomposition $A=U\Lambda U^*$ and note that $\Lambda=|\Lambda|W$, where $W$ is a unitary diagonal matrix (if $\lambda=|\lambda|e^{\iota\phi}$ is an eigenvalue of $A$, the corresponding factor on the diagonal of $W$ is exactly the phase part of $\lambda$). Hence $A=U|\Lambda|(U\bar{W})^*$. But with $\Sigma:=|\Lambda|$ and $V:=U\bar{W}$, $A=U\Sigma V^*$ is the SVD of $A$.
It remains to show that the diagonal entries of the triangular factor $R$ satisfy
$$\tag{1}
\sigma_{\min}(A)\leq |r_{jj}| \leq \sigma_{\max}(A), \quad 1\leq j\leq n.
$$
Let $e_j$ be the $j$th column of the identity matrix. Then
$$\tag{2}
\sigma_{\max}(A)=\max_{\|x\|=1}\|Ax\|_2=\max_{\|x\|=1}\|Rx\|_2\geq\|Re_j\|_2=\sqrt{\sum_{i=1}^j|r_{ij}|^2}\geq|r_{jj}|,
$$
which proves the upper bound.
For the lower bound, if $A$ is singular, then $\sigma_{\min}(A)=0$, and the bound is trivial. If $A$ is invertible, then similarly as before, we have
$$\tag{3}
\frac{1}{\sigma_{\min}(A)}=\sigma_{\max}(A^{-1})=\max_{\|x\|=1}\|A^{-1}x\|_2=\max_{\|x\|=1}\|R^{-1}x\|_2\geq \frac{1}{|r_{jj}|},
$$
where the last equality is due to the fact that the diagonal of $R^{-1}$ is the inverse of the diagonal of $R$. Putting (2) and (3) together gives (1).
