I'm trying to solve this exercise in a Numerical Linear Algebra book but am having difficultly with a specific part of the proof of this statement.
The exercise states:
Let $A \in \mathbb{R}^{n \times n}$ be non-singular and let $A = QR$ be a $QR$ factorisation. Show that
$$ \kappa_{2}(A) \geq \frac{\displaystyle\max_i |r_{ii}|}{\displaystyle\min_i|r_{ii}|} $$
Where $\kappa_2(A) = ||A||_2 \cdot ||A^{-1}||_2$. This is one of a 2 part question, the first part is to show that $||A||_2 \geq max_{i,j} |a_{ij}|$.
Where I have got to
So I know that
$$||A||_2 = ||R||_2$$
As
$$ ||A||^{2}_2 = \rho(A^{T}A) $$
and
$$ A^{T}A = R^{T}Q^{T}QR = R^{T}R $$
due to $Q$ being orthongonal. Where $\rho$ is the spectral norm.
From the first part of the question, I have $||A||_2 \geq max_{i,j}|a_{ij}|$, true for any $A \in \mathbb{R}^{n \times n}$. So by substituting for $R$ we have $||A||_2 = ||R||_2 \geq max_{i,j}|r_{ij}| \geq max_{i}|r_{ii}|$. Hence
$$ ||A||_2 \geq \max_{i}|r_{ii}| $$
Now all that is left to show is that $||A^{-1}||_2 \geq \frac{1}{min_{i}|r_{ii}|}$.
I have managed to derive the relationship in earlier exercises:
$$ \min_{x \neq 0}\frac{||Ax||}{||x||} = \frac{1}{||A^{-1}||}$$
which shows
$$ \kappa (A) = \frac{\displaystyle\max_{x \neq 0}\frac{||Ax||}{||x||}}{\displaystyle\min_{x \neq 0}\frac{||Ax||}{||x||}} $$
But I cannot find a relationship between $\min_{i}|r_{ii}|$ and something related to $||A^{-1}||_2$ or vise versa.
I feel as though I am overlooking something fairly obvious. I would appreciate any help.
"You can use the same approach as for $∥A∥$ and using the fact that $R−1$ has $1/r_{ii}$, $i=1,…,n,$ on the diagonal." – Algebraic Pavel
Thank you very much.