# Norms for a Linear Transformation ... Browder, Lemma 8.4 ...

#### Peter

##### Well-known member
MHB Site Helper
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 8: Differentiable Maps and am specifically focused on Section 8.1 Linear Algebra ...

I need some help in fully understanding Lemma 8.4 ...

In the above proof of Lemma 8.4 by Browder we read the following:

" ... ... On the other hand since $$\displaystyle \sum_{ j = 1 }^m ( a_k^j )^2 = \ \mid T e_k \mid \ \le \| T \|^2$$ for every $$\displaystyle k, 1 \le k \le n$$ ... ... "

My question is as follows:

Can someone please demonstrate rigorously that $$\displaystyle \sum_{ j = 1 }^m ( a_k^j )^2 = \ \mid T e_k \mid \ \le \| T \|^2$$ ...

(... ... it seems plausible that $$\displaystyle \sum_{ j = 1 }^m ( a_k^j )^2 = \ \mid T e_k \mid \ \le \| T \|^2$$ but how do we demonstrate it rigorously ... ... )

Help will be much appreciated ...

Peter

#### Opalg

##### MHB Oldtimer
Staff member
Can someone please demonstrate rigorously that $$\displaystyle \sum_{ j = 1 }^m ( a_k^j )^2 = |T e_k|^2 \le \| T \|^2$$ ...
The equality $$\displaystyle \sum_{ j = 1 }^m ( a_k^j )^2 =|T \mathbf{e}_k|^2$$ comes from the definition of the matrix of $T$. In fact, $T \mathbf{e}_k = (a_k^1 \mathbf{e}_1,a_k^2 \mathbf{e}_2,\ldots,a_k^m \mathbf{e}_m).$

The inequality $|T \mathbf{e}_k| \leqslant \|T\|$ comes from the definition of $\|T\|$. In fact, $\|T\| = \sup\{|T\mathbf{v}|:\mathbf{v}\in\Bbb{R}^n, |\mathbf{v}|\leqslant1\}$, which implies that $|T\mathbf{v}| \leqslant \|T\|$whenever $|\mathbf{v}|\leqslant1$. But $|\mathbf{e}_k| = 1$, so we can take $\mathbf{v} = \mathbf{e}_k$ to get $|T \mathbf{e}_k| \leqslant \|T\|$.