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

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
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 ...

Lemma 8.4 reads as follows:



Browder - 1 - Lemma 8.4 ... PART 1 .. .png
Browder - 2 - Lemma 8.4 ... PART 2 ... .png



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
Feb 7, 2012
2,678
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 $|T \mathbf{e}_k| \leqslant \|T\|$.