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The "Operator Norm" for Linear Transfomations ... Browder, page 179, Section 8.1, Ch. 8 ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,916
The "Operator Norm" for Linear Transfomations ... Browder, page 179, Section 8.1, Ch. 8 ... ...

I am reader 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 some remarks by Browder concerning the "operator norm" for linear transformations ...

The relevant notes form Browder read as follows:


View attachment 7451



My questions regarding the above text by Browder are as follows:



Question 1

In the above notes from Browder we read the following:

" ... ... A perhaps more natural way to define the distance between linear transformations is by using the so-called "operator norm" defined by

\(\displaystyle \lvert \lvert T \rvert \rvert = \text{ sup} \{ \lvert Tv \rvert \ : \ v \in \mathbb{R}^n , \ \lvert v \rvert \le 1 \}
\)


It is not hard to verify that this definition is equivalent to


\(\displaystyle \lvert \lvert T \rvert \rvert = \text{ inf} \{ C \ge 0 \ : \ \lvert Tv \rvert \le C \lvert v \rvert \text{ for all } v \in \mathbb{R}^n \} \)

... ... "



Can someone please demonstrate rigorously exactly why/how

\(\displaystyle \lvert \lvert T \rvert \rvert = \text{ sup} \{ \lvert Tv \rvert \ : \ v \in \mathbb{R}^n , \ \lvert v \rvert \le 1 \}
\)

is equivalent to

\(\displaystyle \lvert \lvert T \rvert \rvert = \text{ inf} \{ C \ge 0 \ : \ \lvert Tv \rvert \le C \lvert v \rvert \text{ for all } v \in \mathbb{R}^n \}
\)

... ... ... ?




Question 2


In the above notes from Browder we read the following:

" ... ... The finiteness of \(\displaystyle \lvert \lvert T \rvert \rvert\) is easy to see ... "


Can someone please rigorously demonstrate why/how \(\displaystyle \lvert \lvert T \rvert \rvert\) is necessarily finite ... ?






Help will be much appreciated ... ...

Peter
 
Last edited:

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,896
Re: The "Operator Norm" for Linear Transfomations ... Browder, page 179, Section 8.1, Ch. 8 ... ...

Hi Peter ,


Question 1

In the above notes from Browder we read the following:

" ... ... A perhaps more natural way to define the distance between linear transformations is by using the so-called "operator norm" defined by

\(\displaystyle \lvert \lvert T \rvert \rvert = \text{ sup} \{ \lvert Tv \rvert \ : \ v \in \mathbb{R}^n , \ \lvert v \rvert \le 1 \}
\)


It is not hard to verify that this definition is equivalent to


\(\displaystyle \lvert \lvert T \rvert \rvert = \text{ inf} \{ C \ge 0 \ : \ \lvert Tv \rvert \le C \lvert v \rvert \text{ for all } v \in \mathbb{R}^n \} \)

... ... "



Can someone please demonstrate rigorously exactly why/how

\(\displaystyle \lvert \lvert T \rvert \rvert = \text{ sup} \{ \lvert Tv \rvert \ : \ v \in \mathbb{R}^n , \ \lvert v \rvert \le 1 \}
\)

is equivalent to

\(\displaystyle \lvert \lvert T \rvert \rvert = \text{ inf} \{ C \ge 0 \ : \ \lvert Tv \rvert \le C \lvert v \rvert \text{ for all } v \in \mathbb{R}^n \}
\)

... ... ... ?

If $v\in \Bbb R^n - \{\bf 0\}$, then $\dfrac{v}{\lvert v\rvert}$ has norm $1$, so $\left\lvert T\left(\dfrac{v}{\lvert v\rvert}\right)\right\rvert \le \|T\|$. Since $T$ is linear, $T\left(\dfrac{v}{\lvert v\rvert}\right) = \dfrac{1}{\lvert v\rvert}T(v)$. Thus $\lvert T(v)\rvert \le \|T\|\lvert v\rvert$. Since $T$ is linear, $T(0) = 0$, so the inequality $\lvert T(v)\rvert \le \|T\|\lvert v\rvert$ holds for $v = 0$. Since $\lvert Tv\rvert \le \|T\|\lvert v\rvert$ for all $v\in \Bbb R^n$, then $\|T\|$ is an element of the set $\{C\ge 0 : \text{$\lvert Tv\rvert \le C\lvert v\rvert$ for all $v\in \Bbb R^n$}\}$. Therefore,
$$\|T\| \ge \inf\{C\ge 0: \text{$\lvert Tv\rvert \le C\lvert v\rvert$ for all $v\in \Bbb R^n$}\}\tag{1}\label{eq1}$$
On the other hand, if $C > 0$ such that $\lvert Tv\rvert \le C\lvert v\rvert$ for all $v\in \Bbb R^n$. In particular, for all $v\in \Bbb R^n$ with $\lvert v\rvert \le 1$, $\lvert Tv\rvert \le C$. Hence, $\sup_{\lvert v\rvert \le 1} \lvert Tv\rvert \le C$, i.e., $\|T\| \le C$. This shows that $\|T\|$ is a lower bound for the set $\{C\ge 0: \text{$\lvert Tv\rvert \le C\lvert v\rvert$ for all $v\in \Bbb R^n$}\}$. Hence, $$\|T\| \le \inf\{C\ge 0 : \text{$\lvert Tv\rvert \le C\lvert v\rvert$ for all $v\in \Bbb R^n$}\}\tag{2}\label{eq2}$$ By inequalities \eqref{eq1} and \eqref{eq2}, the result follows.



Question 2


In the above notes from Browder we read the following:

" ... ... The finiteness of \(\displaystyle \lvert \lvert T \rvert \rvert\) is easy to see ... "


Can someone please rigorously demonstrate why/how \(\displaystyle \lvert \lvert T \rvert \rvert\) is necessarily finite ... ?
Let $\mathbf{e}_1,\ldots, \mathbf{e}_n$ be the standard basis of $\Bbb R^n$. For all $\mathbf{v} = (v_1,\ldots, v_n)\in \Bbb R^n$, $$T(\mathbf{v}) = T\left(\sum_{i = 1}^n v_i \mathbf{e}_i\right) = \sum_{i = 1}^n v_i T(\mathbf{e}_i) = \mathbf{v}\cdot (T(\mathbf{e}_1),\ldots, T(\mathbf{e}_n))$$ The Cauchy-Schwarz inequality gives $\lvert Tv\rvert \le C\lvert \mathbf{v}\rvert$ where $C$ is the norm of the vector $(T(\mathbf{e}_1),\ldots, T(\mathbf{e}_n))$. Note the $C$ a constant independent of $\mathbf{v}$. Since $\mathbf{v}$ was arbitrary, $\|T\| \le C$. Therefore, $\|T\|$ is finite.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,916
Re: The "Operator Norm" for Linear Transfomations ... Browder, page 179, Section 8.1, Ch. 8 ... ...

Hi Peter ,





If $v\in \Bbb R^n - \{\bf 0\}$, then $\dfrac{v}{\lvert v\rvert}$ has norm $1$, so $\left\lvert T\left(\dfrac{v}{\lvert v\rvert}\right)\right\rvert \le \|T\|$. Since $T$ is linear, $T\left(\dfrac{v}{\lvert v\rvert}\right) = \dfrac{1}{\lvert v\rvert}T(v)$. Thus $\lvert T(v)\rvert \le \|T\|\lvert v\rvert$. Since $T$ is linear, $T(0) = 0$, so the inequality $\lvert T(v)\rvert \le \|T\|\lvert v\rvert$ holds for $v = 0$. Since $\lvert Tv\rvert \le \|T\|\lvert v\rvert$ for all $v\in \Bbb R^n$, then $\|T\|$ is an element of the set $\{C\ge 0 : \text{$\lvert Tv\rvert \le C\lvert v\rvert$ for all $v\in \Bbb R^n$}\}$. Therefore,
$$\|T\| \ge \inf\{C\ge 0: \text{$\lvert Tv\rvert \le C\lvert v\rvert$ for all $v\in \Bbb R^n$}\}\tag{1}\label{eq1}$$
On the other hand, if $C > 0$ such that $\lvert Tv\rvert \le C\lvert v\rvert$ for all $v\in \Bbb R^n$. In particular, for all $v\in \Bbb R^n$ with $\lvert v\rvert \le 1$, $\lvert Tv\rvert \le C$. Hence, $\sup_{\lvert v\rvert \le 1} \lvert Tv\rvert \le C$, i.e., $\|T\| \le C$. This shows that $\|T\|$ is a lower bound for the set $\{C\ge 0: \text{$\lvert Tv\rvert \le C\lvert v\rvert$ for all $v\in \Bbb R^n$}\}$. Hence, $$\|T\| \le \inf\{C\ge 0 : \text{$\lvert Tv\rvert \le C\lvert v\rvert$ for all $v\in \Bbb R^n$}\}\tag{2}\label{eq2}$$ By inequalities \eqref{eq1} and \eqref{eq2}, the result follows.





Let $\mathbf{e}_1,\ldots, \mathbf{e}_n$ be the standard basis of $\Bbb R^n$. For all $\mathbf{v} = (v_1,\ldots, v_n)\in \Bbb R^n$, $$T(\mathbf{v}) = T\left(\sum_{i = 1}^n v_i \mathbf{e}_i\right) = \sum_{i = 1}^n v_i T(\mathbf{e}_i) = \mathbf{v}\cdot (T(\mathbf{e}_1),\ldots, T(\mathbf{e}_n))$$ The Cauchy-Schwarz inequality gives $\lvert Tv\rvert \le C\lvert \mathbf{v}\rvert$ where $C$ is the norm of the vector $(T(\mathbf{e}_1),\ldots, T(\mathbf{e}_n))$. Note the $C$ a constant independent of $\mathbf{v}$. Since $\mathbf{v}$ was arbitrary, $\|T\| \le C$. Therefore, $\|T\|$ is finite.



Thanks so much Euge ...

Really appreciate your help on this matter ...

Just working through what you have written ...

Thanks again ...

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,916
Re: The "Operator Norm" for Linear Transfomations ... Browder, page 179, Section 8.1, Ch. 8 ... ...

Hi Peter ,





If $v\in \Bbb R^n - \{\bf 0\}$, then $\dfrac{v}{\lvert v\rvert}$ has norm $1$, so $\left\lvert T\left(\dfrac{v}{\lvert v\rvert}\right)\right\rvert \le \|T\|$. Since $T$ is linear, $T\left(\dfrac{v}{\lvert v\rvert}\right) = \dfrac{1}{\lvert v\rvert}T(v)$. Thus $\lvert T(v)\rvert \le \|T\|\lvert v\rvert$. Since $T$ is linear, $T(0) = 0$, so the inequality $\lvert T(v)\rvert \le \|T\|\lvert v\rvert$ holds for $v = 0$. Since $\lvert Tv\rvert \le \|T\|\lvert v\rvert$ for all $v\in \Bbb R^n$, then $\|T\|$ is an element of the set $\{C\ge 0 : \text{$\lvert Tv\rvert \le C\lvert v\rvert$ for all $v\in \Bbb R^n$}\}$. Therefore,
$$\|T\| \ge \inf\{C\ge 0: \text{$\lvert Tv\rvert \le C\lvert v\rvert$ for all $v\in \Bbb R^n$}\}\tag{1}\label{eq1}$$
On the other hand, if $C > 0$ such that $\lvert Tv\rvert \le C\lvert v\rvert$ for all $v\in \Bbb R^n$. In particular, for all $v\in \Bbb R^n$ with $\lvert v\rvert \le 1$, $\lvert Tv\rvert \le C$. Hence, $\sup_{\lvert v\rvert \le 1} \lvert Tv\rvert \le C$, i.e., $\|T\| \le C$. This shows that $\|T\|$ is a lower bound for the set $\{C\ge 0: \text{$\lvert Tv\rvert \le C\lvert v\rvert$ for all $v\in \Bbb R^n$}\}$. Hence, $$\|T\| \le \inf\{C\ge 0 : \text{$\lvert Tv\rvert \le C\lvert v\rvert$ for all $v\in \Bbb R^n$}\}\tag{2}\label{eq2}$$ By inequalities \eqref{eq1} and \eqref{eq2}, the result follows.





Let $\mathbf{e}_1,\ldots, \mathbf{e}_n$ be the standard basis of $\Bbb R^n$. For all $\mathbf{v} = (v_1,\ldots, v_n)\in \Bbb R^n$, $$T(\mathbf{v}) = T\left(\sum_{i = 1}^n v_i \mathbf{e}_i\right) = \sum_{i = 1}^n v_i T(\mathbf{e}_i) = \mathbf{v}\cdot (T(\mathbf{e}_1),\ldots, T(\mathbf{e}_n))$$ The Cauchy-Schwarz inequality gives $\lvert Tv\rvert \le C\lvert \mathbf{v}\rvert$ where $C$ is the norm of the vector $(T(\mathbf{e}_1),\ldots, T(\mathbf{e}_n))$. Note the $C$ a constant independent of $\mathbf{v}$. Since $\mathbf{v}$ was arbitrary, $\|T\| \le C$. Therefore, $\|T\|$ is finite.




I am revising question 1 above ... where Euge is proving that


\(\displaystyle \lvert \lvert T \rvert \rvert = \text{ sup} \{ \lvert Tv \rvert \ : \ v \in \mathbb{R}^n , \ \lvert v \rvert \le 1 \}
\)


is equivalent to


\(\displaystyle \lvert \lvert T \rvert \rvert = \text{ inf} \{ C \ge 0 \ : \ \lvert Tv \rvert \le C \lvert v \rvert \text{ for all } v \in \mathbb{R}^n \}
\)





In the above proof by Euge ... Euge writes the following:


" ... ... In particular, for all $v\in \Bbb R^n$ with $\lvert v\rvert \le 1$, $\lvert Tv\rvert \le C$. Hence, $\sup_{\lvert v\rvert \le 1} \lvert Tv\rvert \le C$, i.e., $\|T\| \le C$. ... ... "




Can someone please formally and rigorously demonstrate that ...


if ... for all $v\in \Bbb R^n$ with $\lvert v\rvert \le 1$, $\lvert Tv\rvert \le C$ ... ...


... it then follows that ... $\sup_{\lvert v\rvert \le 1} \lvert Tv\rvert \le C$ ... ...


(It seems highly plausible ... but how do you rigorously prove it ... ? )



... help will be much appreciated ....

Peter
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,696
Re: The "Operator Norm" for Linear Transfomations ... Browder, page 179, Section 8.1, Ch. 8 ... ...

Can someone please formally and rigorously demonstrate that ...


if ... for all $v\in \Bbb R^n$ with $\lvert v\rvert \le 1$, $\lvert Tv\rvert \le C$ ... ...


... it then follows that ... $\sup_{\lvert v\rvert \le 1} \lvert Tv\rvert \le C$ .
The statement "for all $v\in \Bbb R^n$ with $\lvert v\rvert \le 1$, $\lvert Tv\rvert \le C$" says that $C$ is an upper bound for the set $\{|Tv| : v\in \Bbb R^n,|v|\leqslant1\}$. The number $\sup_{\lvert v\rvert \le 1} \lvert Tv\rvert$ is the supremum, or least upper bound, of that same set. By definition, the least upper bound is less than or equal to any other upper bound. Therefore $\sup_{\lvert v\rvert \le 1} \lvert Tv\rvert \leqslant C$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,916
Re: The "Operator Norm" for Linear Transfomations ... Browder, page 179, Section 8.1, Ch. 8 ... ...

The statement "for all $v\in \Bbb R^n$ with $\lvert v\rvert \le 1$, $\lvert Tv\rvert \le C$" says that $C$ is an upper bound for the set $\{|Tv| : v\in \Bbb R^n,|v|\leqslant1\}$. The number $\sup_{\lvert v\rvert \le 1} \lvert Tv\rvert$ is the supremum, or least upper bound, of that same set. By definition, the least upper bound is less than or equal to any other upper bound. Therefore $\sup_{\lvert v\rvert \le 1} \lvert Tv\rvert \leqslant C$.



Thanks Opalg ...

That certainly clears up the issue ...

Peter