Norm of a Linear Transformation .... Junghenn Propn 9.2.3 ....

[Math Amateur]

I am reading Hugo D. Junghenn's book: "A Course in Real Analysis" ...

I am currently focused on Chapter 9: "Differentiation on $\mathbb{R}^n$"

I need some help with the proof of Proposition 9.2.3 ...

Proposition 9.2.3 and the preceding relevant Definition 9.2.2 read as follows:

In the above proof we read the following:

" ... ... If $\mathbf{x} \neq \mathbf{0} \text{ then } \| \mathbf{x} \|^{-1} \mathbf{x}$ has a norm $1$, hence

$\| \mathbf{x} \|^{-1} \| T \mathbf{x} \| = \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le 1$ ... ... "
Now I know that $T( c \mathbf{x} ) = c T( \mathbf{x} )$

... BUT ...

... how do we know that this works "under the norm sign" ...

... that is, how do we know ...$\| \mathbf{x} \|^{-1} \| T \mathbf{x} \| = \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \|$... and further ... how do we know that ...$\| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le 1$

Help will be appreciated ...

Peter

 

[StoneTemplePython]

Now I know that $T( c \mathbf{x} ) = c T( \mathbf{x} )$

... BUT ...

... how do we know that this works "under the norm sign" ...

... that is, how do we know ...$\| \mathbf{x} \|^{-1} \| T \mathbf{x} \| = \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \|$

The idea is homogeneity of the the exponent.

Look at the 2 norm of some vector $\mathbf v$ (it can be any vector, including perhaps $\mathbf v := T \mathbf x$

$\big \Vert \mathbf v \big \Vert_2 = \big(\sum_{i=1}^n v_i^2\big)^\frac{1}{2}$

thus, in your case with some $c \gt 0$

$c \big \Vert \mathbf v \big \Vert_2 = c \big(\sum_i v_i^2\big)^\frac{1}{2} = \big(\sum_i c^2 v_i^2\big)^\frac{1}{2} = \big(\sum_i (c v_i)^2\big)^\frac{1}{2} = \big \Vert c \mathbf v \big \Vert_2$

Keep an eye for this sort of homogeneity -- it comes up all the time in inequalities.

... and further ... how do we know that ...

$\| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le 1$

I either didn't read the question close enough, or something is missing. My inference is they rescaled such that $T$ has an operator norm of 1 here -- but I didn't catch where that was done. If said rescaling wasn't done, the statement in general is not true. For example, consider a diagonal matrix $T$ where $T_{1,1} = 5$ and all other diagonal entries are one -- then apply the argument where $\mathbf x = \mathbf e_1$ i.e. the first standard basis vector -- the result is 5, not 1. But again I think there was a rescaling / Without Loss of Generality assumption that I missed somewhere.

 

[Math Amateur]

The idea is homogeneity of the the exponent.

Look at the 2 norm of some vector $\mathbf v$ (it can be any vector, including perhaps $\mathbf v := T \mathbf x$

$\big \Vert \mathbf v \big \Vert_2 = \big(\sum_{i=1}^n v_i^2\big)^\frac{1}{2}$

thus, in your case with some $c \gt 0$

$c \big \Vert \mathbf v \big \Vert_2 = c \big(\sum_i v_i^2\big)^\frac{1}{2} = \big(\sum_i c^2 v_i^2\big)^\frac{1}{2} = \big(\sum_i (c v_i)^2\big)^\frac{1}{2} = \big \Vert c \mathbf v \big \Vert_2$

Keep an eye for this sort of homogeneity -- it comes up all the time in inequalities.

I either didn't read the question close enough, or something is missing. My inference is they rescaled such that $T$ has an operator norm of 1 here -- but I didn't catch where that was done. If said rescaling wasn't done, the statement in general is not true. For example, consider a diagonal matrix $T$ where $T_{1,1} = 5$ and all other diagonal entries are one -- then apply the argument where $\mathbf x = \mathbf e_1$ i.e. the first standard basis vector -- the result is 5, not 1. But again I think there was a rescaling / Without Loss of Generality assumption that I missed somewhere.

Thanks StoneTemplePython ...

Appreciate your help ...

Peter

 

[WWGD]

I am reading Hugo D. Junghenn's book: "A Course in Real Analysis" ...

I am currently focused on Chapter 9: "Differentiation on $\mathbb{R}^n$"

I need some help with the proof of Proposition 9.2.3 ...

Proposition 9.2.3 and the preceding relevant Definition 9.2.2 read as follows:
View attachment 221315
View attachment 221316
In the above proof we read the following:

" ... ... If $\mathbf{x} \neq \mathbf{0} \text{ then } \| \mathbf{x} \|^{-1} \mathbf{x}$ has a norm $1$, hence

$\| \mathbf{x} \|^{-1} \| T \mathbf{x} \| = \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \| \le 1$ ... ... "
Now I know that $T( c \mathbf{x} ) = c T( \mathbf{x} )$

... BUT ...

... how do we know that this works "under the norm sign" ...

... that is, how do we know ...$\| \mathbf{x} \|^{-1} \| T \mathbf{x} \| = \| T ( \| \mathbf{x} \|^{-1} \mathbf{x} ) \|$..Peter

Aren't we working with the set of x with ||x||=1 ? Seems like subbing this in would show equality, if I did not miss something obvious.