Norm of a Linear Transformation: Proving Homogeneity From Definition - Peter

[Math Amateur]

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

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

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 7904
https://www.physicsforums.com/attachments/7905

Near the end of the above proof we read the following:

" ... ... To see that \(\displaystyle \| T \|\) defines a norm, note that homogeneity follows directly from the definition ... ... "
My question is as follows:

How, exactly, do we demonstrate rigorously that homogeneity follows directly from the definition ... that is how do we show that

\(\displaystyle \| t T \| = \ \mid t \mid \| T \|\) ... ... for \(\displaystyle t \in \mathbb{R}\) ... ... ?

Help will be appreciated ...

Peter

 

[Opalg]

How, exactly, do we demonstrate rigorously that homogeneity follows directly from the definition ... that is how do we show that

\(\displaystyle \| t T \| = \ \mid t \mid \| T \|\) ... ... for \(\displaystyle t \in \mathbb{R}\) ... ... ?

The defining properties for the norm in $\mathcal{L}(\Bbb{R}^n,\Bbb{R}^m)$ follow from those for the norm in $\Bbb{R}^m$.

For the homogeneity property, the definition of $tT$ is given by $(tT)\mathbf{x}) = t(T\mathbf{x})$. So $$\|(tT)\mathbf{x})\| = \|t(T\mathbf{x})\| = |t|\|T\mathbf{x}\|.$$ If you now take the sup over all $\mathbf{x}$ with $\|\mathbf{x}\|=1$, it follows that $\|tT\| = |t|\|T\|.$

 

[Math Amateur]

The defining properties for the norm in $\mathcal{L}(\Bbb{R}^n,\Bbb{R}^m)$ follow from those for the norm in $\Bbb{R}^m$.

For the homogeneity property, the definition of $tT$ is given by $(tT)\mathbf{x}) = t(T\mathbf{x})$. So $$\|(tT)\mathbf{x})\| = \|t(T\mathbf{x})\| = |t|\|T\mathbf{x}\|.$$ If you now take the sup over all $\mathbf{x}$ with $\|\mathbf{x}\|=1$, it follows that $\|tT\| = |t|\|T\|.$

Thanks Opalg ...

That cleared up that issue ...

Thanks again ...

Peter