I need to prove:
Every linear map $L:\mathbb{R}^n \to \mathbb{R}^m$ is bounded.
I've thought about it but I can't really get any solid ideas.
I need to prove:
Every linear map $L:\mathbb{R}^n \to \mathbb{R}^m$ is bounded.
I've thought about it but I can't really get any solid ideas.
Let $A$ be the matrix representation of the linear map. Then $A$ is an $m\times n$ matrix and we can write $A = (a_{ij})$. Also we can write $x\in \mathbb{R^n}$ as $x = (x_i)$ and $Ax = (y_i)$. Using matrix multiplication we get $y_i = \sum_{j=1}^{n} a_{ij}x_j$. So then $\|Ax\|^2 = \sum_{i=1}^{m}[\sum_{j=1}^{n} a_{ij}x_j]^2$. The Cauchy-Schwarz inequality gives us $[\sum_{j=1}^{n} a_{ij}x_j]^2 \leq [(\sum_{j=1}^{n} a_{ij}^2) (\sum_{i=1}^{n} x_i^2).$ $\implies \|Ax\|^2 \leq \sum_{i=1}^{m}\sum_{j=1}^{n}a_{ij}^2\|x\|^2$ and thus we have our $M^2 = \sum_{i=1}^{m}\sum_{j=1}^{n}a_{ij}^2$.
$\left\|\sum\limits_{i=1}^{n}x_iL(e_i)\right\| \le \sum\limits_{i=1}^{n}|x_i|\|L(e_i)\| \le$ $n\max|x_i|\max\|L(e_i)\| \le n\|x\|\max\|L(e_i)\|$
So let $M = n\max\|L(e_i)\|.$
It might be overkill, but actually this is how I think about this fact geometrically.
The singular value decomposition says that, in suitable bases of domain and codomain, $L$ is (rectangular) diagonal. A diagonal matrix is obviously bounded, and changing a basis is a linear isomorphisms so doesn't effect boundedness.