Question on linear algebra mappings If $T:R^m\to R^n$ is a linear transformation, show that there is a number $M$ such that $|T(h)|\leq M|h|$ for $h\in R^m$.
 A: Let $A$ be the matrix of $T$ in the standard basis. $T(h)=Ah$ write the rows of A which are $a_{i1},..,a_{in}$ as vectors $a_i$ that is $ A=  \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix}$
 then $|Ah| =\sum_{i=1}^n\sqrt{\langle a_{i},h \rangle^2} \le \sqrt{\sum_{i=1}^n|a_i|^2|h|^2}=|h|\sqrt{|a_1|^2+\dots |a_n|^2} $. So you get $M=\sqrt{|a_1|^2+\dots |a_n|^2}$
A: Every linear transformation in finite dimensional space is continuous which means the desired inequality. To find by hand this inequality choose a basis $(e_1,\ldots,e_m)$ of $\mathbb R^m$ and use the uniform norm (since all the norms are equivalent)...
A: Note that if $v = \lambda w$, $\lambda \in \mathbb{R}$, with $|w| = 1$ then if $|T(w)| \leq M |w|$ you have also $$|T(\lambda w)| = |\lambda||T(w)| \leq M |\lambda w| = M|v|$$ So we can consider $S^{m-1} \subset \mathbb{R}^{m}$.
On $S^{m-1}$ set $$g (w) = \frac{T(w)}{w}$$
This is continous because $T$ is continous by linearity; $S^{m-1}$ is compact and so there exists $M$ such that $$|\frac{T(w)}{w}| \leq M   \ \ \ \forall w \ \in S^{m-1}$$
Then $$|T(v)| \leq M |v| \ \ \ \forall v \ \in \mathbb{R}^{m}$$
