If $T:R^n \to R^m$ is linear, then there is $C>0$ such that $||T(x)|| \leq C||x||$

Question

I want to understand if my proof is correct.I'm trying to prove this without the hypothesis that $T$ is injective.

$ ||T(x)|| = ||T(\sum_i x_i e_i)||=||\sum_i x_iT(e_i)|| \leq \sum_i |x_i|||T(e_i)||$.

Let $C = \max_i \{||T(e_i)||\} \geq 0$, $||x|| = \sum_i |x_i|$

Then $||T(x)|| \leq C||x||$.

Suppose $T \equiv 0$, then $C=1$ works, because $||T(x)|| = 0 \leq 1.||x||=||x||$.

Suppose $T \neq 0$, then, if $C=0$, $||T(x)|| \leq 0||x|| = 0 \Rightarrow T(x) = 0 \forall x \Rightarrow T \equiv 0. (\unicode{x21af}) $

If it is indeed correct, why in this post: If $T$ is injective then there exists $\alpha>0$ such that $||Tx||\geq \alpha||x||$ injectivity was needed?

Answer 1

Your proof is fine if you are using the $l^1$ norm on $\mathbb R^n$ and any norm you like on $\mathbb R^m.$ But it would be good to specify this at the beginning.

As for injectivity of a linear map $T:\mathbb R^n \to \mathbb R^m$implying $\|Tx\| \ge \alpha \|x\|:$ That inequality goes the other way, as @Arthur noted, hence doesn't conflict with what you're doing. The proof of that result is easy by the way. Since $T(x) \ne 0$ for $x\ne 0,$ the function $x\to \|T(x)\|,$ which is continuous, has a positive minimum $\alpha$ on the unit sphere. The conclusion follows easily from this.