let $T:R^{m}\rightarrow R^{n}$ a linear application such that $T\ne 0$. Proof that $T(R^m)$ is not bounded.

Question

I need to solve "Consider $T:\Re^{m}\rightarrow \Re^{n}$ a linear application such that $T\ne 0$. Proof that $T(R^m)$ is not bounded". I found an exercice where are replaced '$T(\Re^m)$ not bounded' by '$T$ not bounded application' with the follow solution: Given $x\in R^{m}$, if $|T(x)|=c\in\Re_{+}$ then $|T(nx)|=nc>0$. Therefore we conclude that $T$ is not bounded, because $\Re$ is archimedean. The exercice is equivalent, right?

Answer 1

You don't need the Archimedean property. If $T(x)\ne0$, you can assume $c>0$ (otherwise take $-x$). If $b>0$, you have that $$ \left\|T\left(\frac{2b}{c}x\right)\right\|=\frac{2b}{c}\|T(x)\|=2b>b $$ Thus the image of $T$ is not upper bounded. It's not lower bounded as well, with a similar proof.

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This is different from saying that the linear map is bounded, which it is, because $$ \sup_{x\ne0}\frac{\|T(x)\|}{\|x\|} $$ is finite.