# Not bijective isometry

I want to find a linear isometry $T:V\to V$ such that $T$ is not Bijective.

I think that, I need to considere a infinite dimensional space $V$, but I am not sure about these concepts.

Thanks for your help.

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## 2 Answers

Try $T:\ell^2 \to \ell^2$ where $T(x_1,x_2,\dots)=(0,x_1,x_2,\dots)$. It is injective (like all linear isometries), but not surjective.

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thank you very much! I understand, but what do you mean by $\mathcal{l}^2?$ –  Hiperion Nov 5 '12 at 2:04
@Hiperion: If you're working in $\mathbb{R}$, it is the space of all sequences $(x_n)$ for which $\sum_n x_n^2$ converges. –  wj32 Nov 5 '12 at 2:05
@Hiperion Since we have a norm, we have a metric. $\ell^2$ is the completion of the space I described in my solution with respect to that metric. –  Brett Frankel Nov 5 '12 at 2:10
Thank you both! :) –  Hiperion Nov 5 '12 at 2:12

Let $V$ be the real vector space whose basis is indexed by $\mathbb{N}$, and let $T:V\to V$ $v_i\mapsto v_{i+1}$. Then using the norm $||\sum_{i=1}^\infty a_iv_i||=\sum_{i=1}^\infty a_i^2$, $T$ is a linear isometry.

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