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The theorem reads:

Let $T:\mathbb R^n\longrightarrow \mathbb R^n$ be an isometry. Then $T$ is an isomorphism.

The proof:

Let $\mathbf u_1,\mathbf u_2,\ldots,\mathbf u_n$ be an orthonormal basis for $\mathbb R^n$. Then $T\mathbf u_1,T\mathbf u_2,\ldots,T\mathbf u_n$ is an orthonormal basis for $\mathbb R^n$. Then image $T=\mathbb R^n$.

I see that it was proved that $T$ must be onto, but I don't see why $T$ is 1-1. One would have to prove that for each $\mathbf x,\mathbf y \in \mathbb R^n$, $T\mathbf x = T \mathbf y\Longrightarrow \mathbf x = \mathbf y$. I appreciate your help in pointing out a hint to prove the last part.

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In general, a linear map of finite dimensional vector spaces is onto if and only if it is 1-1.

This follows from the fact that if $T:V\to W$ is linear and $V,W$ finite dimensional, then $$\dim V=\dim\ker(T)+\dim \mathrm{im}(T).$$

Hence, once you know it is onto you are done.

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    $\begingroup$ @VladimirVargas You are welcome ! :) $\endgroup$ – Spenser Mar 13 '15 at 20:59
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An isometry $T$ satisfies $\|Tx - Ty\| = \|x - y\|$. Thus $$ Tx = Ty \implies \|Tx - Ty\| = 0 \implies \|x - y \| = 0 \implies x=y$$ so that $T$ is injective.

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Let ${\bf x} \in \ker T$. So $T{\bf x} = 0$. Hence: $$\|T{\bf x}\|=\|{\bf x}\| = 0 \implies {\bf x} = 0,$$ so $\ker T = \{0\}$ and $T$ is injective.

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