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I need to show the following:

Let $V$ be a finite dimensional vector space with inner product, if $T$ is orthogonal, show that T is injective and surjective

I think it is injective because T preserves inner product but i am not so sure if it is the right wya to prove it

Thanks for your help!

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  • $\begingroup$ "I think it is injective because T preserves inner product" – well yes, but you just said "I think it is injective because T is orthogonal", which is what you want to prove! What would you actually do to try to prove injectivity? $\endgroup$ – diracdeltafunk Jun 5 at 5:12
  • $\begingroup$ What is $T$? I assume it's an orthogonal linear transformation $V\to V$? $\endgroup$ – YiFan Jun 5 at 5:15
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Assume $u\in V$ is such that $Tu=0$. The operator being orthogonal one has

$$(u,u)=(Tu,Tu)=0$$

This means $u=0$ and $T$ is injective and we’re done for bijectivity because $T$ is a linear operator in a finite dimensional vector space.

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  • $\begingroup$ How do you infer that surjectivity is trivial $\endgroup$ – Ivan Bravo Jun 5 at 5:20
  • $\begingroup$ @IvanBravo, If $V$ and $W$ are finite-dimensional vector spaces with the same dimension, then a linear map $T : V → W$ is injective if and only if it is surjective. $\endgroup$ – mathvision Jun 5 at 5:24
  • $\begingroup$ One way to see that is to start with $\operatorname{rank}(T)+\operatorname{dim}(\ker{T})=\operatorname{dim}{V}$ and the dimension of the kernel is $0$ when $T$ is injective $\endgroup$ – marwalix Jun 5 at 17:45
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For $u,v\in V$ so that $T(u)=T(v)$, we have that $T(u-v)=0,$ so that $\langle u-v,u-v\rangle=\langle T(u-v),T(u-v)\rangle=0$. This means that $u-v=0$, thus $u=v$, so that $T$ is injective. On the other hand, the image of an injective linear transformation is a subspace of the codomain $V$ with dimension equal to the dimension of the domain, but since the dimension of the domain is just the dimension of the whole of $V$, $T$ is surjective.

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  • $\begingroup$ That has been so clear, thank! $\endgroup$ – Ivan Bravo Jun 5 at 5:23
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    $\begingroup$ @IvanBravo Glad to help! $\endgroup$ – YiFan Jun 5 at 5:27

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