# Are orthogonal operators always isomorphisms?

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

• "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? – diracdeltafunk Jun 5 at 5:12
• What is $T$? I assume it's an orthogonal linear transformation $V\to V$? – YiFan Jun 5 at 5:15

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.

• How do you infer that surjectivity is trivial – Ivan Bravo Jun 5 at 5:20
• @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. – mathvision Jun 5 at 5:24
• 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 – marwalix Jun 5 at 17:45

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.

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