Is it true that $T$ is orthogonal if and only if $T$ is isomorphism?

Question

I want to prove the following:

Let $V$ be a finite dimensional vector space with inner product, then $T$ is orthogonal if and only $T$ is an isomorphism

I think the sufficiency could be true because $T$ is an isometry and then preserves its inner product, ($\langle \alpha \, , \,\beta \rangle = \langle T(\alpha) \, , \, T(\beta) \rangle$ so it follows that $T$ is injective, but I am not so sure if $T$ is surjective. Furtheremore, I do not know if the necessity it's true because I think I can find a counterexample, please help me!

Answer 1

Take $T(x)=2x$ it is an ismorphism defined on $\mathbb{R}$ but it is not orthogonal.

Answer 2

The backward direction of this statement is not true. An isomorphism (i.e. full-rank) linear transformation need not be an isometry.

Meanwhile, the forward direction of this statement is true. Let $A$ be the matrix representing $T$. Since $A$ has the property that $A^T A = I_n$ (where $\dim V = n$), and the rank of the product of two matrices is at most the rank of any of the two matrices multiplied, it follows that $\text{rank}(A) = \text{rank}(A^T) = n$, so $A$ is full rank. Hence, $T$ must be an isomorphism.

Answer 3

There are plenty of $T$'s which are isomorphisms but not orthogonal. To be precise, anything with nonzero determinant not equal to $\pm1$.