$T$ is a transformation that sets $T^2=-I$, is it necessarily unitary?

Question

I need help in an argument over the following question.

Let $V$ be a unitary space and let $T:V\rightarrow V$ a linear transformation that sets $T^2=-I$, is $T$ necessarily a unitary transformation?

We had this question in an exam, and the official answer is:

Since $(T-iI)(T+iI)=0$, the minimal polynomial of $T$ divides the polynomial $(x-i)(x+i)$, hence $T$ must be diagonalizable, so there is a basis $B=\{v_1,\ldots,v_n\}$ such that $[T]_B=\operatorname{diag}(i,i,i,\ldots,-i,-i,-i)$. So $[T]_B^*=\operatorname{diag}(-i,-i,-i,\ldots,i,i,i)$ hence $[T]_B[T]_B^*=\operatorname{diag}(-i^2,-i^2,-i^2,\ldots,-i^2,-i^2,-i^2)=I$, so $T$ is unitary.

But $[T^*]_B=[T]_B^*$ iff $B$ is an orthonormal basis, so something is missing in the proof, am I right?

Answer 1

The statement is false. Take $T(x,y)=(x-y,2x-y)$. Then $T^2(x,y)=(-x,-y)$, but $T$ is not unitary.

And, yes, what you wrote about the fact that the basis needs to be orthogonal in order to reach the desired conclusion is correct.

Answer 2

The argument is proving that $T$ is unitary with respect to some positive definite Hermitian inner product on $V$, not that it is unitary with respect to a pre-existing choice.

More generally, any finite-order automorphism of $V$ is unitary with respect to some choice of positive definite Hermitian inner product structure on $V$ (by essentially the argument that proves Maschke's theorem).