We have a linear operator T. Show $T^2=Id$ implies $T=T^*$

Question

We have a linear operator $T:V\rightarrow V $. V is a finite-dimension inner product space over the field of the complex numbers. Show $T^2=Id$ implies $T=T^*$.

I've tried working with the inner product, trying to get $(Tx,x)=(x,Tx)$ with no luck. Maybe it has something to do with a basis for T (and thus diagonalizability?)

Answer 1

$(Tx, y)=(Tx, T^2 y)=(T x, T (Ty))=(x,T y)$

if $T$ is an isometric operator

Answer 2

I will build a basis on which $T$ is diagonal.

Given a non-zero vector $u$, consider $v=Tu$. We then have $Tu=v$. Let the first two vectors in our basis be $u+v,u-v$. These are eigenvectors of $T$ with eigenvalues $\pm1$ (if either of those two vectors is $0$, then don't include that in the basis).

Repeat the process by picking a $w$ not in the span of $u,v$, look at $Tw$, add $w\pm Tw$ to the basis, and so on.

We get a basis of eigenvectors with real eigenvalues, meaning $T$ is diagonalizable.

Answer 3

You can't prove it, since it is not true. Take$$\begin{array}{rccc}T\colon&\mathbb{R}^2&\longrightarrow&\mathbb{R}\\&(x,y)&\mapsto&\left(2y,\frac x2\right).\end{array}$$Then $T^2=\operatorname{Id}$, but $T^*(x,y)=\left(\frac y2,2x\right)$.