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

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?)

• Suppose $T^2=I$ and $T=T^*$; then $TT^*=I$, so $T$ is unitary and Hermitian. Can you find a nonsymmetric real $2\times2$ matrix having trace $0$ and determinant $-1$? Easy one: $\begin{bmatrix}0&2\\1/2&0\end{bmatrix}$; less easy: $\begin{bmatrix}2&-3\\1&-2\end{bmatrix}$. – egreg Dec 16 '18 at 22:55

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 $$T^*(x,y)=\left(\frac y2,2x\right)$$.

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

if $$T$$ is an isometric operator

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.