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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?)

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    $\begingroup$ 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: $\begin{bmatrix}2&-3\\1&-2\end{bmatrix}$. $\endgroup$ – egreg Dec 16 '18 at 22:55
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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 $T^*(x,y)=\left(\frac y2,2x\right)$.

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

if $T$ is an isometric operator

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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.

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