Prove $T$ is diagonalisable if $T^*=\alpha T$ for some $\alpha\in\mathbb{C}$

Question

Question

Let $V$ be a finite-dimensional vector space over $\mathbb{C}$. Let $T:V\to V$ be a linear map, and $T^*$ its adjoint.

If $T^*=\alpha T$ for some $\alpha\in\mathbb{C}$ show that $T$ is diagonalisable.

My attempt

For all $v,w\in V$ we have $\langle v, T(w) \rangle = \langle T^*(v), w \rangle = \langle \alpha T(v), w \rangle = \bar{\alpha} \langle T(v), w \rangle = \bar{\alpha} \langle v, T^*(w) \rangle = \bar{\alpha}\alpha \langle v, T(w) \rangle$

So by non-degeneracy, $T=|\alpha|^2 T$.

But how do I find a basis of eigenvectors of $T$?

Answer 1

If $\vert\alpha\vert^2\neq 1$, then $T=0$ and $T$ is diagonalizable. If $\vert\alpha\vert^2=1 $, one may write $\alpha=\dfrac{u}{\bar{u}}$ with $u\neq 0$.

Then $(uT)^*=(uT)$. Since $uT$ is self adjoint, it is diagonalizable, and therefore, so is $T$.

Answer 2

Well if you know the spectral theorem then it follows by just one sentence:-

$TT^{*}=\alpha T^{2}=T^{*}T$ . Thus $T$ is normal and hence unitarily diagonalizable.