Given $T(v)=v-a\left<v,w_1\right>w_1-a\left<v,w_2\right>w_2$ where $a\in \mathbb R$ and $||w_1||=||w_2||=1, w_1\perp w_2$ Show that $T=T^*$

Question

Given $T(v)=v-a\left<v,w_1\right>w_1-a\left<v,w_2\right>w_2$ where $a\in \mathbb R$ and $||w_1||=||w_2||=1, w_1\perp w_2$. Show that $T$ is self adjoint.

I Think that completing $w_1,w_2$ to an orthonormal basis is the way to go, but I'm not sure how to show the requirement once there. I got that $T(v)=v-a\lambda_1w_1-a\lambda_2w_2$. But I'm not sure how to continue (if that's the right way at all)

EDIT: Maybe I can show that it multiplies $\lambda_1$ and $\lambda_2$ by $(1-a)$ which is real, so in the orthonormal basis it's diagonal with real eigenvalues? Would that be right?

Answer 1

Hint: just take any two vectors $v,w$ and test $$\langle Tv,w\rangle =\langle v,Tw\rangle$$ by plugging in T directly.

Another way is showing that any projection operator $$P_w:v\mapsto \langle v,w\rangle w$$ is selfadjoint and that linear combinations of selfadjoint operators are selfadjoint

Answer 2

$T=T^{*}$ is equivalent to $\langle Tv,w \rangle =\langle v,Tw \rangle$ for all $v$ and $w$. This is very straightforward. Simply write down the definition of $T$ and use linearity of inner product. You don't even need orthonormality of $w_1$ and $w_2$.