proving that a linear transformation is unitary and adjacent to itself (details inside) could you help me solve this question? found it in an old textbook.
$V$ is an inner product space. let $w_1,w_2 \in V$ vectors so that $||w_1||=||w_2||=1$ and $(w_1,w_2)=0$. we'll define a linear transformation $T:V \to V$, 
$$Tv=v-2(v,w_1)w_1-2(v,w_2)w_2.$$
a)prove that $T$ is unitary and adjacent to itself(i don't know the terminology, but the mathematical property is: $(Tu,v)=(u,T^*v)$.
b)check if $T$ is non negative
what i tried to do:
a)to show that $T$ is unitary is easier, because i only have to show that$(t(w_1),t(w_w))=(w_1,w_2)$, but to show that it is adjacent to itself(i.e to show that  $(Tu,v)=(u,T^*v)$) is harder, and i don't get to it. since $||w_1||=||w_2||=1$ it's easy to show $T$ unitary with what i said, but i can't do  $(Tu,v)=(u,T^*v)$.
b)i don't know how to show that $T$ is an nongative matrix, though it seems that i need to show that in any case, $v$ is bigger than $-2(v,w_1)w_1-2(v,w_2)w_2$ 
don't know how to solve it.
 A: $T$ as a linear operator to be self-adjoint we should have 
$$ \langle Tu,v \rangle =  \langle u,T^* v \rangle = \langle u,T v \rangle,$$
or simply $T = T^*$. 
Then we have, 
$$\langle Tu,v \rangle = \Big\langle u-2\langle u,w_1\rangle w_1 -2\langle u,w_2\rangle w_2 , v\Big\rangle   = \langle u,v \rangle - 2\langle u,w_1\rangle \langle w_1,v\rangle - 2\langle u,w_2\rangle \langle w_2, v\rangle ,$$
$$\langle u,Tv \rangle = \Big\langle u,v-2 \langle v,w_1 \rangle w_1 -2\langle v,w_2\rangle w_2 \Big\rangle   = \langle u,v \rangle - 2 \overline{\langle v,w_1\rangle} \langle u,w_1\rangle - 2 \overline{\langle v,w_2 \rangle} \langle u,w_2\rangle, $$
as it is clear $\langle Tu,v \rangle = \langle u,Tv \rangle$ and hence $T=T^*$ which means it is self-adjoint.
$T$ to be unitary we should have 
$$\|Tv \| = \|v \|, \quad \forall \;v \in V.$$
Then we have, 
$$ \| Tv \|^2 = \langle Tv,Tv \rangle = \Big\langle v-2\langle v,w_1\rangle w_1 -2\langle v,w_2\rangle w_2 , v-2\langle v,w_1\rangle w_1 -2\langle v,w_2\rangle w_2\Big\rangle \\ = \ldots = \langle v,v \rangle  +4 |\langle v,w_1 \rangle|^* +4|\langle v,w_2 \rangle|^2 = \|v\|^2  +4 (|\langle v,w_1 \rangle|^2 +|\langle v,w_2 \rangle|^2),$$
it is not necessarily $\langle v,w_2 \rangle = 0$, $\langle v,w_1 \rangle = 0$ and hence $\|Tv \| \neq \|v \|$ and so it is not an unitary operator. 
$T$ to be nonnegative (positive semidefinite) we should have 


*

*$T$ is self-adjoint,

*$\langle Tv, v \rangle \geq 0 \quad \forall \; v \in V$.


Now, it needs to check whether $\langle Tv, v \rangle \geq 0$ or not. 
\begin{aligned}
\langle Tv, v \rangle &= \big \langle v-2 \langle v,w_1 \rangle w_1 -2\langle v,w_2\rangle w_2, v\rangle 
\\
&= \|v\|^2 -2 |\langle v,w_1 \rangle|^2 -2 |\langle v,w_2 \rangle|^2, \\
&\geq \|v\|^2 -2 \|v\|^2 \|w_1\|^2 -2 \|v\|^2 \|w_2\|^2, \quad (\text{by cauchy-schwarz inequality}) \\  
&= -3\|v\|^2 ,
\end{aligned}
which implies $\langle Tv, v \rangle \ngeq 0, \; \forall v\neq 0$.
