Proof regarding unitary self-adjoin linear operators I'm suck on how to do the following Linear Algebra proof:
Let $T$ be a self-adjoint operator on a finite-dimensional inner product space $V$. 
Prove that for all $x \in V$, $||T(x)\pm ix||^2=||T(x)||^2+||x||^2.$
Deduce that $T-iI$ is invertible and that $[(T-iI)^{-1}]^{*}=(T+iI)^{-1}.$
Furthermore, show that $(T+iI)(T-iI)^{-1}$ is unitary.

My attempt at a solution (to the first part):
$||T(x)\pm ix||^2=\left< T(x)\pm ix, T(x)\pm ix\right>$
$=\left< T(x), T(x) \pm ix\right>\pm \left<ix,  T(x)\pm ix \right>$
...
$=\left<T(x), T(x) \right>+ \left<x,x \right>$
$=||T(x)||^2+||x||^2$
The ... is the part I'm stuck on (I know, it's the bulk of the first part).
I have yet to consider the next parts since I'm still stuck on this one.
Any help would be appreciated! Thanks.
 A: we have $$(Tx+ix,Tx+ix)=(Tx,Tx)+(ix, Tx)+(Tx, ix)+(ix,ix)=|Tx|^{2}+i(x,Tx)-i(x,Tx)+|x|^{2}$$where I assume you define the inner product to be Hermitian. I think for $Tx-ix$ it should be similar. The rest should be leave as an exercise for you; they are not that difficult. 
To solve the last one, notice we have $(T-iI)^{-1}$'s adjoint to be $(T+iI)^{-1}$, and $(T+iI)$'a adjoint to be $T-iI$. The first one can be proved by expanding $(T-iI)^{-1}$ as $(I-iT^{-1})^{-1}T^{-1}$, then use geometric series. 
The second one follows by $(Tx+ix,y)=(Tx,y)+(x,-iy)=(x,Ty)+(x-iy)=(x,(T-I)y)$. 
If we want $(T+iI)(T-iI)^{-1}$ to be unitary, then we want 
$$((T+iI)(T-iI)^{-1}x,(T+iI)(T-iI)^{-1}x)=(x,x)$$ for all $x$. Moving around this is the same as 
$$((T^{2}+I)(T-iI)^{-1}x,(T-iI)^{-1}x)=(x,x)$$ but this is the same as 
$$((T+iI)x, (T-iI)^{-1}x)=(x,x)$$
and the result follows because we know $(T-iI)^{*}=(T+iI)^{-1}$. 
A: I think this is should be a correct sketch of the proof...I will leave the details to you
1). $||T(x) \pm ix||^2 = <T(x) \pm ix,T(x) \pm ix>$ 
by definition

2).$<T(x) \pm ix,T(x) \pm ix> = <T(x), T(x) > + <ix, ix> \pm <T(x), ix> \pm <ix, T(x)>$

3). But.. we know that $<T(x) , ix> = -<ix,T(x)> $ since $T = T^*$  and $ix = (-ix)^*$ which means the last two terms of the RHS of step two cancel each other out leaving you with the result. $<T(x), T(x)> + <ix, ix> = ||T(x)||^2 + ||x||^2$ as desired. 
