If $T$ is a unitary operator, and $-1$ is not an eigenvalue for $T$, then there exists a $V$ such that $V^*=-V$ and $T=(I-V)(I+V)^{-1}$

Question

The original question is as follows:

I'm stuck at d(i). I have managed to show uniqueness, but I don't know how to prove the existence of such an operator.

Here's my attempt so far. Since $\zeta$ is unitary, there exists an orthonormal basis $B$ such that it is diagonal. That is,

$$ [\zeta]_B=\left(\begin{array}{rrrr} a_1 & 0 & 0 & 0 \\ 0 & a_2 & 0 & 0 \\ 0 & 0 & \ddots & 0 \\ 0 & 0 & 0 & a_n \end{array}\right) $$

Since the objective is to find a linear operator, $\eta$, such that $\zeta= (I-\eta)(I+\eta)^{-1}$, I thought that if $c_i$ was an eigenvalue of $\eta$, then $a_i$ will be related to $c_i$ by $a_i=(1-c_i)/(1+c_i)$. I thought of proposing: $$ [\eta]_B=\left(\begin{array}{rrrr} c_1 & 0 & 0 & 0 \\ 0 & c_2 & 0 & 0 \\ 0 & 0 & \ddots & 0 \\ 0 & 0 & 0 & c_n \end{array}\right) $$ as the operator required, where $a_i=(1-c_i)/(1+c_i)$, but I don't think that $\eta^* = -\eta$.

Any ideas?

Edit: I made some progress on the question:

I think this is close to the answer, but this method requires me to prove $T=-T^*$, which is I think too strong a condition. We let $T=\zeta$ and $V=\eta$ because laziness.

We want to find $T=(I-V)(I+V)^{-1}$. That is, $(I+V)T=I-V \implies T+VT=I-V \implies VT+V=I-T \implies V(T+I)=I-T$. So $V$ will exist provided $T+I$ is invertible, and this is the criterion given in the question, since $-1$ is not an eigenvalue of $T$.

So we can just take $V=(I-T)(T+I)^{-1}$. We want to show that $V^* = -V=(T-I)(T+I)^{-1}=(T+I-2I)(T+I)^{-1}=I-2(T+I)^{-1}$.

$V^*=[(I-T)(T+I)^{-1}]^*=[(I+T-2T)(T+I)^{-1}]^*=[I-2T(T+I)^{-1}]^*=I-[(T+I)^{-1}]^*2T^*$

but after this I'm stuck.

Answer 1

Alright, I think I finally did it. Continuing from the question. We want to show $V^*=(T-I)(T+I)^{-1}=I-2(T+I)^{-1}$.

The first identity is that $(I+T^*)^{-1}=[(I+T)^{-1}]^*$. The proof is that $(I+T^*)[(I+T)^{-1}]^*=(I+T)^*[(I+T)^{-1}]^*=[(I+T)^{-1}(I+T)]^*=I$.

Now $V=(I-T)(T+I)^{-1}=(I+T-2T)(T+I)^{-1}=I-2T(T+I)^{-1}$.

Then $V^* = I-2[(T+I)^{-1}]^*T^*=I-2(I+T^*)^{-1}T^{-1}=I-2[T(I+T^*)]^{-1}=I-2[T+I]^{-1}$, and we are done.

Answer 2

Your proof is correct.

Alternatively, your proposed operator $\eta$ will indeed satisfy $\eta^* = -\eta$ if we take $c_i = \frac{1 - a_i}{1 + a_i}$. In particular, it suffices to note that $\operatorname{Re}[c_i] = 0$ for all $i$, and that because $B$ is an orthonormal basis we have $\eta = W[\eta]_BW^*$ for a unitary (change-of-basis) operator $W$.