Let $\left\{ q_1, q_2, q_3, q_4 \right\}$ be an orthonormal basis of an inner product space $H$, and let
$$H_1=\text{span} [q_1, q_2], \quad H_2=\text{span} [q_3, q_4]. $$
Let $T$ be an operator on $H$ that satisfies
$$ Tw= \alpha w \quad \forall w\in H_1, \quad \quad Tw'=I\alpha w' \quad\forall w'\in H_2, $$
where $\alpha \neq 0$ is a complex number and $I$ is the imaginary unit. Show that $T^4=\alpha^4 I$. Use this to give a closed form for $T^{-1}$.
Here is what I have done:
Since $\left\{ q_1, q_2, q_3, q_4 \right\}$ is an orthonormal basis of $H$, then $H=H_1 \oplus H_2$ $\,$ ($H$ is the direct sum of $H_1$ and $H_2$), since every vector in $H_1$ is orthogonal to every vector in $H_2$, and vice versa.
Let $v\in H$. Then we can uniquely express $v=w+w'$, where $w \in H_1$ and $w'\in H_2$. Then $$Tv=Tw+Tw'=\alpha(w+iw'),$$ $$T^2v= \alpha T(w+iw')=\alpha^2(w-w'),$$ $$T^3v=\alpha^2T(w-w')=\alpha^3(w-iw'),$$ $$T^4v=\alpha^3T(w-iw')=\alpha^4(w+w')=\alpha^4v=\alpha^4Iv.$$
Hence, $T^4=\alpha^4 I$.
How can I use this to find a closed form for $T^{-1}$? Since $T^4=\alpha^4 I$ I know that $T^4$ is invertible, but I am not sure where to go from there. I appreciate any help.