# Finding the inverse of an operator $T$ such that $T^4=\alpha^4 I$, where $\alpha\neq 0$ is complex

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.

• Your equality says that $ST=TS=I$ where $S=\alpha^{-4}T^3$. – Pedro Tamaroff Aug 17 '16 at 2:22
• @PedroTamaroff Oh my goodness! How silly of me! This is the result of little sleep. Thank you – Macarena Aug 17 '16 at 2:33
• Hint: You can derive it pairwise on the sets $H_1, H_2$, $w = \alpha T^{-1}w$, $w' = I\alpha T^{-1}w'$ and rewrite it as a direct sum of those results with some scalar juggling. – mathreadler Nov 29 '16 at 10:19