Show that $A$ is a difference between two orthogonal projections.

Question

Let $V$ be a finitedimensional complex vector space. Linear operator $A \in L(V) $ is hermitian and unitary. Show that $A$ is a difference between two orthogonal projections.

The questions seems easy enough, I just have to find the right projectors, but I am stuck in doing so. $A$ is hermitian so $ {A} ^{*}=A$ and unitary $A{A} ^{*}={A} ^{*}A =I$. I just have to find two operators $P$ and $Q$ so that $P^2=P={P} ^{*}$ and $Q^2=Q={Q} ^{*}$ and $A= P - Q$. If anyone could give me a hint that would be great.

Answer 1

HINT: Note that $A$ is a zero of $X^2-1=(X-1)(X+1)$. For each of the factors $$(A\pm I)^2=A^2\pm 2A+I=2(\pm A+I).$$ Do you see how to construct the two projections from here?

Answer 2

Using Spectral decomposition of $A$ write $A=P^*\Delta P$ where $\Delta$ is diagonal with diagonal entries real (since it is self adjoint) and modulus 1 (since it is unitary). Thus the diagonal entries can only be $1$ or $-1$. The rest is easy.