TITLE: If $A^{2016} = I_n$, show that $A^{576} - A^{288} + I_n$ is invertible, and calculate it's inverse in terms of $A$.
QUESTION [3 upvotes]: Let $A$ be a real valued $n \times n$ matrix,where $n \geq 2$, such that $A^{2016} =I_n.$ Show that the matrix $B = A^{576} - A^{288} + I_n$ is invertible, and calculate it's inverse in terms of $A$.
Well, I was able to prove that matrix is invertible, however it is a bit long-winded. Call $p(x) = x^{576} - x^{288} + 1$. The eigenvalues of $B$ are of the form $p(\lambda)$, where $\lambda \in Spec A \subset U_{2018},$ the 2018th roots of unity. We want to show that $B$ can't have a zero eigenvalue; That is, no $2018$th root of unity is a root of $p(x)$. 
Since $p(x) = \left( x^{288} - e^\frac{2\pi}{5} \right) \left( x^{288} - e^\frac{8\pi}{5} \right)$, it's enough to show that there is no k such that
$$2016\left(\frac{2\pi}{5\cdot288}+\frac{2k\pi}{288}\right) = 2l\pi$$
or
$$2016\left(\frac{8\pi}{5\cdot288}+\frac{2k\pi}{288}\right) = 2l\pi,$$
where $l$ is another integer.
This follows from the fact that $5$ does not divide $2016$.
However, I have no idea how to find $B$ in terms of $A$. I think that there is a way to show that $B$ is invertible which also gives an expression for $B$, but I can't figure that out. Any ideas?

REPLY [4 votes]: Although Donald is not entirely correct, his idea took me to a nice answer: Again $B = A^{288},$ so $B^7=I$ and we want the inverse of $B^2-B+I$. Turns out that
$$(B^2-B+I)(B^{12} + B^{11} -B^9 - B^8 + B^6 -B^4 -B^3 + B + I) = B^{14} - B^7 + I = I$$
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