Find the characteristic polynomial, given the square of a matrix

Question

I've been studying linear algebra, and I came across this problem. We're told that A $\in$ $M_n$ ($K$), $A^2$ = $I_n$, with A ≠ $I_n$ and A ≠ -$I_n$ , and then we're supposed to find the characteristic polynomial. I've been trying for a bit, but just can't seem to find the right way to do this :( Any help is appreciated!

Answer 1

If $\lambda$ is eigenvalue of $A$, then $\lambda^2$ is eigenvalue of $A^2$. Characteristic polynomial of $A^2$ gives you $c(A^2) = det(\lambda^2 I_n - A^2) = (\lambda^2 - 1)^n = 0$. This means all eigenvalues of $A$ are $\pm 1$.

Next, note that minimal polynomial of $A$ is $p(x) = x^2 - 1$ and since minimal polynomial divides characteristic polynomial, characteristic polynomial of $A$ will be $c(x) = (x-1)^k*(x+1)^{n-k}$ where $1\leq k < n$.

Answer 2

The minimal polynomial is $X^2-1=(X+1)(X-1)$, since $A^2-I_n=0$ and $A-I_n\neq0$ and $A+I_n\neq0$ (so the minimal polynomial divides $(X+1)(X-1)$, but is not $X+1$ or $X-1$). The characteristic polynomial now has degree $n$ and has exactly the same irreducible factors as the minimal polynomial, so it's of the form $(X+1)^k(X-1)^{n-k}$ for $1\leq k\leq n-1$.