Showing that the matrix $A+I$ is invertible

Question

If I have the matrix equation:

$A^3=KA$ , k≠1

How to show that $A+I$ is invertible? I'd like to do this problem using eigenvalues, so how can I prove that $-1$ is an eigenvalue of the matrix $A$?

Answer 1

The idea is inspired by the Neumann series $(1-x)^{-1} = \sum_{k=0}^\infty x^k$ for $|x|<1$. We want to use this idea with $x=-A$. Since $A^3=kA$, this series will contain powers of $A$ only up to order $2$. So we guess the inverse to be of the form $I+xA+yA^2$. Then we get $$ (A+I)(I + xA + yA^2) = I + xA + yA^2 + A+xA^2+yA^3 =I + (x+ky+1)A + (x+y)A^2. $$ This is equal to the identity if and only if $$x=-y, \quad x(1-k)=1.$$ Hence setting $x = \frac1{1-k}$, $y=-\frac1{1-k}$ we get $$ (A+I)^{-1} = I + \frac1{1-k}(A-A^2). $$

Answer 2

The equation $A^3=kA$ tells you that $f(A)=0$ for the polynomial $f(x)=x^3-kx=x(x^2-k)$, so the minimal polynomial (a divisor of $f$) can only have the zeroes $0$ and $\pm \sqrt{k}$ (provided $k\ge 0$). The characteristic polynomial has the same zero set as the minimal polynomial, so $-1$ is not an eigenvalue of $A$. Hence $A+I$ is invertible.