Assume that $A^3=A$. For which values of $w$ is the matrix $wI-A$ invertible?

Question

I am interested in the following problem:

Assume that $A$ is a $2\times 2$ matrix such that $A^3=A$, and that $A\neq 0,\pm I$. For which values of $w\in{\mathbb R}$ is $B_w=wI-A$ invertible? In that case, determine $B_w^{-1}$.

It is fairly easy using the minimal polynomial. But, since I have to explain to some high school students, the only tools I can use are basics on matrices and determinant (usual PreCalculus material), as well as the characteristic polynomial and Cayley-Hamilton Theorem (but no diagonalisation).

The first part is simple: $B_w$ is invertible if it is not a root of the characteristic polynomial.

*The second part seems more tricky. Any idea?

What I did so far:

1. We have $\det(A)\in\{-1,0,1\}$ since $A^3=A$.
2. Under some conditions on $A$ and/or $v\in{\mathbb R}$, $I-vA$ is invertible and $\displaystyle (I-vA)^{-1}=\sum_{k\geqslant 0}(vA)^k$. By considering the $3$ different cases for the determinant, I can determine simplify the series. For example, if $\det(A)=0$, then $A^k=t^{k-1}A$ (with $t$ the trace of $A$).

The second part seems convoluted and I wish to simplify it...

Answer 1

The characteristic polynomial of $A$ is one of the three: $$ \det(wI-A)=w^2-1, \ \mathrm{or} \ w(w-1), \ \mathrm{or} \ w(w+1). $$ By Cayley-Hamilton, $$ A^2=I, \ \mathrm{or} \ A(A-I)=O, \ \mathrm{or} \ A(A+I)=O. $$ In the first case $A^2=I$, $$ (A-wI)(A+wI)=A^2-w^2I = A^2-I+(w^2-1)I= (w^2-1)I. $$ Then we have $$ B_w^{-1}=-\frac1{w^2-1}(A+wI). $$ The other two cases can be treated similarly.

Answer 2

Actually you can solve this without the characteristic polynomial.

$A=wI-B$ implies $A^3=w^3I-3w^2B + 3 wB^2 -B^3$ and from here $wI-B=w^3I-3w^2B + 3 wB^2 -B^3$ equivalent $I=B[(\frac {1}{w - w^3})(1 - 3w^2)I + 3wB - B^2]$

Therefore $B^{-1}=[(\frac {1}{w - w^3})(1 - 3w^2)I + 3wB - B^2] \tag1$

Now just replace $B = wI - A$ in (1)