I've been asked to either prove or disprove that for a 2 by 2 matrix $A$ and the 2 by 2 identity matrix $I$, if $A(I-A) = I$ then $A^2-I$ is invertible.
To detail what I've attempted so far, if necessary:
I started by trying to find a counterexample to no avail, so I tried to prove it; but that also didn't quite work out. The things I did show (assuming no silly mistakes were made) are:
- $A$ is invertible, and $A^{-1}=I-A$, which results directly from $A(I-A)=I$ by the definition of matrix invertibility (so $I-A$ is also invertible).
- $A^2 = A-I = -A^{-1}$ through left distributivity.
- $A^3 = -I$, stemming directly from #2 by right-multiplying both sides by $A$.
- $A^2-I = A-2I = (A+I)(A-I)$, using #2 (left side to middle) and left & right distributivity (right side to left side).
As for #4, which feels significant because it involves the expression I'm investigating, I already know that $(A-I) = A^2$ which is invertible (because $A$ is, and also because of #2 above), so I could "reduce" the problem to proving that $A+I$ is invertible because then $(A^2-I)\cdot((-A)\cdot(A+I)^{-1})=I$ and we're done; but I couldn't find a way to do that, either.
Would greatly appreciate some assistance or a nice hint :)
If $(A+I)x=0$ then $A^{2}x=A(-x)=-Ax=x$ and $A(I-A)x=Ax-A^{2}x=-x-x=-2x$. By hypothesis this implies that $x=0$ so the kernel of $A+I$ is $\{0\}$. Hence $A+I$ is invertible. Also $det (A(I-A))=1$ so $det (A) \neq 0$ and $det (I-A) \neq 0$. It follows now that $A-I$ and $A+I$ are both invertible. Hence $A^{2}-I=(A-I)(A+I)$ is also invertible.