Generalization of nilpotent matrix property: I-A has an inverted matrix

Question

It can be proven that if $A$ is a nilpotent square matrix then $I-A$ can be inverted and the inverted matrix takes the form $I+A+A^2+A^3...+A^{(k-1)}$

where $k$ is the $k$ in $A^k=0$

I am to find the generalization of the property.

Prove that if square matrix A is nilpotent, then $aI-bA$ can be inverted, where $a \ne o$ and $a$ and $b$ are real numbers.

We are still in the beginning of the course, so no eigenvalues or upper triangles. I tried to prove it in the same fashion as the basic property was proven, but couldn't figure it out. Later realized that it doesn't seem to have the same kind of formula as above and that I don't even need to find the formula for the inverted matrix. I just need to prove that for $aI-bA$ an inverted matrix can be found.

Answer 1

I will assume that both $a$ and $b$ are non-zero. In this case, we can write $$ aI-bA = a(I - \frac ba A) $$ Note that $\frac ba A$ is a nilpotent matrix of the same order as $A$ (that is, $[\frac ba A]^k= 0$ if and only if $A^k=0$). So, we have a formula for $(I - \frac ba A)^{-1}$.

From there, note that $$ [a(I - \frac ba A)]^{-1}= a^{-1} (I - \frac ba A)^{-1} $$

Answer 2

You only have to prove that $\ker bA-aI=\{0\}$. Let $v\in \ker bA-aI$, that is to say $bAv=av$.

It's easy to prove by induction that $\forall m, b^mA^mv=a^mv$.

With $m=k$ you get $0=a^kv$. Since $a\neq 0, v=0$.