Prove that $(I_n-A)^{-1}$ exists

Question

Let $A \in M_{nxn}(\mathbb{C})$, and assume $A^5=\mathbb{O}_{nxn}$ (A zero matrix). Prove that $(I_n-A)^{-1}$ exists and what does it equal to? So the property that $A^5=\mathbb{O}$ is useful here, and to prove that the inverse exists, should I use the determinant method (calculate that the determinant cannot be $0$?). To find what it is equal to, I have an intuition that it is the identity matrix + $A^1 + A^2$ and such. What would be the exact answer?

Answer 1

\begin{align*} (I - A^5) &= (I - A)(I + A + A^2 + A^3 + A^4) \\ \Rightarrow |(I - A^5)| &= |(I - A)||(I + A + A^2 + A^3 + A^4)| \\ \Rightarrow 1 &= |(I - A)||(I + A + A^2 + A^3 + A^4)| \\ \Rightarrow |(I - A)| &\neq 0 \\ \end{align*} Qed.

Answer 2

If $A^5=0_n$ then $\mu\mid x^5$ where $\mu\in\mathbb{C}[x]$ is the minimal polynomial of $A$. Then $\mu=x^t$ for some $t\in\{1,2,3,4,5\}$. It is known that the minimal polynomial share the same roots as the characteristic polynomial of $A$. Hence $\left|xI_n-A\right|=x^n$ and $\left|I_n-A\right|=1$.

There is a result, easy to prove, garanting us that $$\sum_{k=0}^\infty{B^k}=(I_n-B)^{-1}$$ whenever $\rho(B)<1$ where $\rho(B)$ is the spectral radius of $B$.

Using that in our case, since $\rho(A)=0$, we have the following: $$(I_n-A)^{-1}=\sum_{k=0}^4{A^k}.$$

Answer 3

Let $A \in \mathcal{M}_{n}(\Bbb{C})$ be a $n \times n$ matrix such that $A^5 = \mathcal{O}_n$. Note that

\begin{align*} (I_n - A)(I_n + A + A^2 + A^3 +A^4) & = I_n + A + A^2 +A^3 + A^4 - A - A^2 - A^3 -A^4 - A^5 \\ & = I_n + A^5 \\ & = I_n + \mathcal{O}_n \\ & = I_n \end{align*}

Then, $I_n - A$ is invertible, and $A^{-1} = I_n + A + A^2 + A^3 + A^4$.