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I have been asked to prove the following:

$$(I_n - A)^{-1} = I_n + A + A^2 + A^3\,,$$

where $A$ is a square matrix of size $n\times n$, and provided that $A^4 = \mathcal{O}_n$, where $\mathcal{O}_n$ is the $n \times n$ zero matrix.

However, I have no idea from where I can start. It occurred to me to try and multiply $(I_n - A)^{-1}$ with $(I_n - A)$, however I do not think this is the case.

ADi
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Entangled Being
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2 Answers2

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What is $(I_n - A)(I_n + A + A^2 + A^3)$? What is $A^n$ for $n \ge 4$?

marty cohen
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Let $n \in \mathbb{N}$ and $A \in \mathcal{M}_{n}(\mathbb{K})$. Suppose that $A^4 = \mathcal{O}_n$. To show that $(I_n - A)^{-1} = I_n + A + A^2 + A^3$ is to show that $I_n + A + A^2 + A^3$ is the inverse matrix of $I_n - A$, i.e. $$(I_n - A)(I_n + A + A^2 + A^3) = I_n = (I_n + A + A^2 + A^3)(I_n - A)\,.$$ It follows from the properties of the addition and multiplication of matrices that

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

Similar computations shows that $(I_n + A + A^2 + A^3) (I_n - A) = I_n$. Hence $I_n + A + A^2 + A^3$ is the inverse of $I_n - A$. Therefore, $(I_n - A)^{-1} = I_n + A + A^2 + A^3$. $\square$

ADi
  • 302