# Computing the inverse of a matrix function

I'm having trouble completing this exercise: given the following matrix: $$\begin{bmatrix} a & 0 & 0 & 1\\ 0 & a & 0 & 0\\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & a \end{bmatrix}$$ I need to compute the inverse of $$B = A^n, n \ge 2$$ Now I began by noticing that $$A = aI + C, C = \begin{bmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$ Now It's easy to see that $$C^n = [0], n \ge 2 \tag{1}$$ , where $$[0]$$ denotes the null matrix.
Therefore we get: $$A^n = (aI+C)^n = I\sum_{k = 0}^{n} \frac{n!}{k!(n-k)!}a^{n-k}C^{k}$$ where I used the fact that the identity matrix is idempotent. From property $$(1)$$, it follows that the sum above consists of only the $$K = 0, k = 1$$ terms. I end up getting, after some easy computations, the following expression: $$B = A^n = a^nI+ na^{n-1}C$$ Hence: $$B^{-1} = a^{-n}[I + \frac{n}{a}C]^{-1}$$ And here's where the trouble begins: I can't seem to compute the inverse of the term in square brackets. I've been trying to come up with a way but I ended up with nothing. Two possibilities lie in front of me: either I'm missing something very trivial, or there's some sort of very subtle trick to compute the inverse of that quantity, impossible to spot by the untrained eye (namely myself).Any help or answer is much appreciated, as always.

P.S. There's some problem with the latex code, it seems not to recognize the matrix pattern. I don't know whether it's a problem of mine only or it affects the reader. I used the standard latex notation begin{bmatrix} end{bmatrix} as per usual but it doesn't work somehow. If some editor could help and fix it because I cannot identify the mistake right now

• The trick is to guess that $(I+A)^{-1}=I-A+A^2-\cdots$ Commented Jul 2, 2023 at 20:59
• I've never encountered a geometric series of matrices up until now. Aren't there some extra conditions to be satisfied or assumptions to make in order for me to use this, because, in a standard geometric series, we have $$\sum_{k = 0}^{\infty}x^k = \frac{1}{1-x}, |x|<1$$. Anyways, I finally got: $$B^{-1} = a^{-n}\frac{1}{I-(-\frac{nC}{a})} =a^{-n}( I - \frac{nC}{a})$$ because of property $(1)$ Commented Jul 2, 2023 at 21:31
• I found out that this condition needs to be satisfied in order for the sum to be convergent to our result, $|\lambda_i| < 1$ for every eigen value. And our matrix has a single eigenvalue $\lambda = 0$ with algebraic multiplicity $4$ so everything's fine Commented Jul 2, 2023 at 21:45
• just do it formally. you know the series converges since it terminates. thats why I said "guess". You can check that the result is indeed correct by direct multiplication. Commented Jul 3, 2023 at 3:37

As you noticed, $$B= \begin{bmatrix} a^n & 0 & 0 & na^{n-1}\\ 0 & a^n & 0 & 0\\ 0 & 0 & a^n & 0 \\ 0 & 0 & 0 & a^n \end{bmatrix}$$
Hence $$B^2 = \begin{bmatrix} a^{2n} & 0 & 0 & 2na^{2n-1}\\ 0 & a^{2n} & 0 & 0\\ 0 & 0 & a^{2n} & 0 \\ 0 & 0 & 0 & a^{2n} \end{bmatrix} = 2a^nB - a^{2n}I$$
So $$B(2a^n I-B) = a^{2n} I$$
which gives that $$\boxed{B^{-1} = \dfrac{1}{a^{2n}}(2a^n I-B)}$$