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Please help me with this. Let $n>2$, I try to calculate the $n$th power of the $n\times n$ matrix with $0$ in diagonal and $-1$ in other places. Can someone give me a hint?

TShiong
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smulslippy
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Jan 10 '23 at 08:48
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    Hint:Do experiment for n=3,4,5 then you will realize the pattern and pattern do exist.Follow up that(I.e the pattern) try to generalize. And dear, plz follow the rules of MSE of uploading question and answers. –  Jan 10 '23 at 08:51
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    This is a rank-one update of the identity matrix. Its powers can be obtained using the binomial theorem. – user1551 Jan 10 '23 at 09:06
  • This post about matrices of this form is helpful – Ben Grossmann Jan 10 '23 at 18:44

1 Answers1

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Here's a solution using the suggestion user1551's comment. We can write you matrix in the form $A = I - nP$ where $I$ is the identity matrix and $P$ is the matrix whose entries are all equal to $1/n$. Importantly, $I^k = I$ and $P^k = P$ for all integers $k$. So, with the binomial theorem, you can easily expand $(I - nP)^n$.

In particular, we have \begin{align} (I - nP)^n &= \sum_{j=0}^n \binom nj I^{n-j}(-nP)^{j} \\&= \binom n0 I^n P^0 + \sum_{j=1}^n \binom nj I^{n-j}(-n)^j P^{j} \\ & = I + \sum_{j=1}^n \binom nj (-n)^j P \\&= I + \left[\sum_{j=1}^n \binom nj (-n)^j \right]P \\ & = I + \left[(-1) + \sum_{j=0}^n \binom nj (1)^{n-j}(-n)^j \right]P \\ & = I + \left[(-1) + (1-n)^n \right]P = I + \left[(1-n)^n - 1\right]P. \end{align} Interestingly, the entries of $\left[(1-n)^n - 1\right]P$ are all equal to $$ \left[(1-n)^n - 1\right] \cdot \frac 1n = - \frac{(1-n)^n - 1}{(1-n) - 1} = - \sum_{k=0}^{n-1} (1-n)^k. $$

Ben Grossmann
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