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Problem: Let $A$=[$a_{ij}$]$_{n x n}$ where $a_{ii}$=$1$, $i=\overline {1,n}$, $a_{ij}=a\not=1, i\not=j$. Find $A^n$, $n\in \mathbb N$.

I need a little help solving this problem. Now, I know how to find $A^n$ if this was matrix 2 by 2, or 3 by 3. I know there is a way to find the eigenvalues and the eigenvectors and form the matrices $P$ (made of eigenvectors) and $D=diag(\lambda_1,...,\lambda_n)$ using $A^n$=$PD^n$$P^{-1}$ (correct me if I am wrong). However, I do not know how to find the eigenvalues and eigen vectors. So, I tried this:

My matrix $A$ would have this form:$$A= \begin{bmatrix} 1 & a & a & \cdots &a \\ a & 1 & a &\cdots &a \\ \vdots & \vdots &\vdots &\ddots&\vdots\\ a & a & a &\cdots &1 \\ \end{bmatrix} $$

So, that is symmetric matrix. If I look at my rows, I see that sum of all elements in each row is $(n-1)a+1$. Could I then say that my eigenvalue is $(n-1)a+1$ ? I was thinking about using $A^n\overrightarrow v=\lambda^n \overrightarrow v$, where $\overrightarrow v$ is my eigenvector, to get the result but I am really not sure how to go from here.

Any help is greatly appreciated.

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  • $\begingroup$ Yes you can indeed say that is an eigenvalue, using the fact that for $v=\begin{bmatrix} 1\\1\\\vdots\\1\end{bmatrix}$ $$Av=((n-1)a+1)v$$. Think of other eigenvectors to find them all. But, in my opinion, the best way would be solving the equation $$\det(A-\lambda I)=0$$, as the roots $\lambda$ would be the eigenvalues. In particular, this new matrix looks a lot like yours, only the diagonal entries are different. $\endgroup$
    – Emre
    Jun 2, 2016 at 1:03

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Continuing from the comment I provided:

$$A-\lambda I=\begin{pmatrix} 1-\lambda&a&\cdots&a\\ a&1-\lambda&\cdots&a\\ \vdots&\vdots&\ddots&\vdots\\ a&a&\cdots&1-\lambda \end{pmatrix} $$

We want this matrix to be singular. We already know that making the diagonals $-(n-1)a$ makes it singular. Trying out for $n=2$, one can immediately see that making all the diagonals equal to $a$ also works. This second possibility corresponds to the eigenvalue $1-a$.

These are the only eigenvalues, as $v_i=(1,0,\ldots,0,\overbrace{-1}^\text{i-th place},0,\ldots,0)$ are for $i=2,\ldots, n$ are the eigenvectors for $\lambda=1-a$.

So, we have $$P=\begin{pmatrix} 1&1&1&\ldots&1\\ 1&-1&0&\ldots&0\\ 1&0&-1&\ldots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&0&0&\ldots&-1 \end{pmatrix}\qquad D=\begin{pmatrix} (n-1)a+1&0&0&\ldots&0\\ 0&1-a&0&\ldots&0\\ 0&0&1-a&\ldots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\ldots&1-a \end{pmatrix}$$

After some calculation, one can see that $$P^{-1}=\frac1n\begin{pmatrix} 1&1&1&\ldots&1\\ 1&1-n&1&\ldots&1\\ 1&1&1-n&\ldots&1\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&1&1&\ldots&1-n \end{pmatrix}$$

Let $c=(1-a)^n$ and $b=((n-1)a+1)^n$. Then,

$$A^n=PD^nP^{-1}=\frac1n\begin{pmatrix} b+(n-1)c&b-c&b-c&\ldots&b-c\\ b-c&b+(n-1)c&b-c&\ldots&b-c\\ b-c&b-c&b+(n-1)c&\ldots&b-c\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ b-c&b-c&b-c&\ldots&b+(n-1)c \end{pmatrix}$$

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  • $\begingroup$ Thank you very much! I have one more question. After I have $P$ and $D$, do I continue to find $P^{-1}$, and then calculate $P$ $D^n$ $P^{-1}$ or is there some better way to find $A^n$? $\endgroup$
    – Asleen
    Jun 2, 2016 at 17:45
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    $\begingroup$ My first attempt would be to calculate $A^n$ for $n=1,2,3,4$ and to figure out the pattern. But, calculating $D^N$,$P^{-1}$ and multiplying them is not that hard either. $\endgroup$
    – Emre
    Jun 2, 2016 at 17:50
  • $\begingroup$ I edited the solution, hope this helps. $\endgroup$
    – Emre
    Jun 2, 2016 at 18:27

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