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I am learning linear algebra and now I'm in eigenvalues and eigenvectors part of it. there is a question that I can't solve it or any idea that I have is hard and nasty. I think this question must have a trick that I am not familiar with it because I'm new to eigenvalues and eigenvectors.

the question is this:

Let $M \in \Bbb R^{n\times n}$ and real numbers $a_1$ to $a_n$ and every $m_{ij} = \frac{a_i}{a_j}$, so: $$ M = \begin{pmatrix}1&\cdots&\frac{a_1}{a_n}\\\vdots&\ddots&\vdots\\\frac{a_n}{a_1}&\cdots&1\end{pmatrix} $$ find all eigenvalues.

any help would be appreciated.

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  • $\begingroup$ But you are saying from the start that the eigenvalues are the $\lambda_i$'s. Are you sure about that? $\endgroup$ – José Carlos Santos Dec 20 '18 at 19:51
  • $\begingroup$ A matrix of size $n\times n$ can't have more than $n$ distinct eigenvalues. So your question makes no sense as it is written right now. $\endgroup$ – Mark Dec 20 '18 at 19:52
  • $\begingroup$ sorryyyy! I'm going to edit it $\endgroup$ – Peyman mohseni kiasari Dec 20 '18 at 19:53
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    $\begingroup$ Note that all columns are scalar multiples of the first column. Thus this matrix has rank $1$, and there is only one nonzero eigenvalue. $\endgroup$ – Robert Israel Dec 20 '18 at 19:57
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    $\begingroup$ @Damien Seems more fruitful to look at the vector $(a_1, a_2, \ldots, a_n)$. $\endgroup$ – Bungo Dec 20 '18 at 20:18
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Another way to do this is by noting that if $\mathbf{a} = (a_1, \dots, a_n)^\top$ and $\mathbf{b} = (1/a_1, \dots, 1/a_n)^\top$, then $$ M = \mathbf{a} \mathbf{b}^\top, $$ where $\mathbf{b}^\top$ denotes the transpose of $\mathbf{b}$. The rank of $M$ is therefore 1 (can you see why?), meaning that only one eigenvalue is non-zero. This eigenvalue is found by considering $$ M \mathbf{a} = (\mathbf{a} \mathbf{b}^\top) \mathbf{a} = \mathbf{a} (\mathbf{b}^\top \mathbf{a}) = n \mathbf{a}, $$ i.e. the final eigenvalue is $n$.

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  • $\begingroup$ short and nice. sorry if my question is silly but how do you mult two vectors in this way? I just know that mult of two vectors is a number. this way of mult is not in the book yet. $\endgroup$ – Peyman mohseni kiasari Dec 20 '18 at 20:28
  • $\begingroup$ @Peyman, you may interpret a vector as an $n\times 1$ matrix. $\endgroup$ – Decaf-Math Dec 20 '18 at 20:30
  • $\begingroup$ You multiply them by normal matrix multiplication. Note that the number of columns of $\mathbf{a}$ (i.e. one) is equal to the number of rows of $\mathbf{b}^\top$ (also one). $\endgroup$ – ekkilop Dec 20 '18 at 20:32
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To find the eigenvalues, we are calculating the zeroes of the characteristic polynomial of $M$.

$$0= \det(M - \lambda I) = \begin{vmatrix} 1-\lambda &\frac{a_1}{a_2} & \cdots & \frac{a_1}{a_n} \\ \frac{a_2}{a_1} & 1-\lambda & \cdots & \frac{a_2}{a_n} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{a_n}{a_1} & \frac{a_n}{a_2} & \cdots & 1-\lambda \end{vmatrix}$$

Since $a_1, \ldots, a_n \ne 0$, we can multiply $j$-th column by $a_j$ for $j = 1, \ldots, n$ to obtain:

$$0 = \begin{vmatrix} a_1(1-\lambda) & a_1 & \cdots & a_1 \\ a_2 & a_2(1-\lambda) & \cdots & a_2 \\ \vdots & \vdots & \ddots & \vdots \\ a_n & a_n & \cdots & a_n(1-\lambda) \end{vmatrix}$$

Now divide $i$-th row by $a_i$ for $i =1, \ldots, n$ to obtain

\begin{align} 0 &= \begin{vmatrix} 1-\lambda & 1 & \cdots & 1 \\ 1 & 1-\lambda & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & 1-\lambda \end{vmatrix} \\ &= \begin{vmatrix} 1-\lambda & 1 & \cdots & 1 \\ \lambda & -\lambda & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ \lambda & 0 & \cdots & -\lambda \end{vmatrix} \\ &= \begin{vmatrix} n-\lambda & 1 & \cdots & 1 \\ 0 & -\lambda & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0& 0 & \cdots & -\lambda \end{vmatrix} \\ &= (n-\lambda)(-\lambda)^{n-1} \end{align}

so the eigenvalues are $0$ and $n$.

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    $\begingroup$ I think it is wrong. let a1=1 and a2 =1 then M is [[1,1],[1,1]]. but the eigenvalues of M are 0 and 2 $\endgroup$ – Peyman mohseni kiasari Dec 20 '18 at 20:10
  • $\begingroup$ @Peyman Whoops, you are right. The eigenvalues are $0$ and $n$. $\endgroup$ – mechanodroid Dec 20 '18 at 20:15
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We can write your matrix as $M = DJD^{-1}$, where $$ D = \pmatrix{a_1\\ & \ddots \\ && a_n}, \quad J = \pmatrix{1 & \cdots & 1\\ \vdots & \ddots & \vdots \\ 1 & \cdots & 1} $$ So, $M$ is similar to $J$. $J$ is a rank $1$ symmetric matrix, so it's only non-zero eigenvalue will be $\operatorname{tr}(J) = n$.

We could also recognize that $M$ has rank $1$ as Robert did in his comment.

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