What are the eigenvalues of the given matrix $M$

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.

• But you are saying from the start that the eigenvalues are the $\lambda_i$'s. Are you sure about that? – José Carlos Santos Dec 20 '18 at 19:51
• 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. – Mark Dec 20 '18 at 19:52
• sorryyyy! I'm going to edit it – Peyman mohseni kiasari Dec 20 '18 at 19:53
• Note that all columns are scalar multiples of the first column. Thus this matrix has rank $1$, and there is only one nonzero eigenvalue. – Robert Israel Dec 20 '18 at 19:57
• @Damien Seems more fruitful to look at the vector $(a_1, a_2, \ldots, a_n)$. – Bungo Dec 20 '18 at 20:18

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 $$n$$.

• 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. – Peyman mohseni kiasari Dec 20 '18 at 20:28
• @Peyman, you may interpret a vector as an $n\times 1$ matrix. – Decaf-Math Dec 20 '18 at 20:30
• 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). – ekkilop Dec 20 '18 at 20:32

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 $$n$$.

• 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 – Peyman mohseni kiasari Dec 20 '18 at 20:10
• @Peyman Whoops, you are right. The eigenvalues are $0$ and $n$. – mechanodroid Dec 20 '18 at 20:15

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.