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The problem asks for the eigenvalues of the matrix

$$\left(\begin{array}{cccc} 0 & \cdots & 0 & 1 \\ \vdots & & \vdots & \vdots \\ 0 & \cdots & 0 & n-1 \\ 1 & \cdots & n-1 & n \end{array}\right).$$

Since it was an $n \times n$ matrix, and I can't think of something immediatly that says this is a diagonalisable matrix so that I can use the trace and the product of the diagonal elements to deduce the eigen values, I thought about applying the general rule $\operatorname{det}(A)=\sum_{\sigma \in S_{n}}\left(\operatorname{sgn}(\sigma) \prod_{i=1}^{n} a_{i, \sigma_{i}}\right)$

So it was something along: $ -\lambda \times \det\left(\begin{array}{cccc} -\lambda & \cdots & 0 & 2 \\ \vdots & \ddots & \vdots & \vdots \\ 0 & \cdots & -\lambda & n-1 \\ 2 & \cdots & n-1 & n - \lambda \end{array}\right) \pm \det\left(\begin{array}{cccc} 0 & -\lambda & 0 & \cdots \\ 0 & & \ddots & 0 \\ \vdots & & 0 & -\lambda \\ 1 & \cdots & 0 & n-1 \\ \end{array}\right) $

I didn't continue along this it just didn't look like the best or practical way to approach it. Are there any particular properties I can use to shortcut that I couldn't see here?

The solution to the problem is (which moreover looked a bit far from where I was heading, although I don't see where the 6 is coming from and how it makes sense) :

$ 0 \text { and the roots of } 6 \lambda^{2}-6 n \lambda-n(n-1)(2 n-1)=0 \text {. }$

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  • $\begingroup$ Hello :) One can see, that the rank of $A$ is $n-2$, because the first $n-1$ columns are multiples of $(0,\ldots,0,1)^t$. Hence, zero is an eigenvalue, which has multiplicity $n-2$. Further, the sum of all eigenvalues is $n$ (the trace of $A$). From both observations we get $\chi(x)=x^{n-2}(x^2-n\cdot x+a)$, where $a$ is an unknown integer. It is left to prove, that $a=-\binom n3$. $\endgroup$
    – Jochen
    Oct 14 at 20:44
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Let's call your matrix $A$. The matrix $A$ is real and symmetric and so it is similar to diagonal matrix with entries $\lambda_1, \dots, \lambda_n$. Also, $A$ has rank two so the diagonal matrix must also have rank two so let's say $\lambda_3 = \dots = \lambda_n = 0$. Now you need two equations to find $\lambda_1,\lambda_2$. A possible set of such equations is:

  1. $\operatorname{tr}(A) = \lambda_1 + \lambda_2 + \dots + \lambda_n = \lambda_1 + \lambda_2 = n$.
  2. $\operatorname{tr}(A^2) = \lambda_1^2 + \lambda_2^2 + \dots + \lambda_n^2 = \lambda_1^2 + \lambda_2^2.$

One readily computes $\operatorname{tr}(A^2)$ to be $$ \operatorname{tr}(A^2) = 1^2 + 2^2 + \dots + (n-1)^2 + 1^2 + 2^2 + \dots + n^2 = \frac{n(2n^2+1)}{3}. $$ By plugging $\lambda_2 = n - \lambda_1$ into $\lambda_1^2 + \lambda_2^2 = \operatorname{tr}(A^2)$ you get a quadratic equation for $\lambda_1$ whose solutions are $$ \frac{n}{2} \pm \sqrt{\frac{8n^3 - 6n^2 + 4n}{12}} $$ and those are the eigenvalues.

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Two observations to start:

  1. The matrix (call it $M$) has rank 2, so it has (at most) two linearly independent eigenvectors with nonzero eigenvalues.

  2. If $\vec{x} = (x_1, \ldots, x_n)$, then $M \vec{x} = (x_n, 2x_n, 3x_n, \ldots, (n-1) x_n, x_1 + 2x_2 + \cdots + n x_n)$. Suppose $\vec{x}$ satisfies $M \vec{x} = \lambda \vec{x} \neq 0$. Then $x_n \neq 0$, because if $x_n = 0$, then the first $n-1$ components of $M\vec{x}$ and (therefore) $\vec{x}$ must be zero as well. It follows that $\vec{x}$ must have all nonzero components.

Now suppose $M \vec{x} = \lambda \vec{x} \neq 0$. Wlog suppose $x_n = 1$ (we know that $\vec{x}$ can't have zero components, so we can scale it arbitrarily to give any component whatever nonzero value we want). Then $\lambda x_k = k x_n$ for $1 \leq k \leq n-1$ (i.e. $x_k = k/\lambda$), and the last component of $M\vec{x} = \lambda \vec{x}$ gives the equation $x_1 + 2 x_2 + \cdots + n x_n = \lambda x_n \implies \lambda^{-1} \sum_{k=1}^{n-1} k^2 + n = \lambda$. The formula $1^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6}$ gives the necessary quadratic equation for $\lambda$, which has negative constant term and positive leading coefficient and (therefore) two distinct solutions for all values of $n$.

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