# Characteristic polynomial of an $n \times n$ matrix with only one column and one row non zero.

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) :

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

• 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 $a=-\binom n3$. Oct 14 at 20:44

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.

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 $$n$$.