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Is there a general formula for the eigenvalues of this type of (symmetric) matrix?

$A = \left[ \begin{array}{ccccc} 0 & 0 & \cdots & 0 & a_1 \\ 0 & 0 & \cdots & 0 & a_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & a_{n-1}\\ a_1 & a_2 & \cdots & a_{n-1} & a_n \end{array} \right]$

Here, $a_i \in \mathbb{R}$.

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    $\begingroup$ Try calculating the characteristic polynomial $\endgroup$ – Alex Aug 29 '17 at 20:03
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Observation. The first $n-1$ rows are multiples of $(0,0,\ldots,0,1)$, and hence the rank of $A$ is at most $2$, and hence the eigenvalue $\lambda=0$ has multiplicity at least $n-2$.

Next, observe that if $\lambda\ne 0$ is an eigenvalue, then there is exists an eigenvector $(c_1,\ldots,c_n)$ $$A = \left[ \begin{array}{ccccc} 0 & 0 & \cdots & 0 & a_1 \\ 0 & 0 & \cdots & 0 & a_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & a_{n-1}\\ a_1 & a_2 & \cdots & a_{n-1} & a_n \end{array} \right]\left[\begin{array}{c}c_1 \\ c_2 \\ \vdots \\ c_{n-1} \\ c_n\end{array}\right]=\left[\begin{array}{c}a_1c_n \\ a_2c_n \\ \vdots \\ a_{n-1}c_n \\ \sum a_jc_j\end{array}\right]=\lambda\left[\begin{array}{c}c_1 \\ c_2 \\ \vdots \\ c_{n-1} \\ c_n\end{array}\right] $$ Clearly, $c_n\ne 0$, otherwise $c_1=\cdots=c_n=0$, and hence $$ c_j=\frac{c_n}{\lambda}a_j,\quad j=1,\ldots,n-1 \quad\text{and}\quad \lambda c_n=\sum a_jc_j $$ and thus $$ \lambda c_n=a_1c_1+\cdots a_{n-1}c_{n-1}+a_nc_n=\frac{c_n}{\lambda}(a_1^2+\cdots a_{n-1}^2)+a_nc_n $$ or $$ \lambda^2-a_n\lambda-(a_1^2+\cdots a_{n-1}^2)=0 $$ and thus the two remaining eigenvalues are obtained from the quadratic equation above.

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As a generic matrix, $A$ is a rank-2 matrix of the form $uv^T+vu^T$, where $u^T=(0,\ldots,0,1)$ and $v^T=(a_1,\ldots,a_{n-1},\frac{a_n}2)$. The eigenvalues of this kind of matrix are known to be $\lambda=u^Tv\pm\sqrt{(u^Tu)(v^Tv)}$. In your case, we get $\lambda=\frac{a_n}2\pm\sqrt{a_1^2+\cdots+a_{n-1}^2+\frac{a_n^2}4}$. Since eigenvalues vary continuously with matrix entries, the same answer also agrees with the true answer in the degenerate case where $A$ is a diagonal matrix of rank 1.

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Experimenting with low values of $n$, I obtained the following ansatz (which has to be proved by induction) for the characteristic polynomial of $A$:

$p(\lambda) = (-1)^{n+1} \lambda^{n-2} (-\lambda^2 + a_n \lambda + s)$,

with $s = a_1^2 + \cdots + a_{n-1}^2$. Now just apply Bhaskara.

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  • $\begingroup$ Thus you have answered your own question. Btw, what is Bhaskara ? $\endgroup$ – Jean Marie Aug 29 '17 at 20:50
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    $\begingroup$ @JeanMarie it is the quadratic formula, discovered by that guy. $\endgroup$ – Eduardo Longa Aug 29 '17 at 21:03
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Here's a push in the right direction. Noting that each symmetric pair of off-diagonal terms can be written as $a_k e_k e_n^T+a_k e_n e_k^T$, we may write the matrix as \begin{align} A &=\sum_{k=1}^{n-1} (a_k e_k e_n^T+a_k e_n e_k^T)+a_n e_n e_n^T \\ &=\left(\sum_{k=1}^n a_k e_k\right)e_n^T+e_n\left(\sum_{k=1}^n a_k e_k\right)^T - a_n e_n e_n^T \\ &=ae_n^T+e_n a^T-a_n e_n e_n^T \end{align} where $a=(a_1,a_2,\cdots a_n)^T$. To see the use of this expression, consider how $A$ acts on any vector orthogonal to $a$ and $e_n$, and what the dimension of this subspace will be. For the remaining eigenvalues, consider vectors in the span of $a,e_n$.

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