Is there an explicit decomposition of the form $A = \lambda_1v_1v_1^T +\lambda_2v_2v_2^T$ for the matrix
$$A=\begin{bmatrix} 0 & 0 & 0 & \dots & x_{1} \\ 0 & 0 & 0 & \dots & x_{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{1} & x_{2} & x_{3} & \dots & x_{n} \end{bmatrix}$$
That is, the symmetric rank-$2$ matrix consisting of identical row and column, and the rest of the entries are zero.
I guess that maybe OP can refer to ``Eigenvalues of a matrix with only one non-zero row and non-zero column.''
Alternatively, we can directly compute the determinant of $|\lambda I - A|$. we can rewrite the matrix $\lambda I - A$ as block matrix: \begin{equation} \lambda I - A = \begin{bmatrix} \lambda & 0 & 0 & \ldots & x_{1} \\ 0 & \lambda & 0 & \ldots & x_{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{1} & x_{2} & x_{3} & \ldots & \lambda-x_{n}\end{bmatrix} = \begin{bmatrix} D & u \\ v & \lambda-x_n \end{bmatrix} \end{equation} where $v=u^T = (x_1,x_2,\dots,x_{n-1})$, and $D=\operatorname{diag}(\lambda, \dots, \lambda)$. And the determinant is \begin{equation} \begin{aligned} \operatorname{det}(\lambda I - A) &= \operatorname{det}(D)\operatorname{det}(\lambda-x_n-vD^{-1}u) \\ &= \lambda^{n-1}(\lambda-x_n-\lambda^{-1}\sum_{i=1}^{n-1}x_i^2) \end{aligned} \end{equation} Let the determinant equal to zero, we can compute two non-zero eigenvalues: \begin{equation} \lambda = \frac{x_n \pm \sqrt{x_n^2+4\sum_{i=1}^{n-1}x_i^2}}{2} \end{equation} The eigenvectors are (before normalization) \begin{equation} v_1 = \begin{bmatrix} \frac{u}{\lambda_1} \\ 1 \end{bmatrix}, \quad v_2 = \begin{bmatrix} \frac{u}{\lambda_2} \\ 1 \end{bmatrix} \end{equation}
