I was looking into how perturbation changes eigenvalues of a given symmetric matrix, and I came across lot of results regarding off-diagonal perturbation. But how does diagonal perturbation affect eigenvalues.
Suppose we have the following matrices $A$ and $D$ where $A$ is our original matrix and $D$ denotes the diagonal perturbation.
$$A = \begin{bmatrix}\frac{1}{N} & \frac{1}{N} & \ldots & \frac{1}{N}\\ \frac{1}{N} & \frac{1}{N} & \ldots & \frac{1}{N}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{1}{N} & \frac{1}{N} & \ldots & \frac{1}{N} \end{bmatrix}$$ $$D = \begin{bmatrix}x_1 & 0 & \ldots & 0\\ 0 & x_2 & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \ldots & x_{N} \end{bmatrix}$$
Here, all the $x_i$'s are identical and independent random variables sampled from a normal distribution $\mathcal N(0, \sigma^2)$.
Now, how do we find the eigenvalues of the matrix $A' = A - D$? Note that, the eigenvalues of the matrix $A$ are $1, \underbrace{0, \ldots, 0}_{N - 1 \text{ times}}$ respectively.