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I have a symmetric matrix like the following:$$beginbmatrixa&a&a&a\a&b&b&b\a&b&b&b\a&b&b&bendbmatrix$$It's a symmetric real matrix with only 3 unique eigenvalues. Given it's highly symmetric nature, I was wondering how much the eigenvalues would change if I add another row and column keeping it's symmetric property intact. Specifically adding $[a, b, b, b, b]$ as a column and a row at the end.

Is there any bound for the change in eigenvalues given these sort of highly symmetric matrices?

Let $mathbf1$ denote that all-ones column vector of length $n$, and $I$ and $J$ the identity and all-ones matrices of order $n$ respectively. $newcommandonemathbf 1$

Theorem

Let $M$ be the $(n + 1) times (n + 1)$ matrix of the form
$beginbmatrixa & aone^T\ aone & bJendbmatrix$, where $a ne 0$ and $b$ are distinct real numbers. Then the eigenvalues of $M$ are:

1. 
   $0$ with multiplicity $n - 1$.


2. The two roots of the equation $lambda^2 - (a + nb)lambda - na(a - b) = 0$, each with multiplicity $1$.

Proof. Since $M$ is symmetric and has rank $2$, i.e., nullity $n - 1$, it has $0$ as an eigenvalue with multiplicity $n - 1$.

Now, let $lambda$ be a root of
beginalign
lambda^2 - (a + nb)lambda - na(a - b) = 0 tag1labeleq:lambda
endalign
and define the vector $x = beginbmatrixlambda - nb \ a oneendbmatrix$ of length $n + 1$. Then

beginalign*
Mx & = beginbmatrix
(lambda - nb)a + a^2 one^T one\
(lambda - nb)a one + ab J one
endbmatrix\
&= beginbmatrix
lambda a + na(a - b)\
lambda a one
endbmatrix
endalign*
where the last step follows from $one^T one = n$ and $J one = n one$. Now observe that on rearranging eqrefeq:lambda, we get $lambda(lambda - nb) = lambda a + na(a - b)$, which shows that $Mx = lambda x$. Thus, $x$ is an eigenvector of $M$ corresponding to the eigenvalue $lambda$, for each root $lambda$ of eqrefeq:lambda. $quadsquare$