I have a symmetric matrix like the following:[aaaaabbbabbbabbb]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?
Answer
Let 1 denote that all-ones column vector of length n, and I and J the identity and all-ones matrices of order n respectively. \newcommand{\one}{\mathbf 1}
Theorem
Let M be the (n + 1) \times (n + 1) matrix of the form
\begin{bmatrix}a & a\one^T\\ a\one & bJ\end{bmatrix}, where a \ne 0 and b are distinct real numbers. Then the eigenvalues of M are:
- 0 with multiplicity n - 1.
- 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
\begin{align} \lambda^2 - (a + nb)\lambda - na(a - b) = 0 \tag{1}\label{eq:lambda} \end{align}
and define the vector x = \begin{bmatrix}\lambda - nb \\ a \one\end{bmatrix} of length n + 1. Then
\begin{align*} Mx & = \begin{bmatrix} (\lambda - nb)a + a^2 \one^T \one\\ (\lambda - nb)a \one + ab J \one \end{bmatrix}\\ &= \begin{bmatrix} \lambda a + na(a - b)\\ \lambda a \one \end{bmatrix} \end{align*}
where the last step follows from \one^T \one = n and J \one = n \one. Now observe that on rearranging \eqref{eq: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 \eqref{eq:lambda}. \quad\square
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