I have a symmetric matrix like the following:$$\begin{bmatrix}a&a&a&a\\a&b&b&b\\a&b&b&b\\a&b&b&b\end{bmatrix}$$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 $\mathbf{1}$ 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}. $\quad\square$
No comments:
Post a Comment