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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?

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    $\begingroup$ In this case you know the eigenvalues of both matrices (i.e., the before and after matrices). So what exactly would the question be? $\endgroup$ – M. Vinay Apr 2 at 2:20
  • $\begingroup$ I was looking for a generalization of the change in eigenvalues given this situation. For any dimension. $\endgroup$ – Hasan Iqbal Apr 2 at 2:25
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    $\begingroup$ You know the eigenvalues of such a matrix for any dimension, right? Or is your question indeed what its eigenvalues would be? $\endgroup$ – M. Vinay Apr 2 at 2:32
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    $\begingroup$ Hm. I think it's easier to compute it explicitly (for such a matrix of any dimension), but let's see if we can avoid that, if you really want to… $\endgroup$ – M. Vinay Apr 2 at 2:36
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    $\begingroup$ Look up Cauchy expansion of a bordered matrix and that should help. $\endgroup$ – Justin Stevenson Apr 2 at 2:38
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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:

  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 \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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    $\begingroup$ If $a = b$, then the matrix is $M = aJ_{n + 1}$, whose eigenvalues are $0$ with multiplicity $n$ and $(n + 1)a$ with multiplicity $1$. This can also be seen from the above, since the quadratic equation would then be $\lambda^2 - a(n + 1)\lambda = 0$. In the other special case, where $a = 0$ (but $b \ne a$), we have $M = 0 \oplus bJ_n$ (direct sum), whose eigenvalues are $0$ with multiplicity $n$ and $nb$ with multiplicity $1$. This also follows from the quadratic, which becomes $\lambda^2 - nb \lambda = 0$. $\endgroup$ – M. Vinay Apr 2 at 13:37
  • $\begingroup$ The reason for excluding these two special cases in the general statement is that the statement "$0$ is an eigenvalue with multiplicity $n - 1$" will not be true if they are included, and it would make some awkward rephrasing necessary. $\endgroup$ – M. Vinay Apr 2 at 13:40
  • $\begingroup$ Hi @M. Vinay, thank you so much for your answer. Could I please ask one more question? What would happen if the principal diagonal of this matrix is all zeros? Would this theorem still apply? $\endgroup$ – Hasan Iqbal Apr 2 at 13:59
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    $\begingroup$ This result certainly will not apply in that case. But a similar analysis could be done there as well. The matrix there could be written as $A + B$, where $A$ is the matrix consisting all the $a$-entries and zeroes; $B$ is the one consisting of all the $b$-entries and zeroes (i.e., $0 \oplus bJ_n$). I think they'll have some common eigenvectors, so those eigenvalues would get added up to give the eigenvalues of $M$. But this is probably not true for all eigenvalues. Note that the matrix structure is $\begin{bmatrix}0 & a\mathbf{1}^T\\a\mathbf{1} & b(J - I)\end{bmatrix}$. $\endgroup$ – M. Vinay Apr 2 at 14:06
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    $\begingroup$ @HasanIqbal No, that's it. I'm just pointing out the structure of the modified matrix. Using that the eigenvalues can be obtained in a very similar way. That is, you assume that $\begin{bmatrix}x\\ y\end{bmatrix}$ is an eigenvector, use block-matrix multiplication to get two equations, and solve (looking at some special case solutions usually helps you guess what the eigenvector should be like). $\endgroup$ – M. Vinay Apr 2 at 14:25

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