# Change in eigenvalues if row and column added to highly symmetric matrix

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?

• In this case you know the eigenvalues of both matrices (i.e., the before and after matrices). So what exactly would the question be? – M. Vinay Apr 2 at 2:20
• I was looking for a generalization of the change in eigenvalues given this situation. For any dimension. – Hasan Iqbal Apr 2 at 2:25
• You know the eigenvalues of such a matrix for any dimension, right? Or is your question indeed what its eigenvalues would be? – M. Vinay Apr 2 at 2:32
• 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… – M. Vinay Apr 2 at 2:36
• Look up Cauchy expansion of a bordered matrix and that should help. – Justin Stevenson Apr 2 at 2:38

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}. $$\quad\square$$

• 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 $\lambda^2 - nb \lambda = 0$. – M. Vinay Apr 2 at 13:37
• 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. – M. Vinay Apr 2 at 13:40
• 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? – Hasan Iqbal Apr 2 at 13:59
• 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 $\begin{bmatrix}0 & a\mathbf{1}^T\\a\mathbf{1} & b(J - I)\end{bmatrix}$. – M. Vinay Apr 2 at 14:06
• @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). – M. Vinay Apr 2 at 14:25