Eigenvalues and eigenvectors of a symmetric matrix whose rows and columns sum to $1$

Question

Given an $n \times n$ symmetric matrix $A$, where

$$\sum_{i=1}^{n} A(i,j) = 1,\qquad j=1,2,\dots,n$$

and

$$\sum_{j=1}^{n} A(i,j) = 1,\qquad i=1,2,\dots,n$$

is there any property of its eigenvalues and eigenvectors? Or is there any method to get its eigenvalues and eigenvectors? Or does anyone know what to call this kind of matrices?

Answer 1

We have that for $x=(1,1,...,1)$

$$Ax=\begin{bmatrix}\sum_j A_{1j} \\ \sum_j A_{2j} \\ \vdots \\ \sum_j A_{nj}\end{bmatrix} =x$$

thus for this kind of matrix, $x$ is an eigenvector with eigenvalue $1$.

Answer 2

As noted, 1 is an eigenvalue, and since the matrix is assumed symmetric, all eigenvalues are real. Beyond that, maybe not much more can be said. In the $2\times2$ case, your matrix is of the form $$\pmatrix{a&1-a\cr1-a&a\cr}$$ and its eigenvalues are 1 and $2a-1$.

Answer 3

Suppose that $$ \mathbf{x} = \left[ \begin{matrix} x_1 \\ \vdots \\ x_n \end{matrix} \right] $$ is an eigenvector of this matrix corresponding to the eigenvalue $\lambda$. Then we have the equation $$ \mathbf{A} \mathbf{x} = \lambda \mathbf{x}, $$ which is equivalent to the following system of equations $$ \begin{align} \sum_{j=1}^n a_{1j} x_j &= \lambda x_1, \\ \cdots &= \cdots, \\ \sum_{j=1}^n a_{nj} x_j &= \lambda x_n. \end{align} $$ And this system has a non-trivial solution if and only if $$ \det ( \mathbf{A} - \lambda \mathbf{I} ) = 0. $$ So far there is nothing new.

Now we can come up with the following:

Step 1.

Adding up the remaining columns of $\mathbf{A} - \lambda \mathbf{I}$ to the first we end up with a new matrix that has the entry $1-\lambda$ throughout the first column, but the new matrix has the same determinant as $\det ( \mathbf{A} - \lambda \mathbf{I} )$. So $$ \det \left[ \begin{matrix} 1-\lambda \ & a_{12} \ & a_{13} \ & a_{14} \ & \ldots \ & a_{1n} \\ 1- \lambda \ & a_{22} - \lambda \ & a_{23} \ & a_{24} \ & \ldots \ & a_{2n} \\ 1- \lambda \ & a_{32} \ & a_{33} - \lambda \ & a_{34} \ & \ldots \ & a_{3n} \\ a_{41} \ & a_{42} \ & a_{43} \ & a_{44} - \lambda \ & \ldots \ & a_{4n} \\ \vdots \ & \vdots \ & \vdots \ & \vdots \ & \vdots \ & \vdots \\ 1- \lambda \ & a_{n2} \ & a_{n3} \ & a_{n4} \ & \ldots \ & a_{nn} - \lambda \end{matrix} \right] = 0. $$

Step 2.

Now subtracting the first row from each of the remaining ones and then expanding the determinant of the resulting matrix, we obtain $$ \det \left[ \begin{matrix} a_{22} - a_{12} - \lambda \ & a_{23} - a_{13} \ & a_{24} - a_{14} & \ldots \ & a_{2n} - a_{1n} \\ a_{32} - a_{12} \ & a_{33} - a_{13} - \lambda \ & a_{34} - a_{14} & \ldots \ & a_{3n} - a_{1n} \\ a_{42} - a_{12} \ & a_{43} - a_{13} \ & a_{44} - a_{14} - \lambda \ & \cdots \ & a_{4n} - a_{1n} \\ \vdots \ & \vdots \ & \vdots \ & \vdots \ & \vdots \\ a_{n2} - a_{12} \ & a_{n3} - a_{13} \ & a_{n4} - a_{14} \ & \ldots \ & a_{nn} - a_{1n} - \lambda \end{matrix} \right] = 0. $$

Now repeat Steps 1 and 2 with this $(n-1) \times (n-1)$ matrix.

Hope this helps.

Answer 4

One special kind of these matrices is where all rows and collumns have the same figures but permuted. And a special kind of those is symmetric circulant matrices, https://en.wikipedia.org/wiki/Circulant_matrix, which can be subclassed to be of the kind in this question.