Eigenvalues of $aI + bJ$

Question

Just as context, this problem arose when I was looking at the adjacency matrices of moore graphs of diameter 2.

Given that $I$ is the identity matrix and $J$ is the all 1 matrix. I have constructed some matrix $aI + bJ$. I am completely stuck in terms of how to find the eigenvalues of this matrix. I have considered trying to find the characteristic polynomial, but I have no idea how I would find the determinant of such a matrix.

Answer 1

Write $A = aI+ bJ$. Note that $A-aI = bJ$ has rank 1, so the eigenvalue $a$ has geometric multiplicity $n-1$.

So all that’s left to do is find the other eigenvalue. Trying $\lambda=a+nb$ gives that $A-\lambda I$ has diagonal entries $(1-n)b$ and all other entries $b$. In particular, the sum of all the rows is the zero vector, so this matrix is singular.

Edit: I should note it is assumed here that $b\ne 0$. If $b=0$, then $A-aI=bJ$ is the zero matrix, which has rank 0, so $A$ has eigenvalue $a$ with multiplicity $n$.

Answer 2

here is a matrix that shows a basis of the eigenvectors as columns, these being perpendicular to each other as well. Below is the 10 by 10 case. In smaller dimension, take the upper left square corner.

$$ \left( \begin{array}{rrrrrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9 \end{array} \right). $$