I'm not sure how to better phrase the title of the question, because I don't know the specific name of the matrix I am after, but I want to consider matrices of the form $$ \begin{align*} \begin{pmatrix} a & b & b& b&b & \cdots & b\\ b & a & b& b&b&\cdots & b\\ \vdots\\ b&b&b&b&b&\cdots & a \end{pmatrix}, \end{align*} $$ that is, the diagonal entries are identical, and all the off diagonals are all the same. The most obvious example for when this occurs is when $a=1, b=0$ and you obtain the identity matrix. I just want to know if there is anything we can see about the eigenvalues of this matrix, or if there are any special properties about this matrix itself.

EDIT: The matrix size itself is $\mathbb{R}^{m\times n}$ where $m$ is not necessarily equal to $n$.

  • $\begingroup$ Assuming the matrix is $n\times n$, clearly $\left(a+(n-1)b, \begin{bmatrix} 1\\ \vdots \\ 1\end{bmatrix}_{n\times 1}\right)$is an eigenpair of the matrix. $\endgroup$ – Git Gud Apr 18 '14 at 10:52
  • $\begingroup$ Do you know its multiplicity? $\endgroup$ – user61038 Apr 18 '14 at 10:54
  • $\begingroup$ Did you try with $x = \alpha \begin{bmatrix} 1\\ \vdots \\1\end{bmatrix} + x'$ ? with $x'$ orthogonal to $\begin{bmatrix} 1 \\ \vdots \\1\end{bmatrix}$? $\endgroup$ – Sylvain Biehler Apr 18 '14 at 10:55
  • 2
    $\begingroup$ See Determinant of a specially structured matrix or Prove determinant of $n \times n$ matrix is $(a+(n-1)b)(a-b)^{n-1}$?. Although both questions are about determinant. The eigenvalues are evident from the answers. $\endgroup$ – user1551 Apr 18 '14 at 10:59

Assuming that the matrix you mentioned has dimension $N \times N$, then its eigenvalues are

$\lambda_{1}=\lambda_{2}= \cdots =\lambda_{N-1}=(a-b)$ and


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