0
$\begingroup$

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 $n$.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\lambda_{N}=(a+(N-1)b)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.