# Eigenvalues of block matrices related

Let,

$C= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & ... & 1 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & ... & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & ... & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & ... & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & ... & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & ... & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ \end{bmatrix}_{n \times n}$

What are the eigenvalues of matrix $C$. Also how to find eigenvalues of following block matrix.

$A=\begin{bmatrix} B & C \\ C^T & O \end{bmatrix}$

Where $B$ is a square matrix of order $n$ and $O$ is zero matrix of order $n$.

I know that one eigenvalue of matrix $C$ is $2$ as addition of entries in each row is $2$.But how to find other eigenvalues?

Also for block matrix,if we can convert the matrix in terms of knonecker product or direct-sum of two matrices then we can find the eigenvalues easily.But I don't know how to do it.

• I'm sorry, but the bottom of the matrix doen't seem to be in coherence with the top. Could you describe precisely for which values of $i$ and $j$ we have $a_{ij}=1$? Commented Apr 8, 2016 at 9:47
• Except 1st row in all other rows it is in consecutively placed till last rows as shown in matrix Commented Apr 8, 2016 at 9:51
• Consecutive 1s is not possible with this single rule. It can't be a square matrix. Commented Apr 8, 2016 at 9:58
• The rule mentioned in your last comment is not possible for $i=n$. Commented Apr 8, 2016 at 10:00
• I think now its ok Commented Apr 8, 2016 at 10:04

Expand $\det(C-\lambda I)$. You should find $$\chi_C(\lambda)=(1-\lambda)^n+(-1)^{n-1}.$$
$C$ is a circulant matrix, with $c_k=\delta_{0k}+\delta_{1k}$.
Its eigenvalues are $$\lambda_j = 1+\mathrm{e}^{2\pi i \frac{n-1}{n}j}$$