3
$\begingroup$

For a matrix like \begin{bmatrix} A & B \\ B & A \\ \end{bmatrix} which A and B are block matrix and are circulant, is there any simple way to find eigenvalues and eigenvectors? To be clear, following matrix is an example of my described matrix. \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \\ 4 & 3 & 2 & 1 \\ \end{bmatrix}

$\endgroup$
1
$\begingroup$

Your matrix commutes with the matrices $$P_2 = P \oplus P$$ and $$X = \begin{pmatrix} 0 & I \\ I& 0\end{pmatrix}$$ where $P$ is the cyclic permutation matrix on one subblock. Moreover, also $P_2$ and $X$ commute. Thus the three matrices can be simultaneously diagonalized. The combined eigenspaces of $P_2$ and $X$ have the form $$E_{\lambda,\pm} = \mathop{\rm span}\begin{pmatrix}v_\lambda \\ \pm v_\lambda \end{pmatrix}$$ with the eigenvalues $\lambda_j =e^{2\pi i j/n}$ and $$v_\lambda =(1,\lambda, \lambda^2, ...)^t.$$

As the eigenspaces are 1 dimensional, the eigenvectors of your matrix are given by $$\begin{pmatrix}v_{\lambda_j} \\ \pm v_{\lambda_j} \end{pmatrix}$$ and the eigenvalues are easily found.

In particular, the unitary transformation $$U=\frac{1}{\sqrt{2}} \begin{pmatrix} U_\text{DFT} & U_\text{DFT} \\ U_\text{DFT} & -U_\text{DFT}\end{pmatrix}$$ with $U_\text{DFT}$ the discrete Fourier transform brings your matrix onto diagonal form.

$\endgroup$

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.