# Eigenvalues and Eigenvectors of a block matrix

I need to find the eigenvalues and eigenvectors of the following matrix: $$D = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -i & 0 & 0 \\ 0 & 0 & i & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 \\ \end{bmatrix}$$ Now I'm sure my professor does not want me to make some sterile computations, therefore I've tried analyzing the matrix first. First of all, I noticed the matrix is Hermitian, which means all its eigenvalues must be real numbers. I've looked up the properties of Hermitian matrices but nothing useful came out. Then I noticed I can rewrite $$D$$ as a block matrix composed of $$2\times2$$ blocks: $$\begin{bmatrix} A & [0] & B \\ [0] & \sigma_2 & [0] \\ C & [0] & F \\ \end{bmatrix}$$ where $$\sigma_2$$ denotes the second Pauli matrix (I do not believe this to be a mere coincidence). The upside is that the eigenvalues of all the $$2 \times 2$$ matrices are trivial or I have already computed them (same thing for the eigenvalues). The main problem is that I've never actually worked with block matrices in any of my courses, so my personal knowledge of the topic is close to zero. I've seen something on Wikipedia about the determinant of matrices of the form: $$\begin{bmatrix} A & B \\ C & F \end{bmatrix}$$ But this does not apply to my case, so maybe the block matrix route is unlikely the right one. Is there some trick to compute the eigenvalues and eigenvectors of such a peculiar matrix that I'm most likely missing? I'm honestly at a dead end this time so any input is much appreciated.

• The matrix is unitary (each column is orthogonal to all others and has unit length). This guarantees the eigenvalues to be on the unit circle. Commented Sep 27, 2023 at 20:56
• whatever your professor wants, you will benefit by also doing basic computations Commented Sep 27, 2023 at 21:00
• @Laray Thanks Larry I didn't notice that, so the possible eigenvalues will be only $\pm 1$ due to the fact that the matrix is also self-adjoint. Commented Sep 27, 2023 at 21:16
• @WillJagy yes, that's always helpful, undoubtedly. I was thinking that maybe this matrix had some physical implications and use since it presents such a peculiar structure, but I'll go on with the basic computations as you suggest Commented Sep 27, 2023 at 21:18
• But that already gets you further - you skip the step of finding the eigenvalues and only need the eigenvectors. You spared solving $\det (D-\lambda I)=0$ :). Commented Sep 27, 2023 at 22:27

The matrix, as you note, is of the form $$\pmatrix{A&0&B\\0&\sigma_2 &0\\C&0&F}.$$ With the block permutation matrix $$P = \pmatrix{0&I_2&0\\I_2&0&0\\0&0&I_2},$$ we find that this matrix is similar to the block-diagonal matrix $$PDP^T = \pmatrix{A&B&0\\C&F&0\\0&0&\sigma_2}.$$ So, the eigenvalues of $$D$$ are equal to those of $$\sigma_2$$ (namely $$\pm 1$$) together with those of the matrix $$M = \left[\begin{smallmatrix}A&B\\C&F\end{smallmatrix}\right]$$. On the other hand, $$M$$ is the matrix $$M = \left[\begin{array}{c|cc|c} 1&0&0&0\\ \hline 0&0&1&0\\ 0&1&0&0\\ \hline 0&0&0&-1\end{array}\right].$$ As the partition lines indicate, this matrix is block-diagonal with diagonal blocks $$1$$, $$\left[\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right]$$, $$-1$$.

With that, you can conclude that the eigenvalues of $$D$$ are equal to $$1$$ and $$-1$$, both with multiplicity $$3$$.

For a more direct approach, the permutation matrix $$P = \pmatrix{ 1&0&0&0&0&0\\ 0&0&0&0&0&1\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&1&0&0&0&0}$$ is such that $$PDP^T = \pmatrix{\sigma_3 &0&0\\0&\sigma_2&0\\0&0&\sigma_1}.$$

To find the eigenvectors of $$D$$, begin by finding the eigenvectors of $$PDP^T$$. We can find these eigenvectors using the eigenvectors of the blocks $$\sigma_1,\sigma_2,\sigma_3$$.

For example, $$\sigma_2$$ has the eigenvector $$v = (1,i)$$ associated with eigenvalue $$1$$. It follows that $$PDP^T$$ has the "block-vector" $$\tilde v = \pmatrix{0_{2 \times 1}\\ v\\ 0_{2 \times 1}}$$ as an eigenvector associated with the eigenvalue 1.

Once all eigenvectors are obtained in this fashion, note that for any eigenvector $$x$$ of $$PDP^T$$, $$P^Tx$$ will be an eigenvector of $$D$$ because $$D(P^Tx) = (DP^T)x = P^T(PDP^T)x = P^T(\lambda x) = \lambda P^T x.$$ So, once the eigenvectors $$v_1,\dots,v_6$$ of $$PDP^T$$ are obtained, $$P^Tv_1,\dots,P^Tv_6$$ will be the eigenvectors of $$D$$.

• This is very helpful @Ben Grossman. I've never encountered permutation matrices: in fact, I honestly wanted to know if there was a way to safely change the structure of my original matrix into a more suitable one. Commented Sep 28, 2023 at 14:50
• @ClaudioMenchinelli Glad to help! If you're interested I can say more about how to get the eigenvectors of the original matrix, but perhaps you already see how this can be done Commented Sep 28, 2023 at 16:27
• Grossman I'll gladly accept your help in determining the eigenvectors if that isn't too much of a hassle. I will most likely come back to this thread in the future as a reference, so the possibility of having your explanations is something I can't deny myself Commented Sep 28, 2023 at 19:35
• @ClaudioMenchinelli See my edit Commented Sep 28, 2023 at 21:48
• @ClaudioMenchinelli if you want to see why this zero-padding approach works, you could use block-matrix multiplication to find the product $D\tilde v$. Commented Sep 28, 2023 at 22:20