Block diagonalization of a symmetric 4$\times$4 matrix.

Question

I have a symmetric 4$\times$4 matrix, A. I need to find it's eigenvalues. The elements are not purely numerical: some elements are exponential functions and so it's extremely difficult to find the eigenvalues using the typical characteristic equation approach.

(Will Jagy): Here is a 4 by 4 matrix. Edit in the actual entries (Thank you, Will!)

The 4 by 4 matrix I'm trying to block diagonalise looks like this: $$ A = \begin{bmatrix}e^{2x+y} & 1 & 1 & e^{-2x+y} \\ 1 & e^{2x-y} & e^{-2x-y} & 1 \\ 1 & e^{-2x-y} & e^{2x-y} & 1 \\ e^{-2x+y} & 1 & 1 & e^{2x+y} \\ \end{bmatrix} $$ It's quite nasty to deal with: directly calculating the eigenvalues through computer algebra software gives a very nasty answer which doesn't lend well to any analytic uses.

I also have a unitary matrix S, which looks like the following: $$ S = \begin{bmatrix}0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{bmatrix} $$

Clearly, SA = AS. I know that somehow, I need to block diagonalise A in order to proceed.

How is this kind of block diagonalization typically performed?

Thanks for reading!

Edit: matrices added!

Answer 1

Take $p = e^{2x}$ and $q = e^y.$ Your matrix now has some fractions, so we multiply through by $pq,$ getting $$ pqA = \left( \begin{array}{cccc} p^2 q^2 & pq & pq & q^2 \\ pq & p^2 & 1 & pq \\ pq & 1 & p^2 & pq \\ q^2 & pq & pq & p^2 q^2 \\ \end{array} \right) $$

The characteristic polynomial of this matrix, variable called $t,$ is $$ \left( t - (p^2 -1) \right) \left(t - q^2 (p^2 -1)\right) \left( t^2 - (p^2 + 1)(q^2 + 1) t + q^2 (p^2 - 1)^2 \right) $$ We need the quadratic formula to get two of the eigenvalues.

One quick way is to notice that adding $(1 - p^2)I$ to the matrix makes the two middle rows equal, same as subtracting $(p^2 - 1)I,$ so $p^2 - 1$ is an eigenvalue. Analogous, adding $(q^2 - p^2 q^2)I$ to the matrix makes the first and fourth rows identical, so $p^2 q^2 - q^2= q^2(p^2-1)$ is an eigenvalue.

Answer 2

Thanks to my dissertation supervisor's much welcomed advice, I've found a really nice way to calculate the eigenvalues of the matrix. Block diagonalization is the way after all! The technique itself is super cool, so I wanted to share it.

The first step is to calculate the eigenvectors of S and use them to form the following matrix: $$ \textbf{T} = \begin{bmatrix}-1 & 0 & 0 & 1 \\ 0 & -1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ \end{bmatrix} $$

Calculating the inverse of T provides the following: $$ \textbf{T}^{-1} = \begin{bmatrix}-\frac{1}{2} & 0 & 0 & \frac{1}{2} \\ 0 & -\frac{1}{2} & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & 0 & \frac{1}{2} \\ \end{bmatrix} $$

By the properties of matrix similarity, it follows that $\textbf{T}^{-1 }\textbf{A}\textbf{T}$ will have the same eigenvalues as just A. A fantastic coincidence is that the result of $\textbf{T}^{-1 }\textbf{A}\textbf{T}$ is also a $4\times4$ block-diagonalisable matrix, which means that calculating the eigenvalues is as simple as solving the simpler characteristic equations of the $2\times2$ matrices compromising the blocks. $$ \textbf{T}^{-1 }\textbf{A}\textbf{T} = \begin{bmatrix} e^{2x+y} - e^{-2x+y} & 0 & 0 & 0 \\ 0 & e^{2x-y} - e^{-2x-y} & 0 & 0 \\ 0 & 0 & e^{2x-y} - e^{-2x-y} & 2 \\ 0 & 0 & 2 & e^{2x+y} + e^{-2x+y} \\ \end{bmatrix} $$

The eigenvalues are now a lot easier to calculate.