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October 6th, 2014, 12:40 AM   #1
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Diagonalization of a big scary matrix

I would need to diagonalize this tridiagonal block matrix $M$:

$$M = \begin{bmatrix}
A & B & & \\
B^T & A & B & \\
& B^T & A & B \\
& & \ddots & \ddots & \ddots \\
& & & B^T & A & B \\
& & & & B^T & A
\end{bmatrix}_{n \times n}$$

where matrices $A$ and $B$ are also $n \times n$ tridiagonal:

$$A = \begin{bmatrix}
C & D & & \\
D & C & D & \\
& D & C & D \\
& & \ddots & \ddots & \ddots \\
& & & D & C & D \\
& & & & D & C
\end{bmatrix}_{n \times n}$$

$$B = \begin{bmatrix}
E & F & & \\
G & E & F & \\
& G & E & F \\
& & \ddots & \ddots & \ddots \\
& & & G & E & F \\
& & & & G & E
\end{bmatrix}_{n \times n}$$

where:

$$C = \begin{bmatrix}
8 & 0\\
0 & 8
\end{bmatrix}
\quad
D = \begin{bmatrix}
-2 & 0\\
0 & 0
\end{bmatrix}
\quad
E = \begin{bmatrix}
0 & 0\\
0 & -2
\end{bmatrix}
\quad
F = \begin{bmatrix}
-1 & 1\\
1 & -1
\end{bmatrix}
\quad
G = \begin{bmatrix}
-1 & -1\\
-1 & -1
\end{bmatrix}
$$

So essentially $M$ is $2n^2 \times 2n^2$ large.

I would need to do this, because I need to solve this system of differential equations $\ddot{\vec{x}} = -M \vec{x}$. When I set $\vec{x} = \vec{u} e^{i \omega t}$, I got the problem of eigenvalues $M \vec{u} = \omega^2 \vec{u}$, where $\omega^2$ are the eigenvalues and $\vec{u}$ are the eigenvectors.

It would be great, if this monstrosity could be solved analytically for an arbitrary $n$.

Last edited by ulrichthegreat; October 6th, 2014 at 12:58 AM.
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October 6th, 2014, 12:51 AM   #2
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