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October 6th, 2014, 12:40 AM  #1 
Newbie Joined: Oct 2014 From: Slovenija Posts: 1 Thanks: 0  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. 
October 6th, 2014, 12:51 AM  #2 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,887 Thanks: 765 Math Focus: Wibbly wobbly timeywimey stuff. 
You could take up nursing... Dan 

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