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Is there an analytical way or a good approximation or any other mathematical method to diagonalise a sparse (symmetric) matrix with elements only onsome diagonals?

For example $$ \begin{bmatrix} B & 0 & 0 & A & 0\\ 0 & B & 0 & 0 & A\\ 0 & 0 & B & 0 & 0\\ A & 0 & 0 & B & 0\\ 0 & A & 0 & 0 & B \\ \end{bmatrix} $$

or similar...

(is there an index notation way of writing the above matrix? Like $A_{m,n} = \cdots$?)

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  • $\begingroup$ I know there are fast solvers for this especially if the matrix is diagonally dominant. $\endgroup$ – lightxbulb Jan 18 at 0:02
  • $\begingroup$ The diagonal is always roughly a factor of $2$ larger than any off diagonal. Have you got any names for these fast solvers? $\endgroup$ – SuperCiocia Jan 18 at 0:03
  • $\begingroup$ Is $A$ a submatrix or a number? $\endgroup$ – Omnomnomnom Jan 18 at 0:12
  • $\begingroup$ Assuming that $A$ and $B$ are square submatrices of (identical) size $n$, we can write your matrix as $$ M = I_5 \otimes B + \pmatrix{0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\\1&0&0&0&0\\0&1&0&0&0} \otimes A $$ where $\otimes$ denotes the Kronecker product $\endgroup$ – Omnomnomnom Jan 18 at 0:15
  • $\begingroup$ If $A$ and $B$ are numbers, it's fairly easy to diagonalize this analytically. $\endgroup$ – Omnomnomnom Jan 18 at 0:17
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By using $Xv_i= \lambda_i v_i$ you can derive the eigenvectors: $(\frac{1}{\sqrt{2}},0,0,\pm\frac{1}{\sqrt{2}},0)$, $(0,0,1,0,0)$, $(0, \frac{1}{\sqrt{2}},0,0,\pm\frac{1}{\sqrt{2}})$, with corresponding eigenvalues $\lambda =(B\pm A, B, B\pm A)$. Let $Q$'s columns be made up of the eigenvectors (in the given order), then: $X= Qdiag(\lambda)Q^T$, where $X$ is your initial matrix. Note that this can be trivially extended to higher dimensions for a matrix with the same structure.

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  • $\begingroup$ I will confirm this as answered, just out of curisioty did you just brute force try for eigenstates or did you use any tricks/known methods? $\endgroup$ – SuperCiocia Jan 18 at 11:01
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    $\begingroup$ @SuperCiocia What do you mean bruteforce? I just wrote the equations $Xv_i = \lambda_i v_i$ and noticed the structure (3rd component doesn't affect the others, 1 and 4 affect each other, and 2 and 5 also). $\endgroup$ – lightxbulb Jan 18 at 11:08

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