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I want to find a decomposition of this matrix in the form of $$ A^{n_1} \cdot B \cdot A^{n_2} \cdot B \cdots B \cdot A^{n_k} $$ where $$ A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} $$ I am aware that $A^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$, but fail at using the Euclidean algorithm as in Example 2.1 to find a decomposition.

I tried using results from here, but that didn´t help me, either.

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I prefer to multiply on the left at each stage, the final "answer," well, you work it out

$$ B \mapsto \left( \begin{array}{cc} 9&-7 \\ -5&4 \\ \end{array} \right) $$ $$ A^2 \mapsto \left( \begin{array}{cc} -1&1 \\ -5&4 \\ \end{array} \right) $$ $$ B \mapsto \left( \begin{array}{cc} -5&4 \\ 1&-1 \\ \end{array} \right) $$ $$ A^{5} \mapsto \left( \begin{array}{cc} 0&-1 \\ 1&-1 \\ \end{array} \right) $$ $$ B \mapsto \left( \begin{array}{cc} 1&-1 \\ 0&1 \\ \end{array} \right) $$ $$ A \mapsto \left( \begin{array}{cc} 1&0 \\ 0&1 \\ \end{array} \right) $$

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  • $\begingroup$ So the answer ist $B^{-1} A^{-2} B^{-1} A^{-5} B^{-1} A^{-1}$. $\endgroup$ – Viktor Glombik May 20 '18 at 16:02

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