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Find a $2 \times 2$ matrix $A$ such that $$A^3 = \begin{pmatrix} -1 & -1 \\ 1 & -1 \\ \end{pmatrix}$$

I'm trying to think about this geometrically. Could this have something to do with a rotation dilation?

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  • $\begingroup$ Hint: We get $A^3+I_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0\\ \end{pmatrix}$. Moreover, we have $\begin{pmatrix} 0 & -1 \\ 1 & 0\\ \end{pmatrix}^2=-I_2$ which implies that $(A^3+I_2)^2=-I_2$. $\endgroup$ – user0410 Apr 7 at 8:17
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There are several ways of tackling this problem, a nice one of which you have already thought of. Since $\ \begin{pmatrix}-1&-1\\1&-1\end{pmatrix}\ $ is indeed a dilated rotation matrix, namely $\ \sqrt{2}\begin{pmatrix}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\end{pmatrix}\ $, where $\ R=\begin{pmatrix}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\end{pmatrix}\ $ is a rotation matrix, then the matrix $\ 2^\frac{1}{6} S\ $ will be an answer for your question if $\ S\ $ is a rotation matrix which rotates vectors through one third of the angle that $\ R\ $ does. The angle through which $\ R\ $ rotates vectors can be found as the angle between the unit vectors $\ \begin{pmatrix}1\\0\end{pmatrix}\ $ and $\ R\begin{pmatrix}1\\0\end{pmatrix}\ $.

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Hint Computing gives $$(A^3)^\top A^3 = 4 I ,$$ so $\lambda A^3$ is an orthogonal matrix for an appropriate scalar $\lambda$. This suggests looking for a solution $A = r S$ which is again a scalar multiple of some orthogonal matrix $S$.

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