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

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?

• 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 $(A^3+I_2)^2=-I_2$. – user0410 Apr 7 at 8:17

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 $$\ R\begin{pmatrix}1\\0\end{pmatrix}\$$.
Hint Computing gives $$(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$$.