$$A=\begin{pmatrix} 1 & -1 \\ -2 & 1 \end{pmatrix}$$
Find a matrix $B$ such that $B^3$ = A
My attempt:
I found $\lambda_1= 1+{\sqrt 2}$ and $\lambda_2= 1-{\sqrt 2}$
I also found their corresponding eigenvectors $\vec v_1 =\begin{pmatrix} \frac{-\sqrt 2}{2} \\ 1 \end{pmatrix}$ and $\vec v_2 = \begin{pmatrix} \frac{\sqrt 2}{2} \\ 1 \end{pmatrix}$
I know the Power function of a matrix formula $A=PDP^{-1}$
Because it's the cubed root I'm looking for I don't know how to get the cubed root of the eigenvaules and keep the maths neat. Is there another way to solve this problem or an I going the wrong way about doing it ?
Best Answer
you find a matrix $P=\left( \begin{array}{cc} \frac{-\sqrt{2}}2 & \frac{\sqrt{2}}2 \\ 1 & 1 \end{array} \right) $ diagonalizes $ A $ that is $P^{-1}AP=\left(\begin{array}{cc} 1+\sqrt{2} & 0 \\ 0 & 1-\sqrt{2} \end{array}\right) $, then we look for a simple matrix $ M =PBP^{-1} $ such that it is diagonal and $ M^3 = D$ an simple solution is $M=\left( \begin{array}{cc} \sqrt[3]{\left( 1+\sqrt{2}\right) } & 0 \\ 0 & -\sqrt[3]{\left( -1+\sqrt{2}\right) } \end{array} \right) $ and so $B=PDP^{-1}$ is a solution. Precisely $$B= \left( \begin{array}{cc} \frac 12\sqrt[3]{\left( 1+\sqrt{2}\right) }-\frac 12\sqrt[3]{\left( -1+\sqrt{% 2}\right) } & -\frac 14\sqrt{2}\sqrt[3]{\left( 1+\sqrt{2}\right) }-\frac 14% \sqrt{2}\sqrt[3]{\left( -1+\sqrt{2}\right) } \\ -\frac 12\sqrt{2}\sqrt[3]{\left( 1+\sqrt{2}\right) }-\frac 12\sqrt{2}\sqrt[3% ]{\left( -1+\sqrt{2}\right) } & \frac 12\sqrt[3]{\left( 1+\sqrt{2}\right) }% -\frac 12\sqrt[3]{\left( -1+\sqrt{2}\right) } \end{array} \right) $$