Find a matrix $B$ such that $B^3 = A$ 
$$A=\begin{pmatrix} 1 & -1 \\ -2 & 1 \end{pmatrix}$$
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 ? 
 A: 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 $$B= \left(
\begin{array}{cc}
\frac 12\sqrt[3]{\left( 1+\sqrt{2}\right) }-\frac 12\sqrt[3]{\left( -1+\sqrt{%
2}\right) } & -\frac 14\sqrt{2}\sqrt[3]{\left( 1+\sqrt{2}\right) }-\frac 14%
\sqrt{2}\sqrt[3]{\left( -1+\sqrt{2}\right) } \\
-\frac 12\sqrt{2}\sqrt[3]{\left( 1+\sqrt{2}\right) }-\frac 12\sqrt{2}\sqrt[3%
]{\left( -1+\sqrt{2}\right) } & \frac 12\sqrt[3]{\left( 1+\sqrt{2}\right) }%
-\frac 12\sqrt[3]{\left( -1+\sqrt{2}\right) }
\end{array}
\right) $$
A: Suppose that there is a matrix $B=\begin{pmatrix} b_1 & b_3 \\ b_2 & b_4\end{pmatrix}$ such that $B^3=A$. Then we obtain two linear equations by applying the Buchberger algorithm to the four equations, namely
$$
b_2=2b_3,\; b_4=b_1.
$$
This yields $b_1= - 4b_3^2 - 2b_3 + 1$, and all equations are satisfied if and only if $b_3$ satisfies 
$$
8b_3^3 - 3b_3 + 1=0.
$$
We obtain exactly one real solution for $b_3$, and hence for $B$ with $B^3=A$.
A: You're almost done.
Take $B = PD^{1/3}P^{-1}$ where $D^{1/3}=\operatorname{diag}(\lambda_1^{1/3},\lambda_2^{1/3})$ and try to compute $B^3$.
A: By the Cayley-Hamilton theorem, any analytic function $f$ of an $n\times n$ matrix can be expressed as a polynomial $p(A)$ of degree at most $n-1$. So, a cube root $B$ of $A$ can be expressed in the form $aI+bA$ for some to-be-determined coefficients $a$ and $b$. Now, if $\lambda$ is an eigenvalue of $A$, then we also have that $f(\lambda)=p(\lambda)$. For this problem, this second property generates a system of two linear equations in the unknown coefficients $a$ and $b$. Solve this system.
