Cfrgtkky

$$A=begin{pmatrix} 1 & -1 \ -2 & 1 end{pmatrix}$$

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 ?

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 $$B= left(
begin{array}{cc}
frac 12sqrt[3]{left( 1+sqrt{2}right) }-frac 12sqrt[3]{left( -1+sqrt{%
2}right) } & -frac 14sqrt{2}sqrt[3]{left( 1+sqrt{2}right) }-frac 14%
sqrt{2}sqrt[3]{left( -1+sqrt{2}right) } \
-frac 12sqrt{2}sqrt[3]{left( 1+sqrt{2}right) }-frac 12sqrt{2}sqrt[3%
]{left( -1+sqrt{2}right) } & frac 12sqrt[3]{left( 1+sqrt{2}right) }%
-frac 12sqrt[3]{left( -1+sqrt{2}right) }
end{array}
right) $$

Suppose that there is a matrix $B=begin{pmatrix} b_1 & b_3 \ b_2 & b_4end{pmatrix}$ 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 $B^3=A$.

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 $B^3$.

By the Cayley-Hamilton theorem, any analytic function $f$ of an $ntimes n$ 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.