I have a matrix $$ \begin{bmatrix} -3.5 & 4.5 \\ 4.5 & -3.5 \\ \end{bmatrix} $$
I am tasked with finding 3 cubic roots of the matrix. I was able to find one using the characteristic equation $(-3.5 - \lambda)^2 - 4.5^2 = 0$, or equivalently $ \lambda^2 + 7 \lambda - 8 = 0$
I found eigenvalues $\lambda_1 = -8, \lambda_2 = 1$, with respective eigenvectors
$$
\begin{bmatrix}
1 \\
-1 \\
\end{bmatrix}
and
\begin{bmatrix}
1 \\
1 \\
\end{bmatrix}
$$
Doing the calculation $$ \begin{bmatrix} 1 & 1 \\ -1 & 1 \\ \end{bmatrix}* \begin{bmatrix} -8 & 0 \\ 0 & 1 \\ \end{bmatrix}^{1/3} * \begin{bmatrix} 1 & 1 \\ -1 & 1 \\ \end{bmatrix}^{-1} $$ Gives me \begin{bmatrix} -.5 & 1.5 \\ 1.5 & -.5 \\ \end{bmatrix}
Which is a correct cubic root, but I am not certain how to get two other cubic roots. I know previously I have taken a $\pm \lambda$ due to the factor $(-\lambda)^2 = \lambda^2$, but those seemed to only work in cases where the power is even. With a cubic root it seems as though the eigenvalue matrix can't be along the lines of \begin{bmatrix} \pm(-8)^{1/3} & 0 \\ 0 & \pm 1 \\ \end{bmatrix} because the negative sign will stick around due to the odd power.