A matrix without cube root

On the final exam of my linear algebra class, the instructor asked us to compute the cube root of matrix $$A = \begin{pmatrix}3 &2 & 0 \\ -4 &-3 &0\\ 0 &0& 8 \end{pmatrix}$$ I calculated that this matrix does have 3 eigenvalues 8, 1 and -1. However, the eigenvectors associated with eigenvalue 8 is a zero vector which means this matrix has less than 3 eigenvectors. So I concluded: this matrix is not diagonalizable. As a result, I can't use A = PDPinverse to calculate the cube root. Based on this fact, can I say this matrix doesn't have cube root?

• By definition of eigenvector, if $8$ is an eigenvalue, then there exists a nonzero vector $v$ such that $Av = 8v$. – Andrew Maurer Dec 6 '12 at 19:30

I checked over your work and got the same characteristic polynomial $$h(x) = (x-8)(x-1)(x+1)$$
When I looked to find the eigenspace of $8$, I looked at $ker(A-8I)$, which is $$ker\begin{pmatrix} -5 & 2 & 0 \\ -4 & -11 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ which is clearly spanned by $(0,0,1)^t$.
In fact $\dfrac{2}{21} I + A - \dfrac{2}{21} A^2$ is a cube root of $A$. This is because the polynomial $f(x) = \dfrac{2}{21} + x - \dfrac{2}{21} x^2$ satisfies $f(1) = 1 = \sqrt[3]{1}$, $f(-1) = -1 = \sqrt[3]{-1}$, $f(8) = 2 = \sqrt[3]{8}$.