Find a matrix $X$ given $X^4$ Find the matrix $X$ such that
$$X^4=\begin{bmatrix}
3&0&0\\
0&3&1\\
0&0&0
\end{bmatrix}$$
This problem I can't work,and I think let the matrix the eigenvalue is $\lambda$,then
$\lambda^4$ is $$\begin{bmatrix}
3&0&0\\
0&3&1\\
0&0&0
\end{bmatrix}$$
eigenvalue?Thank you for your help.
 A: Hint: Write the Jordan Normal Form., then take:
$$X = S.J^{1/4}.S^{-1}$$
So, we have:
$$\begin{bmatrix} 3&0&0\\ 0&3&1\\ 0&0&0 \end{bmatrix}$$
We can write this in Jordan Normal Form as:
$$S.J.S^{-1} = \begin{bmatrix} 0&0&1\\ -1&1&0\\ 3&0&0 \end{bmatrix}.\begin{bmatrix} 0&0&0\\ 0&3&0\\ 0&0&3 \end{bmatrix}.\begin{bmatrix} 0&0&\dfrac{1}{3}\\ 0&1&\dfrac{1}{3}\\ 1&0&0 \end{bmatrix}$$
Now, just take the fourth roots of the diagonal entries of $J$ and then multiply out with the other two matrices.
Spoiler - Do Not Peek (do the work and then look)

 $X = S.J^{1/4}.S^{-1} = \begin{bmatrix} 0&0&1\\ -1&1&0\\ 3&0&0 \end{bmatrix}.\begin{bmatrix} 0&0&0\\ 0&3^{1/4}&0\\ 0&0&3^{1/4} \end{bmatrix}.\begin{bmatrix} 0&0&\dfrac{1}{3}\\ 0&1&\dfrac{1}{3}\\ 1&0&0 \end{bmatrix}=\begin{bmatrix} 3^{1/4}&0&0\\ 0&3^{1/4}&\dfrac{1}{3^{3/4}}\\ 0&0&0 \end{bmatrix}$

A: You can look for a solution of the same form as $X$: calculating $$\begin{pmatrix} a & 0 & 0 \\ 0 & a & b \\ 0 & 0 & 0\end{pmatrix} \begin{pmatrix} a & 0 & 0 \\ 0 & a & b \\ 0 & 0 & 0\end{pmatrix} = a \cdot \begin{pmatrix} a & 0 & 0 \\ 0 & a & b \\ 0 & 0 & 0\end{pmatrix}$$ shows that $$\begin{pmatrix} a & 0 & 0 \\ 0 & a & b \\ 0 & 0 & 0\end{pmatrix}^4 = a^3 \begin{pmatrix} a & 0 & 0 \\ 0 & a & b \\ 0 & 0 & 0\end{pmatrix}.$$ In particular, you can take $a = 3^{1/4}$ and $b = 3^{-3/4}$ for one possibility of $X$.
