Fourth root of a $3\times 3$ matrix I have solved this question by generalizing my assumption about some properties of the given matrix. However, it is not a rigorous proof and I haven't fully understood the background/meaning of it.
Here's the question. Find the matrix X satisfying
$ X^4=
\begin{bmatrix}
3&0&0\\
0&3&1\\
0&0&0\\
\end{bmatrix}
$
I found that
$(X^4)^n=\begin{bmatrix}3^n&0&0\\0&3^n&3^{n-1}\\0&0&0\\\end{bmatrix}$
Therefore I assumed
$X^n=\begin{bmatrix}3^\frac{n}{4}&0&0\\0&3^\frac{n}{4}&3^\frac{n-1}{4}\\0&0&0\\\end{bmatrix}$
Although I am on the entry-level in linear algebra, any approach to this question is welcomed, I will try my best to understand it.

Here is my approach
Notice $X^4$ is an upper-triangular matrix and the multiplication of two upper-triangular matrices is upper-triangular
Let $X^2= \begin{bmatrix} X_{11}&X_{12}\\O&X_{22}\\\end{bmatrix}$
where
$\begin{align*}
X_{11}^2=3I_{2×2}\\
X_{11}X_{12}+X_{12}X_{22}=\begin{bmatrix}0\\1\\\end{bmatrix}\\
X_{22}^2=O
\end{align*}$
therefore
$\begin{align*}
X_{11}=\sqrt{3}I_{2\times2}\\
X_{22}=O\\
X_{12}=\begin{bmatrix}0\\\frac{1}{\sqrt{3}}\end{bmatrix}
\end{align*}$
that is $X^2=\begin{bmatrix}3^{\frac{1}{2}}&0&0\\0&3^{\frac{1}{2}}&3^{-\frac{1}{2}}\\0&0&0\end{bmatrix}$
repeat this once more, then we get (one of) the solution shown above
