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.

Note that $X^4$ is diagonalizable with eigenspaces
\begin{align}
E_3(X^4) &= \langle (1,0,0), (0,1,0) \rangle, \\
E_0(X^4) &= \ker(X^4) = \langle (0,1,-3)\rangle.
\end{align}
Hence, after a change of basis the problem translates to finding $Y$ with
$$
Y^4 = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0\end{pmatrix}.
$$
Since $Y^4$ is singular, $Y$ needs to be singular as well and since $\ker(Y)\subseteq\ker(Y^4)$ we know that $\ker(Y)=\ker(Y^4)=\langle(0,0,1)\rangle$, so the third column of $Y$ must be zero.
Since the dimensions of the kernels of $Y$ and $Y^4$ are equal, the dimensions of the images have to be equal as well and we get $\operatorname{im}(Y)=\operatorname{im}(Y^4)=\langle (1,0,0), (0,1,0)\rangle$.
Hence, we have
$$
Y = \begin{pmatrix} * & * & 0 \\ * & * & 0 \\ 0 & 0 & 0\end{pmatrix},
$$
where $Z=(\begin{smallmatrix}*&*\\*&*\end{smallmatrix})$ is any solution to $Z^4=3I$, where $I$ is the $2\times 2$ identity matrix. Such matrices are of the form
$$
Z = U^{-1} \begin{pmatrix} \eta_1 & 0 \\ 0 & \eta_2\end{pmatrix} U,
$$
where $\eta_1,\eta_2$ are any (possibly equal) solutions to $\eta^4=3$ and $U$ is an invertible $2\times 2$ matrix, so $U=(\begin{smallmatrix}a&b\\c&d\end{smallmatrix})$ with $ad-bc\neq 0$.
Putting things together, we first get
\begin{align}
Z &= \frac{1}{ad-bc} \begin{pmatrix}d&-b\\-c&a\end{pmatrix}\begin{pmatrix} \eta_1 & 0 \\ 0 & \eta_2\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix} \\
&= \frac{1}{ad-bc} \begin{pmatrix} ad\eta_1-bc\eta_2 & bd(\eta_1-\eta_2)\\ac(\eta_2-\eta_1) & ad\eta_2-bc\eta_1\end{pmatrix}.
\end{align}
Then
$$
Y = \frac{1}{ad-bc} \begin{pmatrix} ad\eta_1-bc\eta_2 & bd(\eta_1-\eta_2) & 0 \\ ac(\eta_2-\eta_1) & ad\eta_2-bc\eta_1 & 0 \\ 0 & 0 & 0\end{pmatrix},
$$
and changing back to the original basis finally
$$
X = \frac{1}{ad-bc} \begin{pmatrix} ad\eta_1-bc\eta_2 & bd(\eta_1-\eta_2) & \frac 1 3 b d (\eta_1-\eta_2) \\ ac(\eta_2-\eta_1) & ad\eta_2-bc\eta_1 & \frac 1 3 (ad\eta_2-bc\eta_1)\\ 0 & 0 & 0\end{pmatrix}.
$$
The constraints are $ad-bc\neq 0$ and $\eta_i^4=3$ for $i=1,2$.

The solution obtained by you (with the correction of Henry) is obtained by choosing $\eta_1=\eta_2=3^{\frac 1 4}$ and $U=I$, so $a=d=1$, $b=c=0$.
