How to find $\ker T$ from matrix representation over polynomial finite field? 
Let $\mathbb F$ be a finite field of 5 elements (essentially $\mathbb Z_5$).
$T:\mathbb F_3[x] \to \mathbb F_3[x]$ is a linear map defined by the representation matrix:
  $$
[T]_B=\begin{pmatrix}
1&2&3\\
1&0&4\\
0&1&2
\end{pmatrix}
$$
  with the basis $B=(1, 1+x,1+x+x^2)$. Find $\text{ker}T$.
$\mathbb F_3[x]$ is a space of polynomials of form $ax^2+bx+c$ over $\mathbb F$.

We know that if a vector $v\in \text{ker}T$ then $[v]_B$ is in the nullity space of $[T]_B$.
If we row reduce $[T]_B$ and solve the system for $a_1x_1+a_2x_2+a_3x_3=0$:
$$
\begin{align}
\begin{pmatrix}
1&2&3\\
1&0&4\\
0&1&2
\end{pmatrix}\\Row_2\to Row_2-Row_1 \Rightarrow \begin{pmatrix}
1&2&3\\
0&3&1\\
0&1&2
\end{pmatrix}\\
Row_3\to Row_3-3^{-1}Row_2\Rightarrow \begin{pmatrix}
1&2&3\\
0&3&1\\
0&0&0
\end{pmatrix}\\
Row_2\to 3^{-1}Row_2\Rightarrow \begin{pmatrix}
1&2&3\\
0&1&2
\end{pmatrix}
\end{align}
$$
So: 
$$
x_2=-2x_3=3x_3 \quad(??)\\
x_1=-2x_2-3x_3=3\cdot 3x_3+2x_3=4x_3+2x_3=x_3 \quad(??)
$$
and the general solution is: $(x_3,3x_3,x_3)$ from which follows that $\text{ker}T=span\{1,3,1\}$.
I'm not sure whether my calculations are correct especially when calculating the general solution. 
Also I'm not sure how/whether to take into consideration that a vector belonging to $\text{ker}T$ is actually a polynomial.
 A: Even without reducing, you could write down this:
$\ker(T)=\{v\in V | T(v)=0\}$
So, from your matrix you see that:
$$\begin{pmatrix}
1&2&3\\
1&0&4\\
0&1&2
\end{pmatrix}\begin{pmatrix}x_1\\x_2 \\x_3 \end{pmatrix}=0 \iff \begin{equation} \begin{cases}x_1+2x_2+3x_3=0\\x_1+4x_3=0\\x_2+2x_3=0 \end{cases}\end{equation}$$
From which you obtain that: \begin{equation} \begin{cases}z=\alpha\\y=-2\alpha\\x=-4\alpha \end{cases}\end{equation} So the solution is $\text{Span}(-4,-2,1)=\text{Span}(1,3,1)$
A: The row reduction is
\begin{align}
\begin{pmatrix}
1&2&3\\
1&0&4\\
0&1&2
\end{pmatrix}
&\to
\begin{pmatrix}
1&2&3\\
0&3&1\\
0&1&2
\end{pmatrix}
&& R_2\gets R_2-R_1
\\&\to
\begin{pmatrix}
1&2&3\\
0&1&2\\
0&1&2
\end{pmatrix}
&& R_2\gets 3^{-1}R_2
\\&\to
\begin{pmatrix}
1&2&3\\
0&1&2\\
0&0&0
\end{pmatrix}
&& R_3\gets R_3-R_2
\\&\to
\begin{pmatrix}
1&0&4\\
0&1&2\\
0&0&0
\end{pmatrix}
&& R_1\gets R_1-R_2
\end{align}
This means that the third variable is free, so a basis of the null space of the matrix is given by the single vector
$$
\begin{pmatrix} -4 \\ -2 \\ 1 \end{pmatrix}
=
\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}
$$
This vector gives the coordinates of the generator of the kernel with respect to $B$. Therefore a basis for the kernel is given by the single polynomial
$$
1\cdot 1+3\cdot(1+x)+1\cdot(1+x+x^2)=4x+x^2
$$
