<s>[INST] Basis of Image in a GF(5) matrix with variables consider $A_\alpha := \begin{pmatrix} \alpha & 3 & 4\\ 3 & 1 & 3 \\ 4 & 2 & \alpha \end{pmatrix} \in \mathbb{Z_5}^{3\times3}$. Determine, dependant on $\alpha$, a Basis and the dimension of the Image of the standardinterpretation $F_{A_\alpha}$. What is the dimension of the kernel of $F_{A_\alpha}$? The standardinterpretation is defined as $F_A : K^{n\times1} \rightarrow K^{m \times 1}, x \mapsto F_A(x):=Ax $.
Okay. To determine the Image of a Matrix we just row reduce it and then we will see which columns are in the Image of that matrix.
Let's row reduce the matrix
$\begin{pmatrix}
\alpha &  3 & 4\\ 
 3&  1& 3\\ 
 4&  2& a
\end{pmatrix} \underset{\overset{R1 \leftarrow R_1+2\cdot R_2}{\longrightarrow}}{\overset{R_3 \leftarrow R_3 + 3 \cdot R_2}{\longrightarrow}} 
\begin{pmatrix}
a+1 &  0& 0\\ 
 3&  1& 3\\ 
 0&  0& a+4
\end{pmatrix}$.
This doesn't look too bad, let's just look at the cases we get:
Let $\alpha = 0$ then we get $\begin{pmatrix}
1 &  0& 0\\ 
 3&  1& 3\\ 
 0&  0& 0
\end{pmatrix} \overset{R_2 \leftarrow R_2+2\cdot R_1}{\longrightarrow}
\begin{pmatrix}
1 &  0& 0\\ 
 0&  1& 3\\ 
 0&  0& 0
\end{pmatrix}
$ then we can see that a basis for $Image(A_\alpha)$ is  $\{\begin{pmatrix} a \\ 3 \\ 4 \end{pmatrix}, \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix}\}$. Since we have 2 elements in our basis for the Image, we know that $dim(Image)=2$.
We also remember that $dim(V) = dim(kernel) + dim(Image)$ therefore $3 = dim(kernel) + 2$ so we get $dim(kernel)=1$.
Do I have to repeat this process for all other 4 cases ($\alpha = 1,2,3,4$) or am I completly wrong about my approach? Thanks for any hints on how to solve this problem.
 [/INST] Your row reduction is indeed correct. Your steps from there are problematic.  We have the reduced matrix
$$
\pmatrix{
a+1 &  0& 0\\ 
 3&  1& 3\\ 
 0&  0& a+4}.
$$
It is easy to see that for $\alpha \notin \{1,4\}$, further row-reduction brings us to the identity so that $A$ is invertible. In the case that $\alpha = 1$, we get
$$
\pmatrix{
2 &  0& 0\\ 
 3&  1& 3\\ 
 0&  0& 0} \leadsto
\pmatrix{
2 &  0& 0\\ 
 0&  1& 3\\ 
 0&  0& 0}.
$$
We deduce that the first two columns of $A$ form a basis of the image. That is, the image has basis
$$
\{(\alpha,3,4),(3,1,2)\} = \{(1,3,4),(3,1,2)\}.
$$
So, the dimension of the image is $2$ and the dimension of the kernel is $3 - 2 = 1$.
Similarly, for $\alpha = 4$, we get
$$
\pmatrix{
0 &  0& 0\\ 
 3&  1& 3\\ 
 0&  0& 3} \leadsto
\pmatrix{ 
 3&  1& 0\\ 
 0&  0& 3\\
0&0&0}.
$$
We deduce that the first and third columns of $A$ form a basis of the image. So, the dimension of the image is $2$ and the dimension of the kernel is $3 - 2 = 1$.
</s>