$v_1=\begin{pmatrix} 1\\ 3\\ 5 \end{pmatrix} , v_2=\begin{pmatrix} 4\\ 5\\ 6 \end{pmatrix}, v_3 = \begin{pmatrix} 6\\ 4\\ 2 \end{pmatrix}$ are vectors of the vector space $\mathbb{Z}^{3}_{7}$, over the field $\mathbb{Z}_{7}$. What's the dimension of $U_{7}= \text{span}\left\{v_1,v_2,v_3\right\}$?
I think the elements of all vectors are alright because they are in $\mathbb{Z}_{7}= \left\{0,1,2,3,4,5,6\right\}$, if that is understood correctly at all by me (?)
Now I form the vectors to a matrix: $\begin{pmatrix} 1 & 4 & 6\\ 3 & 5 & 4\\ 5 & 6 & 2 \end{pmatrix}$ and transpose it: $\begin{pmatrix} 1 & 3 & 5\\ 4 & 5 & 6\\ 6 & 4 & 2 \end{pmatrix}$
Multiply first line with $4$:
$\begin{pmatrix} 4 & 12 & 20\\ 4 & 5 & 6\\ 6 & 4 & 2 \end{pmatrix}$
But $12$ and $20 \notin \mathbb{Z}_{7}$. So we have to do $12 \text{ mod } 7= 5$ and $20 \text{ mod } 7= 6$
Insert that in the matrix: $\begin{pmatrix} 4 & 5 & 6\\ 4 & 5 & 6\\ 6 & 4 & 2 \end{pmatrix}$
Now subtract first line second line, we get: $\begin{pmatrix} 4 & 5 & 6\\ 0 & 0 & 0\\ 6 & 4 & 2 \end{pmatrix}$
And thus the dimension of $U_{7}= \text{span}\left\{v_1,v_2,v_3\right\}$ is $2$?
Another question, I had to check at the beginning if the vectors are linearly independent ?
We could have simply reduced the matrix by rows (with rows = the given vectors):
$$\begin{pmatrix} 1&3&5\\ 4&5&6\\ 6&4&2\end{pmatrix}\stackrel{R_2-4R_1,\,R_3-6R_1}\longrightarrow\begin{pmatrix} 1&3&5\\ 0&0&0\\ 0&0&0\end{pmatrix}\implies$$
The second and third vectors are linearly dependent on the first one , and in fact:
$$\begin{pmatrix}4\\5\\6\end{pmatrix}=4\begin{pmatrix}1\\3\\5\end{pmatrix}\;,\;\;\begin{pmatrix}6\\4\\2\end{pmatrix}=6\begin{pmatrix}1\\3\\5\end{pmatrix}$$
and thus the dimension of the span of these three vectors over $\;\Bbb F_7\cong\Bbb Z/7\Bbb Z\;$ is one .
Keeping track of which vector is represented by what row (in case we interchange some of them) we can say, at the end of the row reduction, exactly which ones are lin. dependent on which ones, and we can even write down exactly what is the linear dependency. I think that by far this is the best method.