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$v_1=\begin{pmatrix} 1\\ 3\\ 5 \end{pmatrix} , v_2=\begin{pmatrix} 4\\ 5\\ 6 \end{pmatrix}, v_3 = \begin{pmatrix} 6\\ 4\\ 2 \end{pmatrix}$ 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 ?

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    $\begingroup$ Without checking deeply it looks fine...but rather messy and without a clear, fixed method: why won't you simply reduce the original matrix by means of elementary (Gauss) operations? That's all we need: at the end, the number of non-all-zero rows is the dimension of the space spanned by the rows. Period. What you did seems to have worked by chance in this very particular case. Reduction works always $\endgroup$ – DonAntonio Mar 21 '17 at 11:14
  • $\begingroup$ @DonAntonio Does that mean the transposition is completely useless? Is it also not required if I wanted know basis / image? $\endgroup$ – cnmesr Mar 21 '17 at 11:21
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    $\begingroup$ @cn Indeed. Transposition only makes you waste time... unless it is a requirement from your instructor, of course. Now, for a basis of the kernel of a given matrix: you have to do exactly as above: rows, reduction...and what is left is a basis for the kernel. For the image though you need to work with the columns...but you then can indeed take the transpose, reduce again and etc., and what is left in rows is a basis for the image. $\endgroup$ – DonAntonio Mar 21 '17 at 11:24
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    $\begingroup$ @DonAntonio Thank you very much! :)) $\endgroup$ – cnmesr Mar 21 '17 at 11:25
  • $\begingroup$ @cn My pleasure. $\endgroup$ – DonAntonio Mar 21 '17 at 11:28
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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.

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