The subspace $S \subseteq \mathbb{R}^{4}$ is spanned by the vectors $v_{1}=\begin{pmatrix} 1\\ 2\\ 3\\ 0 \end{pmatrix}, v_{2}=\begin{pmatrix} -1\\ 5\\ 7\\ -1 \end{pmatrix}, v_{3}=\begin{pmatrix} 5\\ -4\\ -5\\ 2 \end{pmatrix}$. From these vectors, choose a basis $B$ of $S.$
$I: x-y+5z=0$
$II: 2x+5y-4z=0$
$III: 3x+7y-5z=0$
$IV: -y+2z=0 \Leftrightarrow y=2z$
insert $IV$ in $I: x-2z+5z=0 \Leftrightarrow x+3z=0 \Leftrightarrow x=-3z$
insert $I$ and $IV$ in $II: 2(-3z)+5(2z)-4z=0 \Leftrightarrow -6z+10z-4z=0 \Leftrightarrow 0=0$
insert $I$ and $IV$ in $III: 3(-3z)+7(2z)-5z=0 \Leftrightarrow -9z+14z-5z=0 \Leftrightarrow 0=0$
So $y=2z$ and $x=-3z$
$\begin{pmatrix} -3z\\ 2z\\ z \end{pmatrix}= z\begin{pmatrix} -3\\ 2\\ 1 \end{pmatrix} \Rightarrow \text{Basis } B= \left\{ \begin{pmatrix} -3\\ 2\\ 1 \end{pmatrix} \right \}$
But the problem is the basis isn't of same size as the given vectors.. : /
I think everything I did is wrong but how can I do such a task correctly?
I'm looking for solutions, not for hints as I have tried this all day long with no real success. It's no homework anyway.
First, put your three vectors as rows of a matrix. Next, reduce by rows the matrix. Finally, all the rows that become all-zero at the end of the reduction you can throw them away, and what you're left with is a basis for the span of the original vectors. In our case$${}$$
$$\begin{pmatrix} 1&2&3&0\\ \!-1&5&7&\!-1\\ 5&\!-4&\!-5&2\end{pmatrix}\stackrel{R_2+R_1,\,R_5-5R_1}\longrightarrow \begin{pmatrix} 1&2&3&0\\ 0&7&10&\!-1\\ 0&\!-14&\!-20&2\end{pmatrix}\stackrel{R_3+2R_2}\longrightarrow$$$${}$$
$$\rightarrow\begin{pmatrix} 1&2&3&0\\ 0&7&10&\!-1\\ 0&0&0&0\end{pmatrix}$$
Thus, the first two original vectors span the subspace:
$$\;\text{Span}\{\,(1,2,3,0)^t,\,(-1,5,7,-1)^t\,\}=\text{Span}\{\,(1,2,3,0)^t,\,(-1,5,7,-1)^t,\,(5,-4,-5,2)^t\}\;$$
or you can instead take the two vectors you got in the last step above:
$$\text{Span}\{\,(1,2,3,0)^t,\,(-1,5,7,-1)^t\,\}=\text{Span}\{\,(1,2,3,0)^t,\,(0,7,10,-1)^t\,\}$$
as you wish.