Find the number of dimensions and a basis $W$ of a generating set of $U$ Given the system of vectors $U= \left \{ \left ( 1, 2, 3 \right ), \left ( 0, -1, -2 \right ), \left ( 2, 3, 4 \right ), \left ( 1, 0, -1 \right ) \right \}.$ Find the number of dimensions and a basis $W$ of a generating set of $U.$
We use the rows reduction
$$\begin{pmatrix} 1 & 0 & 2 & 1\\ 2 & -1 & 3 & 0\\ 3 & -2 & 4 & -1 \end{pmatrix}\xrightarrow{2R_{1}- R_{2}}\begin{pmatrix} 1 & 0 & 2 & 1\\ 0 & 1 & 1 & 2\\ 3 & -2 & 4 & -1 \end{pmatrix}\xrightarrow{3R_{1}- 2R_{2}- R_{3}}\begin{pmatrix} 1 & 0 & 2 & 1\\ 0 & 1 & 1 & 2\\ 0 & 0 & 0 & 0 \end{pmatrix}$$
then $\operatorname{rank}= 2.$
I don't have a way of thinking to continue. What is the relationship between $W$ and $U,$ I need to the help ?!
Edit. Find the value of $k$ so that $u= \left ( 2, 3, k^{2}+ 1 \right )$ is a linear combination of $W,$ and what is $\left [ u \right ]_{W}\!.$
 A: Actually, you should work with the matrix$$\begin{bmatrix}1&2&3\\0&-1&-2\\2&3&4\\1&0&-1\end{bmatrix}$$instead. Note that if you add to the third row the first row times $-2$ and you add to the fourth row the first row times $-1$, then you get$$\begin{bmatrix}1&2&3\\0&-1&-2\\0&-1&-2\\0&-2&-4\end{bmatrix}.$$Now, if you add to the third row the second row times $-1$ and if you add to the fourth row the second row times $-2$, then you get$$\begin{bmatrix}1&2&3\\0&-1&-2\\0&0&0\\0&0&0\end{bmatrix}.$$So, a basis of the space spanned by $U$ is $\{(1,2,3),(0,-1,-2)\}$ and its dimension is $2$.
A: I'll use $M$ to stand for RREF form of $W$ you calculated.
The key idea is that the "relationship" between the columns of $M$ will also hold for the columns of $W$. Here it should be pretty easy to see that the first two columns of $M$ (1,0,0) and (0,1,0) form a basis for the column space of $M$. So the first two columns of $W$ will also form a basis for the column space of $W$. So the basis your looking for is just the first two columns of $W$.
Let me know if you need more details.
