1. ## Basis and dimension

Find the basis for and the dimension of $\displaystyle W=span(\mathbf{v_1, v_2, v_3})$, where

$\displaystyle \mathbf{v_1}=\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}, \mathbf{v_2}=\begin{pmatrix}1\\ 1\\ -1\end{pmatrix}, \mathbf{v_3}=\begin{pmatrix}-1\\ 0\\ 5\end{pmatrix}$

I'm not too sure how to start this question off

2. If the three given vectors are independent, then they are a basis for their own span and the dimension is 3. If they are not, then one of the vectors can be written as a linear combination of the other two and so those two form a basis and the dimension is 2. (It clear that these vectors are not all multiples of one another so the dimension is not 1.)

So start by checking if they are independent or not: do there exist numbers, a, b, and c such that
$\displaystyle av_1+ bv_1+ cv_2= a\begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix}+ b\begin{pmatrix}1 \\ 1 \\ -1\end{pmatrix}+ c\begin{pmatrix}-1 \\ 0 \\ 5\end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$

That is the same as
$\displaystyle \begin{pmatrix}a \\ 2a \\ 3a\end{pmatrix}+ \begin{pmatrix}b \\ b \\ -b\end{pmatrix}+ \begin{pmatrix}-c \\ 0 \\ 5c\end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

$\displaystyle \begin{pmatrix}a+ b- c \\ 2a+ b \\ 3a- b+ 5c\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}$

and that is the same as the three equations a+ b- c= 0, 2a+ b= 0, and 3a- b+ 5c= 0. Obviously a= b= c= 0 is a solution. If it is the only solution, then the three vectors form a basis and the space is three dimensional.

If there exist another solution, we can solve for one of the vectors in terms of the other two. If we had $\displaystyle av_1+ bv_2+ cv_3= 0$ with, at least, c non-zero, then we can write $\displaystyle cv_3= -av_1- bv_2$ so $\displaystyle v_3= -(a/c)v_1- (b/c)v_2$. Since $\displaystyle v_1$, $\displaystyle v_2$, and $\displaystyle v_3$ span the space every vector in it can be written as a linear combination, $\displaystyle v= pv_1+ qv_2+ rv_3$. With $\displaystyle v_3= (-a/c)v_1- (b/c)v_3$ we can replace $\displaystyle v_3$ and get $\displaystyle v= pv_1+ qv_2+ r(-(a/c)v_1- (b/c)v_2)= (p- ar/c)v_1+ (q- br/c)v_2$. Then $\displaystyle \{v_1, v_2\}$ is a basis and the space has dimension 2.

That's the idea. I'll leave the actual calculation to you.