# Math Help - is it possible for a basis to have 1 vector or am I doing something wrong?

1. ## is it possible for a basis to have 1 vector or am I doing something wrong?

question: let W be a subspace spanned by the vectors: { $v_1=(2,0,-1,3) v_2=(1,2,2,-5) v_3=(3,2,1,-2) v_4=(7,2,-1,4)$} Find the basis $W^{\perp}$

work:let u=(a,b,c,d) be in $W^{\perp}$ then $u \cdot v_1=0, u \cdot v_2 = 0, u \cdot v_3 = 0, u \cdot v_4 = 0$
So we've got the following matrix augmented with a column of 0s
$\left [ \begin{array}{cccc} 2 & 0 & -1 & 3 \\ 1 & 2 & 2 & -5 \\ 3 & 2 & 1 & -2 \\ 7 & 2 & -1 & 1 \end{array} \right]$ in rref form is $\left [ \begin{array}{cccc} 1 & 0 & \frac{-1}{2} & 0 \\ 0 & 1 & \frac{5}{4} & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right]$ so that gives me $a=\frac{r}{2} b=\frac{-5r}{4} c=r d=0$ therefore $(\frac{1}{2},\frac{-5}{4},1,0)$ spans $W^{\perp}$

question: did I do that right how d=0 or does d=s?

2. Originally Posted by superdude
question: let W be a subspace spanned by the vectors: { $v_1=(2,0,-1,3) v_2=(1,2,2,-5) v_3=(3,2,1,-2) v_4=(7,2,-1,4)$} Find the basis $W^{\perp}$

work:let u=(a,b,c,d) be in $W^{\perp}$ then $u \cdot v_1=0, u \cdot v_2 = 0, u \cdot v_3 = 0, u \cdot v_4 = 0$
So we've got the following matrix augmented with a column of 0s
$\left [ \begin{array}{cccc} 2 & 0 & -1 & 3 \\ 1 & 2 & 2 & -5 \\ 3 & 2 & 1 & -2 \\ 7 & 2 & -1 & 1 \end{array} \right]$ in rref form is $\left [ \begin{array}{cccc} 1 & 0 & \frac{-1}{2} & 0 \\ 0 & 1 & \frac{5}{4} & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right]$ so that gives me $a=\frac{r}{2} b=\frac{-5r}{4} c=r d=0$ therefore $(\frac{1}{2},\frac{-5}{4},1,0)$ spans $W^{\perp}$

question: did I do that right how d=0 or does d=s?
Since the rank of the row reduced form is 3, the dimension of the solution space of the equations is 1.

CB