# Thread: Which sets of vectors span R^3

1. ## Which sets of vectors span R^3

Which of the following sets of vectors span R^3?
a. {(1,-1,2),(0,1,1)}
b. {(1,2,-1),(6,3,0),(4,-1,2),(2,-5,4)}
c. {(2,2,3),(-1,-2,1),(0,1,0)}
d. {(1,0,0),(0,1,0),(0,0,1),(1,1,1)}

2. Originally Posted by antman
Which of the following sets of vectors span R^3?
a. {(1,-1,2),(0,1,1)}
b. {(1,2,-1),(6,3,0),(4,-1,2),(2,-5,4)}
c. {(2,2,3),(-1,-2,1),(0,1,0)}
d. {(1,0,0),(0,1,0),(0,0,1),(1,1,1)}
What work have you done for yourself on any of these?

I will tell you that the set in part a) does not. You tell us why?

3. Because it does not contain 3 vectors. I think I understand that the other 3 could be if 3 or more of the vectors are linearly independent, but I do not know how to find this.

4. For part a, my work is:

$\displaystyle c_{1}(1,-1,2)+c_{2}(0,1,1)=v=(a,b,c)$

$\displaystyle c_{1}=a$
$\displaystyle -c_{2}+c_{2}=b$
$\displaystyle 2c_{1}+c_{2}=c$

$\displaystyle \begin{pmatrix}1 & 0 & a\\-1 & 1 & b\\2 & 1 & c\end{pmatrix}$

Reduced
$\displaystyle \begin{pmatrix}1 & 0 & a\\0 & 1 & b+a\\0 & 0 & c-3a-b\end{pmatrix}$

The system is inconsistent and has no solution so this set does not span R^3?

Part b: same work resulting in the following reduced matrix

$\displaystyle \begin{pmatrix}1 & 0 & -2 & -{1/3}a+{2/3}b\\0 & 1 & 1 & {-1/9}b+{2/9}a\\0 & 0 & 0 & c-{1/3}a+{2/3}b\end{pmatrix}$
So this would also not span R^3 since there are no solutions?

Part c:

$\displaystyle c_{1}+c_{4}=a$
$\displaystyle c_{2}+c_{4}=b$
$\displaystyle c_{3}+c_{4}=c$
So this set does span R^3?

Does anyone know if this is the way this problem should be done? Thank you for your help.

5. Since two vectors can't span a three dimensional space, and you said that you understood that, I don't know why you did that.

But that method can be used on the others: {(2,2,3),(-1,-2,1),(0,1,0)} spans $\displaystyle R^3$ if and only if there exist x, y, z such that x(2,2,3)+ y(-1,-2, 1)+ z(0, 1, 0)= (a, b, c) for any three numbers a, b, c. We can write that as 2x- y= a, 2x-2y+ z= b, 3x+ y= c. Adding the first and third equations, 5x= a+c or x= (a+c)/5. Then the third equation gives y= c- 3x= c- (3/5)a- (3/5)b. Put that into the second equation to see if you can solve it for z.

Of course that is the same as reducing the augmented matrix
$\displaystyle \begin{bmatrix}2 & -1 & 0 & a \\2 & -2 & 1 & b \\3 & 1 & 0 & c\end{bmatrix}$
Actually, a, b, and c are irrelevant here. such x, y, z will exist for all a, b, c if and only if row reducing the matrix
$\displaystyle \begin{bmatrix}2 & -1 & 0 \\2 & -2 & 1 \\3 & 1 & 0\end{bmatrix}$
does not give a row consisting entirely of 0s. With four vectors you get a four rows and you only need three of them not consisting of 0s.

6. In my text, that is how the work is shown in the examples. When there was a set of 3 vectors for R^4, work was still shown in that format. So that's why I was so confused. But otherwise, my work is sufficient? Simply reduce and make sure there are enough nonzero rows. (ex. for a set of vectors to span R^3, there must be three nonzero rows after the matrix is reduced?)

For part d, the matrix created was already reduced and all rows were nonzero (3 rows x 4 columns), so that would also span a 3-D space? Therefore, a and b are the only sets that would not. I just want to show enough work for my answers. Thank you!