# spanning set

• Sep 10th 2008, 04:49 AM
sunneej
spanning set
Hey, can you guys please help me out with two questions?

1. The following set is a spanning set for R^3.
S = {(1,2,-1),(0,3,4),(2,1,-6),(0,0,2)}
Find a subset of S that is a basis for R^3.

And,

2. Find a basis for the solution space V of the following linear system in four variables x1, x2, x3, x4:
x1+x2=0
-x2+3x3=0

Thanks in advance for great help guys :-).
• Sep 10th 2008, 05:42 AM
mr fantastic
Quote:

Originally Posted by sunneej
Hey, can you guys please help me out with two questions?

1. The following set is a spanning set for R^3.
S = {(1,2,-1),(0,3,4),(2,1,-6),(0,0,2)}
Find a subset of S that is a basis for R^3.

Mr F says: Remove one of the vectors that can be written as a linear combination of the other three vectors.

And,

2. Find a basis for the solution space V of the following linear system in four variables x1, x2, x3, x4:
x1+x2=0
-x2+3x3=0

Mr F says: Where does x4 appear in the above equations?? First you need to solve these two equations for x1, x2, x3. There are an infinite number of solutions. These solutions can be written in parametric form. Let ${\color{red}x_2 = t}$, say, where t can be any real number. You should be able to construct a basis from this solution .....

Thanks in advance for great help guys :-).

If you're still stuck, show your working and state where you're stuck.
• Sep 11th 2008, 04:51 AM
sunneej
Thanks but...
Thanks for the hints.

But for number 1, I couldn't find one of the vectors that can be written as a linear combination of the other three vectors.

And for number 2, I found the solutions, and because it said x4 in question, I included it though it will just be 0. Hence,

x1 = -3t; x2 = 3t; x3 = t; x4 = s.

Could you give me an advice as what to do from here onwards?

Thank you.
• Sep 11th 2008, 05:22 AM
mr fantastic
Quote:

Originally Posted by sunneej
Thanks for the hints.

But for number 1, I couldn't find one of the vectors that can be written as a linear combination of the other three vectors.

And for number 2, I found the solutions, and because it said x4 in question, I included it though it will just be 0. Hence,

x1 = -3t; x2 = 3t; x3 = t; x4 = s.

Could you give me an advice as what to do from here onwards?

Thank you.

1. $(1, \, 2, \, -1) = \frac{1}{2} \, (2, \, 1, \, -6) + \frac{1}{2} \, (0, \, 3, \, 4)$.

2. $(x_1, \, x_2, \, x_3, \, x_4) = t(-3, \, 3, \, 1, \, 0) + s(0, \, 0, \, 0, \, 1)$. A basis should be obvious.
• Sep 11th 2008, 05:47 AM
sunneej
Yeah, I could find one of the vectors that can be written as a linear combination of the other TWO vectors, but not one.

But what happens to (0,0,2)?

Cheers.
• Sep 11th 2008, 04:37 PM
mr fantastic
Quote:

Originally Posted by sunneej
Yeah, I could find one of the vectors that can be written as a linear combination of the other TWO vectors, but not one.

But what happens to (0,0,2)?

Cheers.

The linear combination doesn't have to be with all of the other vectors ..... just so long as there's a vector that can be written as a linear combination of some of the others ....

But if it makes you feel better: $(1, \, 2, \, -1) = \frac{1}{2} \, (2, \, 1, \, -6) + \frac{1}{2} \, (0, \, 3, \, 4) + 0 \, (0, \, 0, \, 2)$