# Spanning set

1. Mar 13, 2014

### negation

1. The problem statement, all variables and given/known data

In R4,

let u = (-1,1,5,-3) and v = (2,-3,-5,-2)

and

let a = (9,-12,-30,3) and b = (2,-1,-14,11)

For each of the vectors a, b you are asked to determine whether it belongs to the subspace of R4 spanned by u, v.

3. The attempt at a solution

Since R4 spans u and v, then, R4 = span(u,v)
This implies also that if a is a spanning set,then,
span(a) = span(u,v)

(x,y,z,u) = λ1u + λ2v =

-λ1 + 2λ2 = x
λ1 -3λ2 = y
5λ1 - 5λ2 = u
-3λ1 -2λ2 = z

(x,y,z,u) = (-λ1 + 2λ2,λ1 -3λ2 ,5λ1 - 5λ2 ,-3λ1 -2λ2 )
(-λ1 + 2λ2,λ1 -3λ2 ,5λ1 - 5λ2 ,-3λ1 -2λ2 ) = γ.a
(-λ1 + 2λ2,λ1 -3λ2 ,5λ1 - 5λ2 ,-3λ1 -2λ2 ) = γ(9,-12,-30,3) = 9γ1 - 12γ2-30γ3+3γ4
This looks very chaotic. Am I on the right track?

EDIT: I think I might have found a much easier way.

Last edited: Mar 13, 2014
2. Mar 13, 2014

### negation

let u = (-1,1,5,-3) and v = (2,-3,-5,-2)

and

let a = (9,-12,-30,3) and b = (2,-1,-14,11)

If a is a spanning set for span(u,v), then, a = span(uv)

check: λ1(-1,1,5,-3) + λ2(2,-3,-5,-2) = (9,-12,-30,3)

-λ1 = 2λ2 = 9
-λ1 -3λ2 = -12
5λ1 -5λ2 = -30
-3λ1-2λ2 = 3

-1 2| 9
1 -3 | -12
5 -5 | -30
-3 -2 | 3

RREF....

-1 2 | 9
0 -11/3 | -11
0 5 | 15
0 -8 | -24

λ2 =3, λ1 = -3
solutions is unique so, a is a spanning set for (u,v)

Check for b:
-λ1 + 2λ2 =2
λ1 -3λ2 = -1
5λ1 -5λ2 = -14
-3λ1 -2λ2 = 11

-1 2 | 2
1 -3 | -1
5 -5 | -14
-3 -2 | 11

RREF...

-1 2 | 2
0 -1 | 1
0 5 | -4
0 -8 | 5

λ2 = -0.625, λ2 = -0.8, λ2 = -1; -λ1 + 2λ2 =2, then, λ1= 2λ2 -2

λ2 has multiple values so b is not a spanning set.

3. Mar 13, 2014

### Staff: Mentor

This is nonsense. A set of vectors can span the vector space they belong to (or not span it), but a space can't span a set of vectors. Also, two vectors can't possibly span a four-dimensional vector space. It takes at least four vectors to make a spanning set.

4. Mar 13, 2014

### Staff: Mentor

This makes absolutely no sense. It's not likely that a single vector will be equal to the span of two vectors, especially two vectors that are (obviously) linearly independent.
The numbers are correct. Your conclusion is wrong though. What you should say is that a is in the span of u and v.

Since you don't understand the terms used here, it would be better to say that a is a linear combination of u and v. That's what a $\in$ Span(u, v) means.

Finally, you can and should check your answers. Just verify that -3u + 3v = <9, -12, -30, 3>.
Again, this makes no sense. b is a vector. The question is whether b can be written as a linear combination of u and v.

Edit: I checked your numbers - they are incorrect.

Last edited: Mar 13, 2014
5. Mar 13, 2014

### negation

I'm still trying to get used to the definition.
Yes, I realised I committee an EROs error. The matrix is not in reduced echelon form.
I'll rework it.

On a tangent, how does the app works? There doesn't seem to be a 'reply to' option.

6. Mar 13, 2014

### Staff: Mentor

Definitions are extremely important in linear algebra. If you don't understand the basic terms being used, you will have major problems, as you're finding out.
EROs error? I don't know what that is.
Phone app? If that's what you're asking about, I don't know. I always use a desktop.

7. Mar 13, 2014

### negation

I've already realized it. In fact, I think having my definitions muddled and with vector space being taught last week coupled with the steep learning curve, it's what giving me the problem. But I'd say I'm getting used to the idea of subspace.

Yes phone app. I just had it installed.
EROs refers to elementary row operations.

8. Mar 13, 2014

### negation

Check for b:
-λ1 + 2λ2 =2
λ1 -3λ2 = -1
5λ1 -5λ2 = -14
-3λ1 -2λ2 = 11

-1 2 | 2
1 -3 | -1
5 -5 | -14
-3 -2 | 11

RREF...

-1 2 | 2
0 -1 | 1
0 5 | -4
0 0| -3

0λ2 = -3
Solution is inconsistent.

Hence. b $\notin span{(u,v)}$

9. Mar 13, 2014

### Staff: Mentor

Correct.
As a completely reduced augmented matrix, I get
$$\begin{bmatrix}1 & 0 & | & -4 \\ 0 & 1 & | & -1 \\ 0 & 0 & | & 1 \\ 0 & 0 & | & 3 \end{bmatrix}$$

Written as equations in x and y, this matrix is equivalent to this system:
x = -4
y = -1
0x + 0y = 1
0x + 0y = 3

All four equations have to be satisfied, but there are no solutions possible for the 3rd and 4th, so the system is inconsistent.