Determining if a & b Belong to R4 Spanned by u & v

[negation]

Homework Statement

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.

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.

 

[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.

 

[Mark44]

Homework Statement

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.

The Attempt at a Solution

Since R4 spans u and v, then, R4 = span(u,v)

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.

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.

 

[Mark44]

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)

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.

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)

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>.

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.

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.

 

[negation]

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.

I'm still trying to get used to the definition.
Yes, I realized 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.

 

[Mark44]

I'm still trying to get used to the definition.

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.

Yes, I realized I committee an EROs error. The matrix is not in reduced echelon form.
I'll rework it.

EROs error? I don't know what that is.

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

Phone app? If that's what you're asking about, I don't know. I always use a desktop.

 

[negation]

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.

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.

 

[negation]

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.

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)}

 

[Mark44]

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)}

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.