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

• negation
In summary: There is no "reply to" option in the sense of a dedicated button. Just click on the "Reply" link in the post you want to quote. It will quote the entire post. You can edit it down to the part you want to quote, if you want.... how does the app works?I assume you're talking about the forum software. There are a lot of features and options and I don't use them all. It would be better to ask a moderator or administrator about it.
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.

Last edited:
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.

negation said:

## 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.
negation said:
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 said:
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.
negation said:
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>.
negation said:
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.

Last edited:
Mark44 said:
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.

negation said:
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.
negation said:
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.
negation said:
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.

Mark44 said:
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.

Mark44 said:
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 $\notin span{(u,v)}$

negation said:
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.

1 person

## 1. How do I determine if a and b belong to R4 spanned by u and v?

To determine if a and b belong to R4 spanned by u and v, you can use the span or linear combination property. This means that if a and b can be written as a linear combination of u and v, then they belong to R4 spanned by u and v.

## 2. What is R4?

R4, also known as R^4, is a four-dimensional vector space. It consists of all possible combinations of four real numbers, represented as (x, y, z, w). It is commonly used in linear algebra and is important in understanding vector spaces.

## 3. What does it mean for two vectors to span a vector space?

When two vectors, u and v, span a vector space, it means that every vector in that space can be written as a linear combination of u and v. This means that u and v are able to reach every point in the vector space through linear combinations.

## 4. How do I know if a vector belongs to a vector space?

A vector belongs to a vector space if it satisfies the properties of that space. These properties may include closure, commutativity, and associativity. In the case of determining if a and b belong to R4 spanned by u and v, the span or linear combination property must be satisfied.

## 5. Can a vector belong to multiple vector spaces?

Yes, a vector can belong to multiple vector spaces. This can happen if the vector satisfies the properties of each of those spaces. For example, a vector can belong to both R3 and R4 if it can be written as a linear combination of three and four-dimensional vectors respectively.

• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
6K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
3K
• Linear and Abstract Algebra
Replies
1
Views
2K