Another vector problem

  • Thread starter needhelp83
  • Start date
  • Tags
    Vector
In summary: The key to showing that something is not a subspace is to find a counterexample. The easiest way to find a counterexample is to find a specific example that does not work. That is, find specific numbers that make the given statement false. For example, for W = { (x1, 0) | x1 > 0 }, you could point out that (1, 0) and (2, 0) are both in W, but their sum (3, 0) is not. Therefore, W is not closed under addition and is not a subspace. Can you find a similar counterexample for W = { (x1, x2) | x1 >
  • #1
needhelp83
199
0
Determine if following subsets of R^2 are subspaces of R^2. If the subset is a subspace show that it is closed under vector addition and scalar multiplication. If the subset is not a subspace show why, indicating property that fails.

[tex] 1) W=\{ \left (x_1,0)\left| x_1\in\Re\} \newline [/tex]
[tex] 2) W=\{ \left (x_1,0)\left| x_1 > 0\} \newline[/tex]
[tex]3) W=\{ \left (2c,-3c)\left| c \in\Re\} \newline[/tex]
[tex]4) W=\{ \left (x_1,x_2)\left| x_1 > 0, x_2>0\} \newline [/tex]

Answers:
[tex] 1) (x,0) + (y,0)= (x+y,0) \in W \} \newline [/tex]
[tex] c(x,0) = (c(x),0) \in W \} \newline [/tex]

2) Not subspace since x1 can't be 0

[tex] 3) (2c,-3c) + (x_1,x_2)= (2c+x_1,-3c+x_2) \in W \} \newline [/tex]
[tex] x(2c,-3c) = (2c(x),-3c(x)) \in W \} \newline [/tex]

4) Not a subspace since x1 and x2 can't be 0.

Am I on the right track with this? Thanks
 
Mathematics news on Phys.org
  • #2
needhelp83 said:
Determine if following subsets of R^2 are subspaces of R^2. If the subset is a subspace show that it is closed under vector addition and scalar multiplication. If the subset is not a subspace show why, indicating property that fails.

[tex] 1) W=\{ \left (x_1,0)\left| x_1\in\Re\} \newline [/tex]
[tex] 2) W=\{ \left (x_1,0)\left| x_1 > 0\} \newline[/tex]
[tex]3) W=\{ \left (2c,-3c)\left| c \in\Re\} \newline[/tex]
[tex]4) W=\{ \left (x_1,x_2)\left| x_1 > 0, x_2>0\} \newline [/tex]

Answers:
[tex] 1) (x,0) + (y,0)= (x+y,0) \in W \} \newline [/tex]
[tex] c(x,0) = (c(x),0) \in W \} \newline [/tex]
Yes, you have shown that the set is closed under addition and scalar multiplication.

2) Not subspace since x1 can't be 0
And what property of a subspace does that violate? It would be sufficient to point out that the 0 vector (0, 0) is not in that set. Since your problem specifically referred to "closed under addition and scalaar multiplication, I think it would be better to show that 0(x_1, 0)= (0, 0) is not in the set so it is not closed under scalar multiplication.

[tex] 3) (2c,-3c) + (x_1,x_2)= (2c+x_1,-3c+x_2) \in W \} \newline [/tex]
[tex] x(2c,-3c) = (2c(x),-3c(x)) \in W \} \newline [/tex]
This is the only one where I think you have lost track of what you are doing- what are x_1 and x_2? To show "closed under addition", you need to look at (2c, -3c)+ (2d, -3d) where c and d are real numbers.

4) Not a subspace since x1 and x2 can't be 0.
Again, since closure under addition and scalar multiplication are specifically mentioned in the problem it would be better to show that 0(x1, x2)= (0, 0) is not in the set so it is not closed under scalar multiplication.

Am I on the right track with this? Thanks
 
  • #3
Okay for the scalar multiplication:


3)Since [tex]c_2 \ in \ R^2[/tex]
[tex] c_2(2c_1,-3c_1) = (2c_1c_2,-3c_1c_2) \in W \} \newline [/tex]

Is this better?
 
  • #4
needhelp83 said:
Okay for the scalar multiplication:


3)Since [tex]c_2 \ in \ R^2[/tex]
[tex] c_2(2c_1,-3c_1) = (2c_1c_2,-3c_1c_2) \in W \} \newline [/tex]

Is this better?
Yes, now what about addition?
 

What is a vector?

A vector is a mathematical object that has both magnitude (size) and direction. It is typically represented by an arrow, where the length of the arrow represents the magnitude and the direction of the arrow represents the direction.

How do you add vectors?

To add two vectors, you must add the corresponding components of each vector. For example, if you have two vectors a = (3,2) and b = (1,4), their sum a + b would be (3+1, 2+4) = (4,6).

What is the difference between a scalar and a vector?

A scalar is a quantity that has only magnitude, while a vector has both magnitude and direction. Examples of scalars include temperature, mass, and time. Examples of vectors include displacement, velocity, and force.

Can vectors be multiplied?

Yes, vectors can be multiplied in two ways: scalar multiplication and vector multiplication. Scalar multiplication involves multiplying a vector by a scalar (a real number), resulting in a new vector with the same direction but a different magnitude. Vector multiplication, also known as the dot product, involves multiplying two vectors to produce a scalar value.

How are vectors used in real life?

Vectors are used in many fields, including physics, engineering, and computer graphics. They can be used to represent physical quantities such as forces and velocities, or to model complex systems and processes. In everyday life, vectors are used in navigation, sports, and video games.

Similar threads

Replies
3
Views
576
  • General Math
Replies
1
Views
690
Replies
9
Views
4K
  • General Math
Replies
1
Views
560
Replies
4
Views
1K
  • General Math
Replies
20
Views
1K
  • General Math
Replies
2
Views
943
Replies
2
Views
1K
Replies
1
Views
779
  • General Math
Replies
3
Views
1K
Back
Top