Determining vector spaces

[burt]

I was given the following question:
I don't understand how to solve it. Can someone solve it, spelling out the steps? I would like to know how to solve problems like this in the future, and seeing a step by step solution to this would be very helpful.

 

[Steven G]

First you need to list the properties of a vector space.
Can you please list them.

For example if V1 and V2 are in W is in V1 + V2 in W?

Note that q = -s/2 + 3p/2 = -s/2 +3/2(-s+3r) = -2s + 9r/2 and p = -s + 3r.

 

[burt]

r

First you need to list the properties of a vector space.
Can you please list them.

For example if V1 and V2 are in W is in V1 + V2 in W?

The properties of a vector space are:
Include the zero vector (this does)
Closed vector addition
Closed scalar multiplication.

Meaning any vector in the space can be added to any other vector in the space and remain in the space, and any vector in the space can be multiplied by any real number and remain in the space.

 

[pka]

I was given the following question: View attachment 20982
I don't understand how to solve it. Can someone solve it, spelling out the steps? I would like to know how to solve problems like this in the future, and seeing a step by step solution to this would be very helpful.

The properties of a vector space are:
Include the zero vector (this does)
Closed vector addition
Closed scalar multiplication.
Meaning any vector in the space can be added to any other vector in the space and remain in the space, and any vector in the space can be multiplied by any real number and remain in the space.

It appears that you want to prove that \(W\) is a subspace of \(\Re^4\)?
Now if \(X~\&~Y\) are vectors in \(W\) and \(\alpha\) is a scalar, can you prove \(X+\alpha Y\in W~?\)
If you can do that then \(W\) is a subspace.

 

[Steven G]

So what might arbitrary vectors X & Y in W look like?

 

[burt]

So what might arbitrary vectors X & Y in W look like?

In order to do that should I pick numbers or just arbitrary letters - like x=[a,b,c,d]?

 

[burt]

So what might arbitrary vectors X & Y in W look like?

Meaning: I can just pick vectors [0,0,0,0]=X and [1,1,2/3,1]=Y. Any scalar multiplied by x and added to y is within the subspace. How can I prove the same thing is true with Y?

 

[Steven G]

Meaning: I can just pick vectors [0,0,0,0]=X and [1,1,2/3,1]=Y. Any scalar multiplied by x and added to y is within the subspace. How can I prove the same thing is true with Y?

Those vectors are not arbitrary! They are very specific vectors.

I will give you an example. Suppose you want to show that W = (a, b, a+b, a-b) is a subspace of R^4.

You can pick X = (a, b, a+b, a-b) and Y = (r, s, r+s, r-s). Let α be in R. All you need to do is verify that αX + Y is in W.

αX + Y = (αa, αb, α(a+b), α(a-b)) + (r, s, r+s, r-s) = (αa+r, αb+s, α(a+b) + (r+s), α(a-b) + (r-s)). Is this result in W?

Let us know if this is clear and if W is a subspace of R^4

 

[burt]

Those vectors are not arbitrary! They are very specific vectors.

I will give you an example. Suppose you want to show that W = (a, b, a+b, a-b) is a subspace of R^4.

You can pick X = (a, b, a+b, a-b) and Y = (r, s, r+s, r-s). Let α be in R. All you need to do is verify that αX + Y is in W.

αX + Y = (αa, αb, α(a+b), α(a-b)) + (r, s, r+s, r-s) = (αa+r, αb+s, α(a+b) + (r+s), α(a-b) + (r-s)). Is this result in W?

Let us know if this is clear and if W is a subspace of R^4

Thanks! This is starting to make sense to me. So I can use arbitrary vectors (a,b,c,d) and (p,q,r,s) and then just check to make sure that my linear combination fits in the given definition of the subspace.

 

[pka]

Thanks! This is starting to make sense to me. So I can use arbitrary vectors (a,b,c,d) and (p,q,r,s) and then just check to make sure that my linear combination fits in the given definition of the subspace.

You can use any two vectors, say \(\left[ {\begin{array}{*{20}{c}} a \\ b \\ c \\ d \end{array}} \right]\quad \left[ {\begin{array}{*{20}{c}} m \\ n \\ s \\ t \end{array}} \right]\) as long as the conditions are correct.
i.e. \(-3a+2b=-d~\&~a=-d+3c\\-3m+2n=-t~\&~m=-t+3s\)