Show the set T is a vector subspace of R^3

[Bronn]

Show the set T is a vector subspace of R^3


T = { Sum( µ_iv_{i ,} i = 1..4 ): µ_i is an element of R, 1<= i <= 4 } where v_1, v_2, v_3, v₄ are fixed vectors in R^3


the solution so far:

so if I expand the summation and call it v

v = µ₁v₁ + µ₂v₂ + µ₃v₃ + µ₄v₄

it is this correct?

so I have to show it fulfills having the zero vector and closure under addition/multiplication by a scalar.

1. since µ_i is of R then setting those to zero give a zero vector.

2. closure under addition and multiplication by a scalar is where I'm stuck. And its probably due to me not being very comfortable with sigma notation.

If I say let u, v be an element of T and u = µ₁u₁ + µ₂u₂ + µ₃u₃ + µ₄u₄ and v = µ₁v₁ + µ₂v₂ + µ₃v₃ + µ₄v₄
then
v + u = µ₁( v₁ + u₁ ) + µ₂( v₂ + u₂ ) + µ₃( v₃ + u₃ ) + µ₄( v₄ + u₄ )

this is where I'm a bit confused (assuming I've even got anything correct to this point)

are all the vectors subscript 1 the same? since it says "where v_1, v_2, v_3, v₄ are fixed vectors in R^3 "

so are u₁ and v₁ the same fixed vector? so if that's the case then v₁ + u₁ would be the same as 2v₁ and then you can pull the factor out showing the addition of the two vectors is n element of T?


I may be way off but that's all I could think of.
cheers for any help.

 

[pka]

Show the set T is a vector subspace of R^3
T = { Sum( µ_iv_{i ,} i = 1..4 ): µ_i is an element of R, 1<= i <= 4 } where v_1, v_2, v_3, v₄ are fixed vectors in R^3
the solution so far:
so if I expand the summation and call it v
v = µ₁v₁ + µ₂v₂ + µ₃v₃ + µ₄v₄ it is this correct?
so I have to show it fulfills having the zero vector and closure under addition/multiplication by a scalar.
1. since µ_i is of R then setting those to zero give a zero vector.
2. closure under addition and multiplication by a scalar is where I'm stuck. And its probably due to me not being very comfortable with sigma notation.
If I say let u, v be an element of T and u = µ₁u₁ + µ₂u₂ + µ₃u₃ + µ₄u₄ and v = µ₁v₁ + µ₂v₂ + µ₃v₃ + µ₄v₄
then v + u = µ₁( v₁ + u₁ ) + µ₂( v₂ + u₂ ) + µ₃( v₃ + u₃ ) + µ₄( v₄ + u₄ )
this is where I'm a bit confused (assuming I've even got anything correct to this point)
are all the vectors subscript 1 the same? since it says "where v_1, v_2, v_3, v₄ are fixed vectors in R^3 "
so are u₁ and v₁ the same fixed vector? so if that's the case then v₁ + u₁ would be the same as 2v₁ and then you can pull the factor out showing the addition of the two vectors is n element of T?

Do you that if \(\displaystyle \mathcal{S}\subseteq\mathbb{R}^3\) to prove that \(\displaystyle \mathcal{S}\) is a vector space you must show:
1) \(\displaystyle 0\in\mathcal{S}\)
2) for any scalar \(\displaystyle \alpha\) and any elements \(\displaystyle \{x,y\}\subset\mathcal{S}\) then \(\displaystyle \alpha x+y \in \mathcal{S}\)
For #1 realize that any four vectors in \(\displaystyle \mathbb{R}^3\) are linearly dependent. That gives us a non-trivial sum equaling zero.
#2 is the easy part. You seem to have already done that part.

 

[Dr.Peterson]

Show the set T is a vector subspace of R^3

T = { Sum( µ_iv_{i ,} i = 1..4 ): µ_i is an element of R, 1<= i <= 4 } where v_1, v_2, v_3, v₄ are fixed vectors in R^3


If I say let u, v be an element of T and u = µ₁u₁ + µ₂u₂ + µ₃u₃ + µ₄u₄ and v = µ₁v₁ + µ₂v₂ + µ₃v₃ + µ₄v₄
then
v + u = µ₁( v₁ + u₁ ) + µ₂( v₂ + u₂ ) + µ₃( v₃ + u₃ ) + µ₄( v₄ + u₄ )

this is where I'm a bit confused (assuming I've even got anything correct to this point)

are all the vectors subscript 1 the same? since it says "where v_1, v_2, v_3, v₄ are fixed vectors in R^3 "

so are u₁ and v₁ the same fixed vector? so if that's the case then v₁ + u₁ would be the same as 2v₁ and then you can pull the factor out showing the addition of the two vectors is n element of T?

I think you've misread the description of T. The vectors v_1, v_2, v_3, v₄ are fixed - that means that they are given to you at the start; you don't get to replace them with some other vectors u_1, u_2, u_3, u₄ . What varies from one element of T to another are the µ_i. That is, the elements of T are all linear combinations of the given v's.

So two elements of T are not u = µ₁u₁ + µ₂u₂ + µ₃u₃ + µ₄u₄ and v = µ₁v₁ + µ₂v₂ + µ₃v₃ + µ₄v4, but a = µ₁v₁ + µ₂v₂ + µ₃v₃ + µ₄v₄ and b = λ₁v₁ + λ₂v₂ + λ₃v₃ + λ₄v₄.

 

[Bronn]

Do you that if \(\displaystyle \mathcal{S}\subseteq\mathbb{R}^3\) to prove that \(\displaystyle \mathcal{S}\) is a vector space you must show:
1) \(\displaystyle 0\in\mathcal{S}\)
2) for any scalar \(\displaystyle \alpha\) and any elements \(\displaystyle \{x,y\}\subset\mathcal{S}\) then \(\displaystyle \alpha x+y \in \mathcal{S}\)
For #1 realize that any four vectors in \(\displaystyle \mathbb{R}^3\) are linearly dependent. That gives us a non-trivial sum equaling zero.
#2 is the easy part. You seem to have already done that part.

I don't know all of that set notation but I know it must satisfy the 3 axioms of having a zero, closure under multiplication by scalar and closure under addition.

I think you've misread the description of T. The vectors v_1, v_2, v_3, v₄ are fixed - that means that they are given to you at the start; you don't get to replace them with some other vectors u_1, u_2, u_3, u₄ . What varies from one element of T to another are the µ_i. That is, the elements of T are all linear combinations of the given v's.

So two elements of T are not u = µ₁u₁ + µ₂u₂ + µ₃u₃ + µ₄u₄ and v = µ₁v₁ + µ₂v₂ + µ₃v₃ + µ₄v4, but a = µ₁v₁ + µ₂v₂ + µ₃v₃ + µ₄v₄ and b = λ₁v₁ + λ₂v₂ + λ₃v₃ + λ₄v₄.

oh, thanks.

so for addition:
if µ,λ are an element of R and v,u are an element of T

where u = µ₁v₁ + µ₂v₂ + µ₃v₃ + µ₄v₄ and v = λ₁v₁ + λ₂v₂ + λ₃v₃ + λ₄v₄

then

v + u = ( µ₁ + λ₁)v₁ + ( µ₂+ λ₂)v₂ + ( µ₃+ λ₃)v₃ + ( µ₄+ λ₄ )v₄

and because µ,λ are an element of R and R is closed under addition then their sums are also an element of R
and therefore v + u is an element of T, and T is closed under addition?





as for closure under multiplication by a scalar:

if v is an element of T and µ,λ are an element of R

where v = λ₁v₁ + λ₂v₂ + λ₃v₃ + λ₄v₄

then µv= µλ₁v₁ + µλ₂v₂ +µλ₃v₃ + µλ₄v₄

so because µ,λ are an element of R their products are also and therefore v is an element of T, and T is closed under multiplication by a scalar.
that right?

 

[Steven G]

I don't know all of that set notation but I know it must satisfy the 3 axioms of having a zero, closure under multiplication by scalar and closure under addition.




oh, thanks.

so for addition:
if µ,λ are an element of R and v,u are an element of T

where u = µ₁v₁ + µ₂v₂ + µ₃v₃ + µ₄v₄ and v = λ₁v₁ + λ₂v₂ + λ₃v₃ + λ₄v₄

then

v + u = ( µ₁ + λ₁)v₁ + ( µ₂+ λ₂)v₂ + ( µ₃+ λ₃)v₃ + ( µ₄+ λ₄ )v₄

and because µ,λ are an element of R and R is closed under addition then their sums are also an element of R
and therefore v + u is an element of T, and T is closed under addition?





as for closure under multiplication by a scalar:

if v is an element of T and µ,λ are an element of R

where v = λ₁v₁ + λ₂v₂ + λ₃v₃ + λ₄v₄

then µv= µλ₁v₁ + µλ₂v₂ +µλ₃v₃ + µλ₄v₄

so because µ,λ are an element of R their products are also and therefore v is an element of T, and T is closed under multiplication by a scalar.
that right?

if µ,λ are an element of R and v,u are an element of T, then what are µ_i and λ_i. There are NO µ and λ. There are scalars µ₁, µ₂, µ₃, µ₄, λ₁, λ₂, λ₃, and λ₄ and they are all in R. Also if µ,λ are in R, they are not an element of R, but rather they are elements of R.

Just clean it up a bit and you'll have it!

 

[Bronn]

if µ,λ are an element of R and v,u are an element of T, then what are µ_i and λ_i. There are NO µ and λ. There are scalars µ₁, µ₂, µ₃, µ₄, λ₁, λ₂, λ₃, and λ₄ and they are all in R. Also if µ,λ are in R, they are not an element of R, but rather they are elements of R.

Just clean it up a bit and you'll have it!

I knew I was being sloppy at the bolded but those subscripts were being a pain in the *** to type out lol


thanks again