• Support PF! Buy your school textbooks, materials and every day products Here!

Basis of a Vector Space

  • Thread starter Izzy
  • Start date
  • #1
6
0

Homework Statement


Let V be a vector space, and suppose that [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] is a basis of V. Let [itex]c\in\mathbb R[/itex] be a scalar, and define [itex]\vec{w}[/itex] = [itex]\vec{v_1} + c\vec{v_2}[/itex]. Prove that [itex]\vec{w}, \vec{v_2}, ... , \vec{v_n}[/itex] is also a basis of V.

Homework Equations


If two of the following conditions hold, the third holds automatically, and the list is a basis for a vector space V.
1. If dimV = the number of vectors in the list.
2. If the list of vectors spans V.
3. If the list of vectors is linearly independent.


The Attempt at a Solution


So, I feel like I did this right, but I have no way to check because we don't use a textbook for the class and I couldn't find it online anywhere. I did the following.

Since [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] is a basis of V, [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] are linearly independent and span V. Also, dimV = n, because n vectors form the basis of V.
[itex]\vec{w}, \vec{v_2}, ... , \vec{v_n}[/itex] is also a list of n vectors, so it will be sufficient to show that this list either spans V or is linearly independent in order for it to be a basis of V.
Since [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] span V, that means there exists some scalars ai (where i is between 1 and n inclusive) such that
a1v1 + a2v2 + ... + anvn = [itex]\vec{x}[/itex], where [itex]\vec{x}[/itex] is in V.
So, for [itex]\vec{w}, \vec{v_2}, ... , \vec{v_n}[/itex] to span V, there must exist some scalars bi such that b1w + b2v2 + ... + bnvn = [itex]\vec{x}[/itex].

Then I basically expand this, and set it equal to the previous expression with all the a's. I show that in order for them to be equal, choose b3 = a3, b4 = a4, ... bn = an.

Then all we have to show is that
b1v1 + (b1c + b2)v2 = a1v1 + a2v2.
Choose b1 = a1, and b2 = a2 - a1c.

Then a1w+ (a2 - a1c)v2 + a3v3 + ... + anvn = [itex]\vec{x}[/itex], so it spans V.

Thus, since the dimension if n and it spans V, it also forms a basis of V.

Is this right, though?

 

Answers and Replies

  • #2
740
13
Looks ok to me.... :rolleyes:
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

Homework Statement


Let V be a vector space, and suppose that [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] is a basis of V. Let [itex]c\in\mathbb R[/itex] be a scalar, and define [itex]\vec{w}[/itex] = [itex]\vec{v_1} + c\vec{v_2}[/itex]. Prove that [itex]\vec{w}, \vec{v_2}, ... , \vec{v_n}[/itex] is also a basis of V.

Homework Equations


If two of the following conditions hold, the third holds automatically, and the list is a basis for a vector space V.
1. If dimV = the number of vectors in the list.
2. If the list of vectors spans V.
3. If the list of vectors is linearly independent.


The Attempt at a Solution


So, I feel like I did this right, but I have no way to check because we don't use a textbook for the class and I couldn't find it online anywhere. I did the following.

Since [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] is a basis of V, [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] are linearly independent and span V. Also, dimV = n, because n vectors form the basis of V.
[itex]\vec{w}, \vec{v_2}, ... , \vec{v_n}[/itex] is also a list of n vectors, so it will be sufficient to show that this list either spans V or is linearly independent in order for it to be a basis of V.
Since [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] span V, that means there exists some scalars ai (where i is between 1 and n inclusive) such that
a1v1 + a2v2 + ... + anvn = [itex]\vec{x}[/itex], where [itex]\vec{x}[/itex] is in V.
So, for [itex]\vec{w}, \vec{v_2}, ... , \vec{v_n}[/itex] to span V, there must exist some scalars bi such that b1w + b2v2 + ... + bnvn = [itex]\vec{x}[/itex].

Then I basically expand this, and set it equal to the previous expression with all the a's. I show that in order for them to be equal, choose b3 = a3, b4 = a4, ... bn = an.

Then all we have to show is that
b1v1 + (b1c + b2)v2 = a1v1 + a2v2.
Choose b1 = a1, and b2 = a2 - a1c.

Then a1w+ (a2 - a1c)v2 + a3v3 + ... + anvn = [itex]\vec{x}[/itex], so it spans V.

Thus, since the dimension if n and it spans V, it also forms a basis of V.

Is this right, though?
There is a theorem that a set of ##n## linearly-independent vectors in an ##n##-dimensional space ##V## does, indeed, span ##V##. If you know and are allowed to use that theorem, you can shorten your argument a bit: just show that ##\vec{w}, \vec{v}_2, \ldots, \vec{v}_n## are linearly independent.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
955
There is a theorem that a set of ##n## linearly-independent vectors in an ##n##-dimensional space ##V## does, indeed, span ##V##. If you know and are allowed to use that theorem, you can shorten your argument a bit: just show that ##\vec{w}, \vec{v}_2, \ldots, \vec{v}_n## are linearly independent.
The OP said in his first post
"
If two of the following conditions hold, the third holds automatically, and the list is a basis for a vector space V.
1. If dimV = the number of vectors in the list.
2. If the list of vectors spans V.
3. If the list of vectors is linearly independent. "
so, yes, he does know that.
 
  • #5
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
The OP said in his first post
"
If two of the following conditions hold, the third holds automatically, and the list is a basis for a vector space V.
1. If dimV = the number of vectors in the list.
2. If the list of vectors spans V.
3. If the list of vectors is linearly independent. "
so, yes, he does know that.
I missed the opening sentence.
 
  • #6
6
0
There is a theorem that a set of ##n## linearly-independent vectors in an ##n##-dimensional space ##V## does, indeed, span ##V##. If you know and are allowed to use that theorem, you can shorten your argument a bit: just show that ##\vec{w}, \vec{v}_2, \ldots, \vec{v}_n## are linearly independent.
I tried it like that first, but I couldn't actually figure out how to show that ##\vec{w}, \vec{v}_2, \ldots, \vec{v}_n## are linearly independent. Can I have a hint for how to do that?

Also, I'm a girl btw. :P
 
  • #7
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
I tried it like that first, but I couldn't actually figure out how to show that ##\vec{w}, \vec{v}_2, \ldots, \vec{v}_n## are linearly independent. Can I have a hint for how to do that?

Also, I'm a girl btw. :P
The message you are quoting above said nothing about your gender!

Anyway, you want to know if there are ##c_1, c_2, \ldots, c_n##, not all zero, such that ##0 = c_1 w + c_2 v_2 + \cdots + c_n v_n = 0##. Take it from there.
 
  • #8
6
0
The message you are quoting above said nothing about your gender!

Anyway, you want to know if there are ##c_1, c_2, \ldots, c_n##, not all zero, such that ##0 = c_1 w + c_2 v_2 + \cdots + c_n v_n = 0##. Take it from there.
I think I got it.
##0 = c_1 w + c_2 v_2 + \cdots + c_n v_n = 0##
Let ##c_3, ... c_n = 0##, and now we'll show that ##c_1w + c_2v_2## = 0.
##c_1(v_1+cv_2) + c_2v_2 = 0##
##c_1v_1+c_1cv_2 + c_2v_2 = 0##
##c_1v_1 + (c_1c+c_2)v_2 = 0##
##c_1## must be 0, because it's the only way to get rid of ##v_1##, so then ##c_2## must also be 0.
Thus call the constants must be zero, so it's linearly independent?
 
  • #9
33,635
5,296
I think I got it.
##0 = c_1 w + c_2 v_2 + \cdots + c_n v_n = 0##
Let ##c_3, ... c_n = 0##, and now we'll show that ##c_1w + c_2v_2## = 0.
##c_1(v_1+cv_2) + c_2v_2 = 0##
##c_1v_1+c_1cv_2 + c_2v_2 = 0##
##c_1v_1 + (c_1c+c_2)v_2 = 0##
##c_1## must be 0, because it's the only way to get rid of ##v_1##, so then ##c_2## must also be 0.
Thus call the constants must be zero, so it's linearly independent?
I don't think you have it. In the equation #c_1v_1 + c_2v_2 + \cdots + c_nv_n = 0##, the fact that all of the constants ci = 0 means absolutely nothing. This happens when the vectors are linearly independent and when they are linearly dependent. The important fact for a set of linearly independent vectors is that there is only a single solution in the constants with all of them being zero.

##c_1 w + c_2 v_2 + \cdots + c_n v_n = 0##
##\Rightarrow c_1(v_1 + cv_2) + c_2 v_2 + \cdots + c_n v_n = 0##

I would get an equation with only the vectors v1, v2, ..., vn in it, and then use the fact that this set of vectors is linearly independent.
 
  • #10
33,635
5,296
Also, I'm a girl btw. :P
You quoted Ray Vickson, but it was HallsofIvy who said "so, yes, he does know that."
 
  • #11
6
0
So, do you mean that I should just say
##c_1(v_1 + cv_2) + c_2 v_2 + \cdots + c_n v_n = 0##
##c_1v_1 + (c_1c+c_2)v_2 + ... + c_nv_n = 0 ## is linearly independent because ##v_1, v_2, ... v_n## are linearly independent?

Also, I know HallsofIvy said it, I just didn't go back and quote him. I figured he'd figure it out lol.
 
  • #12
Dick
Science Advisor
Homework Helper
26,258
618
So, do you mean that I should just say
##c_1(v_1 + cv_2) + c_2 v_2 + \cdots + c_n v_n = 0##
##c_1v_1 + (c_1c+c_2)v_2 + ... + c_nv_n = 0 ## is linearly independent because ##v_1, v_2, ... v_n## are linearly independent?

Also, I know HallsofIvy said it, I just didn't go back and quote him. I figured he'd figure it out lol.
In the second form of the sum all of the coefficients of the v's must be zero since the v's are linearly independent. What does that tell you about ##c_1, c_3, ...##?
 
Last edited:
  • #13
33,635
5,296
So, do you mean that I should just say
##c_1(v_1 + cv_2) + c_2 v_2 + \cdots + c_n v_n = 0##
##c_1v_1 + (c_1c+c_2)v_2 + ... + c_nv_n = 0 ## is linearly independent because ##v_1, v_2, ... v_n## are linearly independent?
You can't describe an equation as "linearly independent". You're trying to show that w, v2, ..., vn are a linearly independent set.
Izzy said:
Also, I know HallsofIvy said it, I just didn't go back and quote him. I figured he'd figure it out lol.
Well, you confused Ray Vickson, whom you quoted, and who thought you were talking about him.
 
  • #14
6
0
In the second form of the sum all of the coefficients of the v's must be zero since the v's are linearly independent. What does that tell you about ##c_1, c_3, ...##?
That they're 0. So the only way to make the second sum 0 is if ##c_2## is also 0. And since the second sum is equivalent to the first sum, all the coefficients must still be 0, so w, ##v_2, ..., v_n## are linearly independent?

Well, you confused Ray Vickson, whom you quoted, and who thought you were talking about him.
Sorry Ray Vickson! Thought you were joking, not confused.
 
  • #15
33,635
5,296
That they're 0. So the only way to make the second sum 0 is if ##c_2## is also 0. And since the second sum is equivalent to the first sum, all the coefficients must still be 0, so w, ##v_2, ..., v_n## are linearly independent?
You're still missing an important concept. Given any vectors v1, v2, ..., vn in the same space, you can always write this equation: ##c_1v_1 + c_2v_2 + \cdot + c_nv_n = 0##. The vectors could be linearly independent or linearly dependent.

If the vectors are linearly independent, then there is one and only one solution, the trivial solution: c1 = c2 = ... = cn = 0. If the vectors are linearly dependent, then there is also a solution for which at least one of the constants is nonzero.
 
  • #16
Dick
Science Advisor
Homework Helper
26,258
618
That they're 0. So the only way to make the second sum 0 is if ##c_2## is also 0. And since the second sum is equivalent to the first sum, all the coefficients must still be 0, so w, ##v_2, ..., v_n## are linearly independent?
Pretty much. Since ##c_1 c + c_2=0## and ##c_1=0## the whole sum reduces to ##c_2 v_2=0##. Since ##v_2## is nonzero (why?) that tells you ##c_2## must also be zero.
 

Related Threads on Basis of a Vector Space

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
11
Views
6K
  • Last Post
Replies
11
Views
7K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
10
Views
2K
Replies
3
Views
3K
Replies
3
Views
816
Replies
1
Views
2K
Replies
2
Views
1K
Top