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

Vector Spaces

  • #1
1,444
0
HI. Okay
Consider
[itex]
\[\left(\begin{array}{c}
0\\
0\\
0\end{array}\right)\] ,

\[\left(\begin{array}{c}
1\\
1\\
0\end{array}\right)\] ,

\[\left(\begin{array}{c}
1\\
0\\
1\end{array}\right)\] ,

\[\left(\begin{array}{c}
0\\
1\\
1\end{array}\right)\] as a subspace of \mathbb{Z}_{2}^{3}

[/itex]
In my notes I've written that this is a 2 dimensional subspace. How?
As far as I can see they are all linealry dependent vectors as if you add 1 of each of them you get back to the zero vector. No?
 

Answers and Replies

  • #2
Defennder
Homework Helper
2,591
5
Firstly note that the zero vector (the first one) doesn't factor into the dimension of the subspace. Just look at the other 3 vectors. Note that any set of vectors containing the zero vector is necessarily linearly dependent.

What happens if you add the 2nd vector to the 3rd one? After considering this, ask yourself if it's possible to express each of the remaining (non-zero) vectors in terms of the others.

And what do you mean by "adding of each of them"? Doing so only shows that the 4 vectors are linearly dependent, but doesn't help much in showing that dim=2.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
956
A set of vectors is linearly independent if and only if the only sum [itex]a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ \a_nv_n= 0[/itex] is if [itex]a_1= a_2= \cdot\cdot\cdot= a_n= 0[/itex]. Obviously you can immediately drop the 0 vector, but you need to look at [itex]a_1(1, 1, 0)+ a_2(1, 0, 1)+ a_3(0, 1, 1)= (a_1+ a_2, a_1+ a_3, a_2+ a_3)= (0, 0, 0)[/itex].
In other words, [itex]a_1+ a_2= 0[/itex], [itex]a_1+ a_3= 0[/itex], [itex]a_2+ a_3= 0[/itex] has the obvious (trivial) solution [itex]a_1= a_2= a_3= 0[/itex]. Does it have any other, non-trivial, solutions?
 
  • #4
1,444
0
yes [itex] a_1 = a_2 = a_3 = 1 [/itex] is also a solution giving linear independence. So if the o vector doesn't contribut to dimension then as the other 3 vectors are linearly dependent, their span is a 1-dimensional subspace as opposed to a 2-dimensional one????
 
  • #5
1,631
4
yes [itex] a_1 = a_2 = a_3 = 1 [/itex] is also a solution giving linear independence. So if the o vector doesn't contribut to dimension then as the other 3 vectors are linearly dependent, their span is a 1-dimensional subspace as opposed to a 2-dimensional one????
As Halls already elaborated in detail, what you ought to do is look if the given vectors are linearly dependent or independent by considering the dependence relation, which Halls perfectly well stated. Now if W is the subspace that is spanned by the given vectors, then in order to be able to find dim(W) we shall eliminate all dependent vectors in the set. That is we shall take only those that are linearly independent. And as far as i can see, the three last vectors, are all lin. independent, which would make dim(W)=3 insdead of 2.
 
  • #6
Defennder
Homework Helper
2,591
5
As Halls already elaborated in detail, what you ought to do is look if the given vectors are linearly dependent or independent by considering the dependence relation, which Halls perfectly well stated. Now if W is the subspace that is spanned by the given vectors, then in order to be able to find dim(W) we shall eliminate all dependent vectors in the set. That is we shall take only those that are linearly independent. And as far as i can see, the three last vectors, are all lin. independent, which would make dim(W)=3 insdead of 2.
Remember that the field has only 2 elements 1 and 0, so that means that 1 is its own additive inverse. So adding vectors 2 and 3 gives 4 so dim=2.
 
  • #7
1,631
4
Remember that the field has only 2 elements 1 and 0,
Hmm, i didn't see this. Well, yeah, then like you said, dim(W) would be 2.
 

Related Threads on Vector Spaces

  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
1
Views
948
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
664
  • Last Post
Replies
22
Views
2K
  • Last Post
Replies
2
Views
686
  • Last Post
Replies
13
Views
3K
  • Last Post
Replies
2
Views
766
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
1K
Top