Vector Spaces, Dimension of Subspace

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Discussion Overview

The discussion revolves around finding the dimension of a subspace spanned by given vectors in the context of linear algebra. Participants explore methods for determining linear dependence and independence of the vectors, as well as the implications for the dimension of the subspace.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant asks how to find the dimension of the subspace spanned by the vectors and seeks clues on the process.
  • Another participant suggests using row echelon form of a matrix constructed from the vectors to determine their dimension and notes that the vectors span either 2 or 3-dimensional subspaces.
  • A different participant expresses uncertainty about the methods taught in class, indicating a lack of familiarity with row echelon form and asking for clarification on determining linear independence.
  • One participant explains that the dimension of a vector space is defined by the number of vectors in a basis, which must span the space and be independent, and emphasizes the importance of determining the independence of the vectors.
  • Another participant proposes a specific method to show the dependence of the first three vectors by manipulating their entries, concluding that the dimensions are 2 and 3.

Areas of Agreement / Disagreement

Participants express varying levels of familiarity with the concepts and methods involved, leading to some disagreement on the best approach to determine the dimension and independence of the vectors. No consensus is reached on the specific dimensions of the subspaces.

Contextual Notes

Participants mention the need for row reduction techniques and the concept of linear dependence, but there is uncertainty regarding the application of these methods and the definitions involved.

ashnicholls
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Find the dimension of the subspace spanned by the vectors u, v, w in each of the following cases:
i) u = (1,-1,2)^T v = (0,-1,1)^T w = (3,-2, 5)^T
ii) u = (0,1,1)^T v = (1,0,1)^T w = (1,1,0)^T

Right, how do I go about this, do I have to find the subspace first then do the dimension.

Can someone give me sum clues.

Cheers
 
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Posting it in linear algebra might be more appropriate.

Anyway, just put the vectors as rows in a matrix, and put it in row echelon form - that tells you the dimension and the space. Just like you ought to have been told in class.Of course, clearly those span 2 or 3 dimensional vector subspaces, so you only need to verify linear dependence/independence, which might be easier for you if you're not familiar with row echelon form. The last three are clearly linearly independent by inspection.
 
Sorry, did not know where to put it, an I did a search on vectors, and it showed up in this section, that's why I did it.

We are not taught anything is class, we are just given work books, and then an assignment, this is one of the final questions in my work book.

And I can not remember row echelon form as such. So what I find whether they are linearly independent or not and then what?

Thanks for your help.
 
I've moved this to the "Linear and Abstract Algebra" forum.

The dimension of a vector space is the number of vectors in a basis. A basis must both span the space and be independent. Since each of these vectors spaces is spanned by three vectors, the dimension cannot be more than 3. Now determine whether or not these vectors are independent. there are many ways to do that but one, as matt grime said, to construct a matrix having the vectors as rows. The number of independent vectors is the number of non-zero rows you have left after row reducing and therefore the dimension of the space. I strongly recommend you review "row reduction"!
 
the only way to ,show the first three are dependent would be to kill off the first entry among the two that have a first entry, so the only possible way is to multiply the first one by -3.

adding that multiple to the last one then does it, since it gives a multiple of the second one.

so the dimensions are 2,3.
 

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