Finding the Dimension of a Vector Space: The Matrix Method

In summary, when finding the dimension of a vector space spanned by a set of vectors, it is necessary to determine the linear independence of the vectors. This can be done by forming a matrix with the vectors as either columns or rows and then row reducing. The resulting linearly independent vectors form a basis for the vector space, and the dimension of this basis is equal to the dimension of the vector space. It does not matter whether the matrix is formed with rows or columns, as the resulting dimension will be the same.
  • #1
hitemup
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Homework Statement



A. Let {t,u,v,w} be a basis for a vector space V. Find dim(U) where

U = span{t+2u+v+w, t+3u+v+2w, 3t+4u+2v, 3t+5u+2v+w}

B. Compute the dimension of the vector subspace V= span{(-1,2,3,0),(5,4,3,0),(3,1,1,0)} of R^4

Homework Equations

The Attempt at a Solution



I know that the dimension is the number of vectors in a basis. Since we're given a span, all we need to do is determine linearly independent vectors. But there's something confuses me in these two questions. I've got the solutions for them and in the first one, it forms a matrix whose columns are the vectors of the span, and in the second one, it constructs a matrix whose rows are the given vectors of the span to get the linearly independent vectors. So my question is simply, when to form the matrix as columns, and as rows?
 
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  • #2
The answer to part A is correct.

If you were given ##U = \text{span} \{ v_1, v_2, v_3, v_4 \}##, and were asked to find ##\text{dim}(U)##, then forming a matrix and row reducing would yield a linearly independent basis for ##U##. Call this basis ##B##, then ##\text{dim}(B) = \text{dim}(U)##.

As for part B, the same logic applies. I don't think it matters how you reduce the vectors, i.e. it doesn't matter if you reduce the row vectors or the column vectors. The dimension of the basis will still be the same because the vectors forming it are linearly independent.
 
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  • #3
Zondrina said:
The answer to part A is correct.

If you were given ##U = \text{span} \{ v_1, v_2, v_3, v_4 \}##, and were asked to find ##\text{dim}(U)##, then forming a matrix and row reducing would yield a linearly independent basis for ##U##. Call this basis ##B##, then ##\text{dim}(B) = \text{dim}(U)##.

As for part B, the same logic applies. I don't think it matters how you reduce the vectors, i.e. it doesn't matter if you reduce the row vectors or the column vectors. The dimension of the basis will still be the same because the vectors forming it are linearly independent.

So there is not any difference between forming the matrix as whether rows or columns and then row reduce, is there?
EDIT:
It turns out we can do it in both ways as you suggest, thank you.
 
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1. What is the "Dimension of a Span" guide about?

The "Dimension of a Span" guide is a comprehensive resource for understanding the concept of dimension, specifically in the context of linear algebra and vector spaces. It covers topics such as basis, linear independence, and spanning sets.

2. Who is the target audience for this guide?

This guide is primarily aimed at students and researchers in mathematics, physics, and engineering, as well as anyone interested in learning about dimensions and vector spaces.

3. What makes this guide different from other resources on dimensions and vector spaces?

This guide provides a clear and concise explanation of dimension, accompanied by numerous examples and exercises to reinforce understanding. It also includes visual representations and interactive demonstrations to aid in conceptual understanding.

4. Do I need any prior knowledge in linear algebra to understand this guide?

While some familiarity with linear algebra may be helpful, this guide is designed to be accessible to anyone with a basic understanding of algebra and mathematical concepts. It starts with the fundamentals and gradually builds upon them, making it suitable for beginners.

5. Can I use this guide for self-study?

Yes, this guide is designed for self-study and can be used as a supplement to textbooks and lectures. It is also a valuable resource for review and practice for those studying for exams or looking to refresh their understanding of dimensions and vector spaces.

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