Finding the Dimension of a Vector Space: The Matrix Method

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The discussion focuses on determining the dimension of vector spaces using matrix methods. For part A, the dimension is found by forming a matrix from the vectors in the span and row reducing to identify linearly independent vectors. In part B, the same principle applies, and it is clarified that whether vectors are arranged as rows or columns does not affect the outcome. The conclusion is that both methods yield the same dimension for the vector space, confirming the correctness of the approach. The overall understanding is that the arrangement of vectors in matrices can be flexible without impacting the result.
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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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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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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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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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