Finding the Dimension of a Vector Space: The Matrix Method

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SUMMARY

The discussion focuses on determining the dimension of vector spaces using the matrix method. For part A, the dimension of the subspace U, defined as the span of the vectors {t+2u+v+w, t+3u+v+2w, 3t+4u+2v, 3t+5u+2v+w}, is found by forming a matrix with these vectors as columns and row reducing to identify linearly independent vectors. In part B, the dimension of the vector subspace V, defined as the span of the vectors {(-1,2,3,0),(5,4,3,0),(3,1,1,0)} in R^4, can be computed similarly, with the vectors forming the matrix as rows or columns yielding the same result due to their linear independence.

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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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