Dimension of a subspace question

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SUMMARY

The dimension of the subspace spanned by the column vectors v1, v2, and v3 is determined to be 2, as established through row reduction of the 3x3 matrix formed by these vectors. The resulting upper triangular matrix indicates a rank of 2, confirming that there are two linearly independent vectors. The basis vectors can be identified as the two nonzero rows of the reduced matrix. This method effectively clarifies the relationship between the rank and the dimension of the span.

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  • Understanding of linear algebra concepts, specifically vector spaces and spans.
  • Familiarity with matrix operations, including row reduction and upper triangular form.
  • Knowledge of the rank-nullity theorem and its implications for vector dimensions.
  • Proficiency in identifying basis vectors from a matrix.
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  • Study the process of row reduction in detail, focusing on Gaussian elimination techniques.
  • Learn about the rank-nullity theorem and its applications in linear algebra.
  • Explore the concept of basis vectors and how to derive them from different types of matrices.
  • Investigate the implications of linear independence in the context of vector spaces.
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de1337ed
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I'm a little confused about some of the matrix terminology.
I have the following subspace:

span{v1, v2, v3} where v1, v2, v3 are column vectors defined as:

v1 = [1 2 3]
v2 = [4 5 6]
v3 = [5 7 9]
(pretend they are column vectors)

How am I supposed to find the dimension of the span?

My Work:

I created a 3x3 matrix using the column vectors, then I performed row operations to get it into upper triangular form. After performing these row operations, I ended up with the resulting matrix:
[1 4 5
0 -3 -3
0 0 0 ]

So because the rank(A) = 2, the dimension is 2. Am I right?
Also, how would I go about finding the basis vectors. Thank you.
 
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If you are going to do row reduction put your vectors into the matrix as rows. So yes, you'll get dimension 2. The two nonzero rows will be basis vectors.
 

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