Really quick question on linear spans

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SUMMARY

The discussion centers on determining the equality of linear spans using reduced row echelon form (RREF). To check if span([4, 0, -3], [2, 2, 1]) equals span([2, -2, -4], [0, 1, 5]), one must compute the RREF of both sets of vectors. If both reduce to the same RREF, they represent the same span, confirming their equivalence. The principles of equivalence in vector spaces are also highlighted, emphasizing that if two matrices have the same RREF, they yield the same solution sets for any vector b.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically linear spans and vector spaces.
  • Familiarity with reduced row echelon form (RREF) and its significance in matrix equivalence.
  • Knowledge of matrix operations and their implications on solution sets.
  • Basic understanding of equivalence relations in mathematics.
NEXT STEPS
  • Learn how to compute the reduced row echelon form (RREF) of matrices using Gaussian elimination.
  • Study the properties of vector spaces, including basis and dimension.
  • Explore the concept of linear independence and dependence among vectors.
  • Investigate the implications of equivalence relations in linear algebra, particularly in relation to solution sets.
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Students and professionals in mathematics, particularly those studying linear algebra, as well as educators seeking to clarify concepts of linear spans and matrix equivalence.

NeonVomitt
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If I want to find if

span ([4, 0, -3], [2,2,1]) = span ([2,-2,-4], [0,1,5]) do I first find their reduced row echelon form, and then see if they match? For instance, if I found both matrices to reduce to:
[ 1 0]
[ 0 1]
[ 0 0]

does that mean that they equal each other? Or do I have to do something else?

Also what is the vector space? Is it R2 in this case?
 
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Essentially, if two matrices are equivalent, then augmenting them with the same vector will give the same solution set. Another way of saying this is that if A ~ B, then in Ax = b and Bx = b, where b can be anything, if x satisfies the first equality, then it satisfies the second, and vice-versa. Also note that when there are no solutions in one, there are no solutions in the other.

If the rref of A is C and the rref of B is also C, then A ~ C and B ~ C. So we have that Ax = b, Cx = b, Bx = b, have the same solutions for x given any vector b (again if there are no solutions for particular b, none of the equations will have a solution). More elaborately, equivalence is transitive and reflexive, so A ~ C and B ~ C implies A ~ B.

That tells us that a vector b that has a solution in Ax = b, also has a solution in Bx = b. And vice-versa. Finding a linear combination of a set {v1, v2, ..., vn} of vectors equal to a vector b, amounts to solving [v1 v2 v3 ... vn]x = b. If {w1, w2, ..., wn} is another set of vectors, we would solve [w1 w2 ... wn]x = b. If [v1 v2 v3 ... vn] ~ C and [w1 w2 ... wn] ~ C then, by the above, one matrix has a solution when the other does. This means when there's a linear combination in {v1, v2, ..., vn} equal to a vector b, there's also one in {w1, w2, ... , wn} and vice-versa. This shows you that {v1, v2, ..., vn} and {w1, w2, ... , wn} have the same span.
 
NeonVomitt said:
If I want to find if

span ([4, 0, -3], [2,2,1]) = span ([2,-2,-4], [0,1,5]) do I first find their reduced row echelon form, and then see if they match? For instance, if I found both matrices to reduce to:
[ 1 0]
[ 0 1]
[ 0 0]

does that mean that they equal each other? Or do I have to do something else?

Remember that row operations transform the row vectors to an equivalent basis, so you look at the rref of

[4, 0, -3]
[2, 2, 1]

etc.
 

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