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Orthogonal complement of a subspace

  • Thread starter bcjochim07
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  • #1
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Homework Statement


Let W be the subspace spanned by the given column vectors. Find a basis for W perp.

w1= [2 -1 6 3] w2 = [-1 2 -3 -2] w3 = [2 5 6 1]
(these should actually be written as column vectors.




Homework Equations





The Attempt at a Solution



So, I put these vectors into a matrix and took its transpose since the orthogonal complement of the column space of a matrix equals the null space of the transpose.

I row reduced the transpose and got null(A transpose) = span{ [-4 1 0 3] , [-3 0 1 0]}
(Again, these should be written as column vectors)

This is the correct answer, but I thought that I should have gotten a null space with dimension one. The three vectors that span W "live" in R4 and the basis for W has dimension three. 4 - 3 =1, so shouldn't the dimension of W perp = 1?
 

Answers and Replies

  • #2
matt grime
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You're sure that w1, w2 and w3 actually span a 3 dimensional space, are you?
 
  • #3
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That would only be true if the three original vectors were linearly independent (i.e., if the space they span had dimension 3). Is this true?
 
  • #4
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Ok, I see. 3*w1 + 4*w2 = w3. I guess I just assumed that when it said a subspace W spanned by the vectors w1,w2,& w3, that those vectors would form a basis for W.
 

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