Calculate dim(U) and company - linear algebra

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Homework Statement


Before I start, I should mention that I'm taking linear algebra in French so I hope I translated all the terms correctly. I'm studying for a test and here is a question i'm having a little trouble with...hope you can help!

U = span{[1 1 1 1]T, [-2 -2 1 1]T, [3 3 3 1]T, [0 0 1 0]T}

a) Calculate dim(U) and give a base of U.
b) Complete the base found in a) to obtain a base of R4.

Homework Equations





The Attempt at a Solution



a) I get a base of (-5/6, -1/3, -1/2, 1) so dim(U) has dim(U) = 1

b) This is where I don't know what to do. How to you complete a base? I don't need the answer, I just want to know how to complete a base. The only thought I had was to use the identity matrix but then again, I really don't know what to do or how to approach this. Any help is appreciated!
 

Answers and Replies

  • #2
Dick
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You seem to be misunderstanding the question since your answer to a) is not at all right. What you want to do is take the 4 vectors in your set and find a minimal set of linearly independent vectors with the same span. The usual technique to do this is called 'row reduction' (sorry I don't know the french!).
 
  • #3
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The dimension is determined by the smallest amount of linearly independent vectors that you can find, which describe entirely U. To do so, reduction is the way out.

marlon
 
  • #4
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hmm I thought thats what I did...

I reduced the matrix from:
1 -2 3 0
1 -2 3 0
1 1 3 1
1 1 1 0

to get..

1 0 0 5/6
0 1 0 1/3
0 0 1 1/2
0 0 0 0 <--- isn't this the "row reduction" you were talking about?
 
  • #5
Dick
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That's sort of it. Though I usually write the vectors as rows. Since you've written them as columns and then (I think) row reduced, it's a little hard for me to interpret the results exactly - but I can clearly see that U has dimension 3. (How many obvious independent columns in your set?). Try the same exercise writing the vectors as rows.
 
  • #6
HallsofIvy
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Yes, that is the "row reduction" but that does NOT mean that the last column gives you the only basis vector. In fact, since you have 3 non-zero rows left, that tells you that the dimension of this subspace is 3.

Use the given vectors as rows, not columns, and row-reduce. The remaining non-zero rows will be the basis vectors.
 
  • #7
radou
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The "row reduction" result doesn't look correct, unless I missed something. Go through it again.

Edit: actually, it's almost correct, you missed a sign (figure out which one!) So, you have shown that the vector (0, 0, 1, 0) can be represented as a linear combination of the other vectors, so the set without the vector is a span too, and linearly independent, which makes it a basis for U.
 
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