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Basis in C-vector space |
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| Jan26-09, 03:54 AM | #1 |
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Basis in C-vector space
1. The problem statement, all variables and given/known data
determine whether or not the given set is a basis for C^3 ( as a C-vector space) (a) {(i,0,-1),(1,1,1),(0,-i,i)} (b) {(i,1,0),(0,0,1)} 2. Relevant equations 3. The attempt at a solution All I did was to put the 3 vectors in part (a) into a matrix as 3 columns. Then I determined that the matrix has 3 leading entries, hence it is a basis. But when I tried using the same method for part (b), it doesnt work. Why is that so? |
| Jan26-09, 09:38 AM | #2 |
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The set in b has only two vectors, which isn't enough for a basis for C^3. There are some vectors in C^3 that aren't any linear combination (i.e., a sum of (complex) scalar multiples of (i, 1, 0) and (0, 0, 1).
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| Jan26-09, 09:41 AM | #3 |
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But what i did for part(a) is right?
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| Jan26-09, 05:44 PM | #4 |
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Basis in C-vector space
Assuming your work is correct, yes. A basis for C^3 has to have three vectors in it. If you have three vectors that are linearly independent, that's a basis.
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