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Math Help - Linear independence and basis for complex vector spaces

  1. #1
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    Linear independence and basis for complex vector spaces

    Determine if the following vectors are linearly independent(1, i, 0); (0, 1+i, 2); (-1+i, 1+i, -1)

    Do they form a basis for R3?

    Do the vectors form an orthogonal or orthonormal basis for C3?

    Thank you.
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  2. #2
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    Re: Linear independence and basis for complex vector spaces

    Hey christianwos.

    Have you put these entries in a 3x3 matrix (each cell has a complex number) and tried to row-reduce the matrix? The results will tell how many linearly independent vectors you have and what a basis for these vectors are (as well as if they are real or complex)
    Thanks from christianwos
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  3. #3
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    Re: Linear independence and basis for complex vector spaces

    You are right. I haven't thought of that. I was confused because I am dealing with complex numbers. To check that they form a basis for C^3 I need to verify that any complex vector in C^3 can be expressed as a combination of these vectors, right? Therefore I should set the 3x3 matrix equal to some complex vector?

    Thanks again.
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  4. #4
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    Re: Linear independence and basis for complex vector spaces

    1. they do NOT form a basis for R3. they aren't even IN R3 (the entries are complex, not real).

    2. to check if they form a basis for C3, it suffices to check linear independence OR spanning, since we have a 3-element set, and dim(C3) = 3.

    personally, i find linear independence easier to check. if you form a 3x3 complex matrix A from these vectors, you can do one of 2 things:

    a) verify rank(A) = 3 (this is the same as checking spanning).

    b) verify det(A) ≠ 0 (this is the same as checking linear independence).

    row-reduction does (a). the equivalence of these two (in this case) is obtained from the rank-nullity theorem. row-reduction may take many steps, the determinant can be found by forming 6 triple products and then summing (with the proper signs), 17 operations at most.

    whichever you do, double-check your calculations.

    3. they obviously do not form an orthonormal set (under the STANDARD complex inner product), as the vector (1,i,0) has norm √2: (√[12 + i(-i) + 02] = √(1+1) = √2).

    they do not even form an orthogonal set, since <(1,i,0),(0,1+i,2)> = (1)(0) + (i)(1-i) + (0)(2) = 1+i ≠ 0.
    Thanks from christianwos
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