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 R^{3}?

Do the vectors form an orthogonal or orthonormal basis for C^{3}?

Thank you.

Printable View

- Oct 24th 2012, 04:38 PMchristianwosLinear 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 R^{3}?

Do the vectors form an orthogonal or orthonormal basis for C^{3}?

Thank you. - Oct 24th 2012, 05:08 PMchiroRe: 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) - Oct 24th 2012, 05:18 PMchristianwosRe: 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. - Oct 25th 2012, 09:26 AMDevenoRe: Linear independence and basis for complex vector spaces
1. they do NOT form a basis for R

^{3}. they aren't even IN R^{3}(the entries are complex, not real).

2. to check if they form a basis for C^{3}, it suffices to check linear independence OR spanning, since we have a 3-element set, and dim(C^{3}) = 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: (√[1^{2}+ i(-i) + 0^{2}] = √(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.