When proving linear independence in a complex vector space can we assume the scalars that we need to show equal to zero are real? i.e then the result would be

If we have $v_1,v_2,...,v_k \in \mathbb{C}$ with scalars $\alpha_1, \alpha_2,...,\alpha_k\in \mathbb{R}$ then $v_1,v_2,...,v_k$ is called linearly independent if $$\sum_{i=1}^{k}\alpha_i v_i=0\Rightarrow \alpha_i=0 $$

The example I am stuck on is this

$v_1=(1, -i, 1+i)^t, \;\;\;v_2=(1-i, 1+i,0)^t,\;\;\;v_3=(-1,i,1+i)^t$

Using the definition I get: $$1) \;\; \alpha+\beta-i\beta-\gamma=0$$ $$2)\; -i\alpha+\beta+i\beta +i\gamma=0$$ $$3)\;\; \alpha+i\alpha+\gamma+i\gamma=0$$ Now if it is possible to assume $\alpha,\beta,\gamma$ real then I can equate real and imaginary parts and receive $\beta=0$ just from the first equation. Is this correct? If not how would I go about solving this?

  • 3
    $\begingroup$ The usual definition of a complex vector space states that the scalars are arbitrary elements of $\Bbb C$, not $\Bbb R$. $\endgroup$ – Omnomnomnom Nov 30 '14 at 15:44
  • $\begingroup$ Your definition treats $\Bbb C$ as a (2-dimensional) real vector space $\endgroup$ – Omnomnomnom Nov 30 '14 at 15:45
  • $\begingroup$ The usual ways of solving a problem like this are Gaussian elimination and computing the determinant. $\endgroup$ – Omnomnomnom Nov 30 '14 at 15:47
  • $\begingroup$ Thanks for clearly it up, I'll try gaussian $\endgroup$ – George1811 Nov 30 '14 at 15:48

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