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This is the problem in my current homework:

"Let V be a $\mathbb{K}$-inner product space and let $U ⊆ V$ be a subset. We define the orthogonal complement $U^⊥$ of $U$ as $ U^⊥ := \{v ∈ V : ∀u ∈ U : \left\langle u, v \right\rangle = 0\}. $

For V = $\mathbb{C^3}$ equipped with the standard dot product calculate $\{z ∈ \mathbb{C^3}: z_1 + iz_2 + 2z_3 = 0\}^⊥$ "

I genuinely have no idea what I'm supposed to do here. Like... at all.

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2 Answers 2

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Use the fact that $z_1 + iz_2 + 2z_3=z_1\cdot 1 + z_2\cdot (-\bar{i}) + z_3\cdot 2=\left\langle (z_1,z_2,z_3),(1,-i,2) \right\rangle$ and you're done.

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  • $\begingroup$ Sorry if I sound stupid, but I think I still don't understand this topic enough to know how to proceed... I'd be really thankful if you could elaborate $\endgroup$
    – pavus
    Jan 21, 2018 at 16:31
  • $\begingroup$ @pavus As $\left\langle (z_1,z_2,z_3),(1,-i,2) \right\rangle=0$, we have that $(z_1,z_2,z_3)$ is orthogonal to $(1,-i,2)$, which means that the orthogonal complement of these $(z_1,z_2,z_3)$ is going to be the span of $(1,-i,2)$. $\endgroup$ Jan 21, 2018 at 16:34
  • $\begingroup$ Ok I understand it now, thank you! $\endgroup$
    – pavus
    Jan 21, 2018 at 16:36
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Hint: $ z_1+i z_2+2z_3 $ is a dot product...

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