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Let $\mathbb{C}^3$ have its standard inner product. Find an orthogonal basis of the subspace $U = ${($x_1, x_2, x_3$) $∈ \mathbb{C}^3$ : $x_1 − x_2 + ix_3 = 0$}.

I understand that I can apply the Gram-Schmidt process to a set of vectors to find an orthogonal basis, but here I'm a bit thrown off by the imaginary terms in the conditions of the subspace. How would I begin this question? How can I find a set of vectors to apply the Gram-Schmidt process to?

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First of all, let's find a basis of linearly independent vectors satisfying the constraint: such a basis is obtained easily if we observe that every solution can be written in terms of $(\lambda,\sigma,i(\sigma-\lambda))$ , with $\lambda,\sigma\in \mathbb{C}$. Thus, fixing two values of $(\lambda,\sigma)\in\mathbb{C}^2$, every other solution will be written as a linear combination of those vectors, and the associated vectors form a basis. $\{(1,0,i);(0,1,-i)\}$ is such a basis (obtained by fixing $\lambda=1;\sigma=1$)

To extract an orthonormal basis (remember that the inner product in $\mathbb{C}^3$ is $\langle z_1,z_2,z_3;\zeta_1,\zeta_2,\zeta_3\rangle=z_1\overline{\zeta_1}+\dots+z_3\overline{\zeta_3}$) we compute $||(1,0,i)||=\sqrt{2}$. The first vector of the orthonormal basis is thus $e_1=\frac{(1,0,i)}{\sqrt{2}}$. The second is obtained by $\frac{(0,1,-i)-\langle (0,1,-i);e_1\rangle e_1}{||(0,1,-i)-\langle (0,1,-i);e_1\rangle e_1||}=\frac{(\frac12,1,-\frac i2)}{\frac{\sqrt{3}}{\sqrt{2}}}$.

The orthonormal basis is $$\left\{\frac{(1,0,i)}{\sqrt{2}},\frac{\sqrt{3}}{\sqrt{2}}\left(\frac12,1,-\frac i2\right)\right\}$$

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  • $\begingroup$ Thanks for the help. I am confused though, is it not true that for n-dimensional space, n vectors must constitute a basis for that space? In this case we only have two vectors for the space of $\mathbb{C}^3$. $\endgroup$ – DPJDPJ Nov 26 '19 at 11:38
  • $\begingroup$ @JamesDebenham we have not found a base for the whole space $\mathbb{C}^3$. We have found a base for the subspace of $\mathbb{C}^3$ made of those vectors that satisfy your equation. This subspace has dimension two, and so all of its basis will be composed of two vectors $\endgroup$ – Caffeine Nov 26 '19 at 11:47
  • $\begingroup$ Of course! Thank you. $\endgroup$ – DPJDPJ Nov 26 '19 at 11:49
  • $\begingroup$ One further confusion, and apologies if I'm being silly, but why is it (λ,σ,i(λ−σ)) and not (λ,σ,i(σ-λ)) ? $\endgroup$ – DPJDPJ Nov 26 '19 at 11:52
  • $\begingroup$ @JamesDebenham That is a typo, I fixed it $\endgroup$ – Caffeine Nov 26 '19 at 12:24
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Pick any nonzero vector in $U$, say $(1,0,i)$. We need to find a nonzero vector $(x_1, x_2, x_3) \in U$ such that $\langle (x_1, x_2, x_3), (1,0,i)\rangle = 0$.

Therefore $(x_1, x_2, x_3)$ satisfies the equations $$\begin{cases} x_1-x_2+ix_3 = 0 \\ x_1 - ix_3 = 0 \end{cases}$$

Adding the equations yields $x_2 = 2x_1$ and the second equation yields $x_3 = -ix_1$. We can set $x_1 = 1$ to obtain $$(x_1, x_2, x_3) = (1,2,-i)$$

Therefore $$\{(1,0,i),(1,2,-i)\}$$ is an orthogonal basis for $U$.

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  • $\begingroup$ Is the second equation not $x_1 + ix_3$? $\endgroup$ – DPJDPJ Nov 26 '19 at 12:41
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    $\begingroup$ @JamesDebenham No, we have $$0 = \langle (x_1, x_2, x_3), (1,0,i)\rangle = x_1 \overline{1}+x_2\overline{0}+x_3\overline{i} = x_1 - ix_3$$ $\endgroup$ – mechanodroid Nov 26 '19 at 12:47
  • $\begingroup$ I see, that's a fundamental lack of understanding on my part. Sorry, and thank you! $\endgroup$ – DPJDPJ Nov 26 '19 at 12:48
  • $\begingroup$ Also, why does the second equation not yield $x_3 = 1/i * x_1$? $\endgroup$ – DPJDPJ Nov 26 '19 at 12:53
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    $\begingroup$ @JamesDebenham It does, we have $\frac1i = -i$. $\endgroup$ – mechanodroid Nov 26 '19 at 12:53

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