# Find the Orthogonal Basis of a subspace in $\mathbb{C}^3$

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?

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

• 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 $\mathbb{C}^3$. – DPJDPJ Nov 26 '19 at 11:38
• @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 – Caffeine Nov 26 '19 at 11:47
• Of course! Thank you. – DPJDPJ Nov 26 '19 at 11:49
• 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)) ? – DPJDPJ Nov 26 '19 at 11:52
• @JamesDebenham That is a typo, I fixed it – Caffeine Nov 26 '19 at 12:24

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 $$U$$.

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