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

Question

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?

Answer 1

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

Answer 2

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 $U$.