Prove that a given set of vectors forms a basis for C3

Let
$A = (1,0,0), \qquad B = (0,i,0), \qquad C = (1,1,i)$

be three vectors in $\mathbb{C}^3$ . Prove that these vectors form a basis for $\mathbb{C}^3$ .
Write the vector $(5,2-i,2i)$ as a linear combination of $A,B,C$ .

Proof. We know that any set of three linearly independent vectors in $\mathbb{C}^3$ will span $\mathbb{C}^3$ , and thus form a basis. (This is from Theorem 12.10, which is valid for $\mathbb{C}^n$ .) Thus, it is sufficient to show that $A,B,C$ are linearly independent. To that end, let $z_1, z_2, z_3$ be scalars in $\mathbb{C}$ , then
$\begin{align*} z_1 A + z_2 B + z_3 C &= O & \implies && z_1 + z_3 &= 0 \\ &&&& iz_2 + z_3 &= 0 \\ &&&& iz_3 &= 0. \end{align*}$

From the third equation we have $z_3 = 0$ , and so the second equation implies $z_2 = 0$ , and finally the third equation implies $z_1 = 0$ . Hence, $z_1 = z_2 = z_3 = 0$ , and $A,B,C$ are linearly independent $. \qquad \blacksquare$
To express $(5,2-i,2i)$ as a linear combination of $A,B,C$ , let $z_1, z_2, z_3 \in \mathbb{C}$ be scalars. Then,
$\begin{align*} z_1 A + z_2 B + z_3 C &= (5,2-i,2i) & \implies && z_1 + z_3 &= 5 \\ &&&& iz_2 + z_3 &= 2-i \\ &&&& iz_3 &= 2i. \end{align*}$

From the third equation we have $z_3 = 2$ . Plugging this into the first and second equations we get $z_1 = 3$ and $z_2 = -1$ . Therefore,

$(5,2-i,2i) = 3A - B + 2C.$