Question

Let f:C3 to C be defined as f(z)=(z1-z2)-i(2z1+z2+z3), where z=(z1,z2,z3) belongs to C3.
Find aw belongs to C3 such that f(z)=<z,w>, where <,> is the standard inner product on C3.

Answer 1

Answer on question #40787 – Math – Linear Algebra

Let $f: \mathbb{C}3$ to $\mathbb{C}$ be defined as $f(z) = (z1 - z2) - i(2z1 + z2 + z3)$, where $z = \{z1, z2, z3\}$ belongs to $\mathbb{C}3$.

Find aw belongs to C3 such that $f(z) = \langle z, w \rangle$, where $\langle , \rangle$ is the standard inner product on C3.

Answer:

The standard inner product defines as

$$\langle z; w \rangle = z_1 \overline{w_1} + z_2 \overline{w_2} + z_3 \overline{w_3},$$

where $z = (z_1, z_2, z_3), w = (w_1, w_2, w_3)$.

So we get

$$\begin{aligned} f(z) &= \langle z, w \rangle \geq z_1 \overline{w_1} + z_2 \overline{w_2} + z_3 \overline{w_3} \\ &= (z_1 - z_2) - i(2z_1 + z_2 + z_3) \\ &= z_1(1 - 2i) + z_2(-1 - i) + z_3(-i), \end{aligned}$$

Therefore we obtain

$$\begin{aligned} \overline{w_1} &= 1 - 2i \Rightarrow w_1 = 1 + 2i; \\ \overline{w_2} &= -1 - i \Rightarrow w_2 = -1 + i; \\ \overline{w_3} &= -i \Rightarrow w_3 = i. \end{aligned}$$

Answer: $w = (1 + 2i; -1 + i; i)$.