For complex vector-space of 2 dimensions, prove that the inner product is conjugate-symmetric, ie:
$$ < \underline{x},\underline{z}> = <\underline{z},\underline{x}>^{*} $$
where:
$$ \underline{x} = \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} $$
$$ \underline{z} = \begin{bmatrix} z_1 \\ z_2 \\ \end{bmatrix} $$
A few things provided by the book:
- Hermitian Conjugate, aka. Conjugate-Transpose:
$$\underline{A}^H = ( \underline{A}^T )^*$$
For Complex Vectors: Inner Product, aka. Dot Product:
$$<\underline{x}, \underline{z}> = \underline{x}^H \underline{z}$$For Real Vectors: Inner Product, aka. Dot Product:
$$<\underline{x}, \underline{z}> = \underline{x}^T \underline{z}$$
$$ \begin{aligned} <\underline{x},\underline{z}> &= \underline{x}^H \underline{z} \\ \\ <\underline{x},\underline{z}> &= \begin{bmatrix} x_1^* & x_2^* \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} \\ \\ <\underline{x},\underline{z}> &= x_1^* z_1 + x_2^* z_2 \end{aligned} $$
$$ \begin{aligned} <\underline{z},\underline{x}> &= \underline{z}^H \underline{x} \\ \\ <\underline{z},\underline{x}> &= \begin{bmatrix} z_1^* & z_2^* \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \\ \\ <\underline{z},\underline{x}> &= z_1^* x_1 + z_2^* x_2 \end{aligned} $$
Now we ask, how can we make these two products equivalent...
$$ <z,x>^* = (z_1^* x_1 + z_2^* x_2)^* = (z_1 x_1^* + z_2 x_2^*) = (x_1^* z_1 + x_2^* z_2 ) = <x,z> $$