# How to prove the inner product of complex vectors is conjugate-symmetric? [closed]

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:
$$<\underline{x}, \underline{z}> = \underline{x}^T \underline{z}$$

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• Did you try to set $x = a + i\cdot b$ and $y = c + i \cdot d$ and apply the inner product? – PackSciences Feb 25 at 22:38
• $\vec{a}$ and $\vec{b}$ being a 2x1 vectors? – John Proxer Feb 25 at 22:49
• Yes, you can do that for vectors, scalars or matrices. – PackSciences Feb 25 at 22:51
• well, i'm told that inner product has both linear and anti-linear properties: $$<\alpha\vec{x} + \beta\vec{y}, \vec{z}> = \alpha^*<\vec{x},\vec{z}> + \beta^*<\vec{y},\vec{z}>$$ $$<\vec{z}, \alpha\vec{x} + \beta\vec{y}> = \alpha<\vec{z},\vec{x}> + \beta<\vec{z},\vec{y}>$$ – John Proxer Feb 25 at 22:52
• Well use linearity and antilinearity on the real and imaginary decomposition and you should get your result. – PackSciences Feb 25 at 22:54

\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}
$$^* = (z_1^* x_1 + z_2^* x_2)^* = (z_1 x_1^* + z_2 x_2^*) = (x_1^* z_1 + x_2^* z_2 ) = $$