Prove that $\mathbf{a}^H\mathbf{b}+\mathbf{b}^H\mathbf{a} = 2\operatorname{Re}(\mathbf{b}^H\mathbf{a})$? Why would the following identity hold for any two complex vectors $\mathbf{a}, \mathbf{b} \in \mathbb{C}^N$
$$\mathbf{a}^H\mathbf{b}+\mathbf{b}^H\mathbf{a} = 2\operatorname{Re}(\mathbf{b}^H\mathbf{a})$$
This is taken from Steven M. Kay's Fundamentals of Statistical Signal Processing - Detection Theory page 475.
Does this identity have a name? Any help is appreciated, thanks.
 A: For any complex number $z=z_r+iz_i$ and its complex conjugate $\overline{z}=z_r-iz_i$ one can see that
$$z+\overline{z}=z_r+iz_i+z_r-iz_i=2z_r=2\operatorname{Re}(z)$$
The scalar product for any two complex vectors $\mathbf{a}, \mathbf{b} \in \mathbb{C}^N$ is
$$\mathbf{a} \cdot \mathbf{b} = \mathbf{a}^{\operatorname{H}}\mathbf{b} = \sum_{k=1}^{N} \overline{a}_kb_k$$ where $\mathbf{a}^{\operatorname{H}}=\overline{\mathbf{a}^{\operatorname{T}}}$ denotes the conjugate transpose.

Thus, one has
$$\mathbf{a}^H\mathbf{b}+\mathbf{b}^H\mathbf{a} \\
=\sum_{k=1}^{N} \overline{a}_kb_k +\sum_{k=1}^{N} \overline{b}_ka_k \\
=\sum_{k=1}^{N}\overline{a}_kb_k + \overline{b}_ka_k \\
=\sum_{k=1}^{N}\overline{\overline{b}_ka_k} + \overline{b}_ka_k \\
=\sum_{k=1}^{N}2\operatorname{Re}(\overline{b}_ka_k) \\
=2\operatorname{Re}\left(\sum_{k=1}^{N}\overline{b}_ka_k\right) \\
=2\operatorname{Re}(\mathbf{b}^H\mathbf{a})
$$

Thank you @Semiclassical for the comment that provided the answer.
References:
complex dot product,
complex identities
