suppose $\underline x,\underline y\in\mathbb C^{n\times 1}$ then because the two vectors are in complex vector field, the definition of their inner product will be:
$$\langle\underline x,\underline y\rangle=\sum_{i=1}^n x_iy_i^*\qquad\text{* represents conjugate}$$
then it seems that the definition for hermitian scalar product of these two vectors is the same $$\langle\underline x|\underline y\rangle=\underline x^T.\underline y^*=\sum_{i=1}^n x_iy_i^*\qquad\text{T represents transpose}$$ Are inner product and hermitian scalar product the same concept in vector spaces defined on complex numbers field?


A scalar product is just another name for inner product. Let V be a complex vector space. A Hermitian inner product on V is any function $$\langle , \rangle : V × V \rightarrow \mathbb{C},$$ that satisfies these axioms:

  1. $$\langle u, v \rangle = \langle v, u \rangle $$
  2. $$\langle u + v, w \rangle = \langle u, w \rangle + \langle v, w \rangle \text{ and}$$

    $$\langle u, v + w \rangle = \langle u, v \rangle + \langle u, w \rangle $$

  3. $$\langle cu, v \rangle = \overline{c} \langle u, v \rangle \text{ and}$$

    $$\langle u, cv \rangle = c\langle u, v \rangle .$$

  4. $\langle u, u \rangle $ is a non-negative real number and $\langle u, u \rangle = 0$ if and only if $\langle u = 0 \rangle $.

What you've defined is just an example of a Hermitian inner product.

  • $\begingroup$ $V\times V$ is cartesian product. right? and you mean hermitian inner product is a broad concept satisfying the above axioms. And what I have defined above as $\sum_{i=1}^n x_iy_i^*$ is an example of inner product. $\endgroup$ – Sepideh Abadppour Aug 6 '15 at 10:10
  • $\begingroup$ here you see I've said that for any inner product space, we should define the inner product $\endgroup$ – Sepideh Abadppour Aug 6 '15 at 10:30
  • 2
    $\begingroup$ Yes, $\langle x, y \rangle=\sum_{i=1}^n x_i \bar{y_i}$ is just one of many inner products that can be defined on $\mathbb{C}^n$. In general, all inner products on $\mathbb{C}^n$ are of the form $\langle x, y \rangle = \bar{y}^T H x$ where $H$ is Hermitian positive-definite matrix. $\endgroup$ – Zoran Loncarevic Aug 6 '15 at 10:40

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