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Find the hermitian inner product $x^*y$ of the following vectors $x$ and $y$:

(a) $\begin{bmatrix} 1+i \\ 5-i \end{bmatrix}$ and $\begin{bmatrix} 8-3i \\ 9+2i \end{bmatrix}$

(b) $\begin{bmatrix} 6-i \\ 1+2i \\ 5-8i \end{bmatrix}$ and $\begin{bmatrix} 3-6i \\ 2+3i \\ 4+6i \end{bmatrix}$

Here are my solutions. Can someone please verify them?

(a) $48+8i$

(b) $4+28i$

Also, is it true that the hermitian inner product is noncommutative? When I take the inner products in the reversed order, I get the conjugate of the original product. Why does it make sense to have a noncommutative inner product?

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The inner products are correct. –  Eleven-Eleven May 9 at 21:46

2 Answers 2

Your calculations are correct. Let me answer the second question:

Why does it make sense to have a non-commutative inner product?

Well, because now we're considering bilinear forms over $\mathbb{C}$, i.e., the bilinear functions from $V \times V \to \mathbb{C}$.

In the good old $\mathbb{R}$, we had $x^2 \geq 0$, but by moving from $\mathbb{R}$ to $\mathbb{C}$ this property is lost because $i^2 < 0$

This can be a source of problem, because in case of $\mathbb{R}$, an inner product could induce a norm on its vector space, but we know that norms are functions that give positive values. So, is it possible to somehow modify our definition of an inner product in someway that when we restrict it to the real case everything seems fine but it still satisfies the property that $<x,x> \text{ } \geq \text{ }0$ for all $x \in V$ ?

Yes. Instead of symmetry, you consider conjugate-symmetry. Then since $x. \bar{x} = |x|^2$ our new definition of inner product satisfies the properties we wished it to have.

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Just to give more information: "Hermitian" is the term used exactly to indicate that $\langle x, y \rangle = \overline {\langle y, x \rangle}$.

Some examples are:

  • In $\mathbb{C}$, $\langle x, y \rangle = \sum_{i = 1}^n x_i \overline{y_i}$

  • In $\mathcal{P}_n (\mathbb{C})$ (and in general $\mathcal{C}^0 ([0,1], \mathbb{C})$ ), we have $\langle f, g \rangle = \int_{0}^{1} f(t) \overline{g(t)} \mathrm{d}t$

  • In $\mathcal{M}_n (\mathbb{C})$, $\langle A, B \rangle = \mathrm{tr} (\overline{B^t} A)$

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