Let us consider two vectors $u$ and $v$ are in the complex vector space. The inner product of these two vectors is defined by

$u\cdot v=(u_1^*\cdot v_1, u_2^*\cdot v_2,\cdots\cdots, u_n^*\cdot v_n)$

How can this definition be proved? How did we even get to this definition?

  • 4
    $\begingroup$ You don't prove a definition, a definition is a definition. The only thing you need to be sure of is that a definition makes sense. For example, I cannot define $\alpha:=\frac{0}{0}\in \mathbb{R}$, that would lead to contradictions. $\endgroup$ – Mathematician 42 May 6 '16 at 12:20
  • $\begingroup$ so why this definition? $\endgroup$ – sajjad islam May 6 '16 at 12:29
  • $\begingroup$ From a historical point of view, people where first interested in lengths of vectors in $\mathbb{R}^n$. Given a vector $v\in \mathbb{R}^3$, people quickly realized that $\sqrt{v\cdot v}$ is the length of $v$. Then it's easy to notice that if $u\cdot v=0$, then $u$ and $v$ are orthogonal, hence the inner product has something to do with angles as well. Then people found this easy connection, and used inner-products to define lengths and angels in more general spaces as well. When you restrict your attention to $\mathbb{R}^2$, calculations involving inner products should be familiar. $\endgroup$ – Mathematician 42 May 6 '16 at 12:37
  • $\begingroup$ In a complex vector space you define $u \cdot v = u_1 \bar{v}_1 + \cdots + u_n \bar{v}_n$ so that the inner product of a vector with itself is real and nonnegative. $\endgroup$ – Ethan Bolker May 6 '16 at 12:49

As was pointed out in a comment above, $\sqrt{v\cdot v}$ is the length of $v$ in a real vector space (say $\mathbb{R}^n$). Maintaining the same relationship in a complex vector space leads to your definition, as follows. First consider the (one-dimensional) complex vector space $\mathbb{C}$. If $v=a+bi\in\mathbb{C}$, then we want the length of $v$ again to be $\sqrt{v\cdot v}$. One (the only?) reasonable definition of the dot product that gives $$\sqrt{v\cdot v} = \sqrt{a^2+b^2}$$ is $v\cdot v = (a+bi)(a-bi) = v\bar{v}$. It's easy to see that this generalizes to $n$-dimensional complex vector spaces.

Of course, one has to go on to show that this is in fact an inner product. But that's where the intuition came from.

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