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Section 1.2 Inner products

The dot product of two vectors in \(\C^n\) is
\begin{equation} x \cdot y = \sum_{i=1}^n \cc y_i x_i.\tag{1.2.1} \end{equation}
Standard notation for the dot product is \(\ip{x}{y}\) and in \(\C^n\) is equivalent to \(y\ad x\text{,}\) where \(\ad\) designates the conjugate transpose of a matrix. The dot product has the following properties:
  1. \(\displaystyle \ip{x}{y} = \cc{\ip{y}{x}} \hspace{.2in} \text{conjugate symmetry}\)
  2. \(\displaystyle \ip{x + y}{z} = \ip{x}{y} + \ip{y}{z} \hspace{.2in} \text{linearity in the first term}\)
  3. \(\displaystyle \ip{x}{x} \geq 0 \hspace{.2in} \text{non-negativity}\)
Once we have the dot product, we can start building the geometry of \(\C^n\text{.}\) First, note that
\begin{equation} \norm{x}^2 = \ip{x}{x} = \sum_{i=1}^n \abs{x}^2.\tag{1.2.2} \end{equation}
Motivated by the real case, we say that two vectors \(x, y\) are orthogonal and write \(x \perp y\) if \(\ip{x}{y} = 0\text{.}\)
Another important inequality is indicated by the relationship between angles and the dot product in \(\R^n\text{,}\) where we have
\begin{equation*} \ip{x}{y} = \norm{x}\norm{y}\cos \theta, \end{equation*}
where \(\theta\) is the angle between the vectors. While the idea of “angle” doesn’t make sense in \(\C^n\) (at least in the same way), we still have the Cauchy-Schwarz inequality
\begin{equation} \abs{\ip{x}{y}} \leq \norm{x}\norm{y}.\tag{1.2.3} \end{equation}
Orthogonality also underlies the vector version of the Pythagorean theorem,
\begin{equation} \norm{x}^2 + \norm{y}^2 = \norm{x+ y}^2 \iff x\perp y.\tag{1.2.4} \end{equation}
Finally, it would be remiss to leave out the single most important inequality in mathematics, our old friend the triangle inequality, which in vector terms can be expressed
\begin{equation} \norm{x + y} \leq \norm{x} + \norm{y}\tag{1.2.5} \end{equation}
Because finite dimensional vector spaces have representations in coordinates as \(\R^n\) or \(\C^n\text{,}\) all finite dimensional vector spaces carry the geometric structure delineated above.