Given the vectors $\vect{u},\,\vect{v}\in\complex{m}$ the inner product of $\vect{u}$ and $\vect{v}$ is the scalar quantity in $\complex{\null}$, \begin{equation*} \innerproduct{\vect{u}}{\vect{v}}= \conjugate{\vectorentry{\vect{u}}{1}}\vectorentry{\vect{v}}{1}+ \conjugate{\vectorentry{\vect{u}}{2}}\vectorentry{\vect{v}}{2}+ \conjugate{\vectorentry{\vect{u}}{3}}\vectorentry{\vect{v}}{3}+ \cdots+ \conjugate{\vectorentry{\vect{u}}{m}}\vectorentry{\vect{v}}{m} = \sum_{i=1}^{m}\conjugate{\vectorentry{\vect{u}}{i}}\vectorentry{\vect{v}}{i} \end{equation*}
This operation is a bit different in that we begin with two vectors but produce a scalar. Computing one is straightforward.
In the case where the entries of our vectors are all real numbers (as in the second part of Example CSIP), the computation of the inner product may look familiar and be known to you as a dot product or scalar product. So you can view the inner product as a generalization of the scalar product to vectors from $\complex{m}$ (rather than ${\mathbb R}^m$).
Note that we have chosen to conjugate the entries of the first vector listed in the inner product, while it is almost equally feasible to conjugate entries from the second vector instead. In particular, prior to Version 2.90, we did use the latter definition, and this has now changed to the former, with resulting adjustments propogated up through Section CB (only). However, conjugating the first vector leads to much nicer formulas for certain matrix decompositions and also shortens some proofs.
There are several quick theorems we can now prove, and they will each be useful later.
Suppose $\vect{u},\,\vect{v},\,\vect{w}\in\complex{m}$. Then
Suppose $\vect{u},\,\vect{v}\in\complex{m}$ and $\alpha\in\complex{\null}$. Then
Suppose that $\vect{u}$ and $\vect{v}$ are vectors in $\complex{m}$. Then $\innerproduct{\vect{u}}{\vect{v}}=\conjugate{\innerproduct{\vect{v}}{\vect{u}}}$.