Just what the title says. I'm reading, from various resources, that the inner product is a generalization of the dot product. However, the only example of inner product I can find, is still the dot product.
Does another "inner product" exist on the $\mathbb{R}^n$ space?
Best Answer
Yes and are all in the form $$ \langle x, y\rangle=x^T\cdot A\cdot y $$ where $x^T$ is the transpose vector of $x$ and $A$ is a $n\times n$ symmetric definite positive matrix.
In fact let $\langle x, y\rangle$ a generic inner product on $\mathbb R^n$ then for every $y$ the function $$ f_y:x\rightarrow \langle x, y\rangle $$ is linear from $\mathbb R^n$ to $\mathbb R$ then exists a vector $\alpha(y)\in\mathbb R^n$ such that $$ \langle x, y\rangle = \alpha(y)^T\cdot x $$
Observe that $$ \langle x, ay+by'\rangle = a\langle x, y\rangle + b\langle x, y'\rangle\Rightarrow \alpha(ay+by')=a\alpha(y)+b\alpha(y') $$ then $\alpha$ is a linear operator from $\mathbb R^n$ in itself then exists an $n\times n$ matrix $A$ such that $$ \alpha(y)=A\cdot y $$ so $$ \alpha(y)^T\cdot x=y^T\cdot A^T\cdot x $$
Now remember that $\langle x, y\rangle=\langle y, x\rangle$ then you can easly prove that $A^T=A$ and $A$ must be symmetric.
Why now $A$ must be definite positive? Because $\langle x, x\rangle\geq 0$ and holds equality if and only if $x=0$. Applying it to the initial formula we obtain the definition of a definite positive matrix.