If $V$ be a vector space over the field $\mathbb K$, the inner product $\langle\cdot, \cdot\rangle: V \times V \rightarrow \mathbb{K}$ will be is hermitically symmetrical: $\langle x, y\rangle=\overline{\langle y, x\rangle}$. In case the field $\mathbb K = \mathbb R$, then the orthogonal projection of the vector $\mathbf{a}$ onto $\mathbf{b}$ is given by
$$\text{oproj}_{\mathbf{b}} \mathbf{a} = \frac{\langle \mathbf{a} , {\mathbf{b}} \rangle} {\|\mathbf{b}\|}=\frac{\langle \mathbf{b} , {\mathbf{a}} \rangle} {\|\mathbf{b}\|}$$
What about the general case, in which $\langle x, y\rangle \neq {\langle y, x\rangle}$ but $\langle x, y\rangle=\overline{\langle y, x\rangle}$? Would the orthogonal projection of $x$ onto $y$ be given by
$\text{oproj}_{y} {x} = \frac{\langle x, y \rangle} {\|y\|}$ or $\frac{\langle y , x \rangle} {\|y\|}$?
Best Answer
So we want to project $x$ onto $y.$ The projected vector $v = \lambda y$ should be in the direction of the vector $y$ up to some scale factor $y.$ Let's try to compute this factor $\lambda\in \mathbb{K}.$ Since we want an orthogonal projection, $$\langle x-v, v\rangle = 0 \implies \lambda \langle x, y\rangle = |\lambda|^2\langle y, y\rangle,$$ and so the valid choice should be the first one, $\frac{\langle x,y \rangle}{||y||}.$ Try the computation with $\langle x - v , v\rangle$ where $v = \frac{\langle y,x \rangle}{||y||}y$ and you will see that this quantity is not equal to $0.$