I'm reading on orthogonal projections from a course's notes and it says the following:
For each x ∈ $R_n$ and each linear subspace U, $\pi_U$(x) exists and is unique. Moreover, $\pi_U$(x) is the only vector in U such that $\pi_U$(x) − x ∈ $U^{\bot}$ (i.e. $\pi_U$(x) is orthogonal).
I can't understand what does it mean by projection - x is an element of the orthogonal projection.
$\mathbb R^n$ equipped with the dot product is a Euclidan vector space, i.e. on where you have a notion of distance and angle. As a consequence, if $U$ is any subspace of $\mathbb R^n$, then there is a "projection map" $\pi_U\colon \mathbb R^n \to U$. For a vector $v \in \mathbb R^n$ the projection $\pi_U(v)$ is the vector in $U$ which is closest to $U$. This forces $v-\pi_U(v)$ to be orthogonal to $U$, that is $v-\pi_U(v) \in U^{\perp}$.
To see why, note that because $\mathbb R^n = U \oplus U^{\perp}$, if we write $v = u_1+u_2$ with $u_1\in U$ and $u_2\in U^{\perp}$, the since, for any $w \in U_1$ we have $u_1-w\cdot u_2=0$,
$$ \|v-w\|^2 = \|u_1-w\|^2+\|u_2\|^2 \geq \|u_2\|^2, $$ with equality if and only if $u_1=w$.