Euclidean Projection

Description:

• Corresponds to the problem of finding a point on a given set that is closest (in Euclidean length) to a given point
• Given a vector $x$ in ${\mathbb{R}}^n$ in a closed set $S\subseteq {\mathbb{R}}^n$, the project of $x$ onto $S$, denoted as ${\sqcap }_S(x)$, is defined as the point in $S$ at minimal distance from $x$
  • S can be a line, a plane, hyperplane
• ${\sqcap }_S(x)=\text{arg }{\text{min }}_{y\in S}||y-x|{|}_2$
  • arg refers to the unique minimizer

Euclidean Projection on a line:

• Denotes $L=x_0+span(u)$
  • $u$ is the direction of the line. WLOG, let $||u|{|}_2=1$
• Let $p\in {\mathbb{R}}^n$ be a given point
  • $p^{\ast }$ be the point on the line that is project of $p$
• $p^{\ast }=\text{arg }{\text{min }}_{x\in S}||x-p|{|}_2$
  • $p^{\ast }$ such that it minimizes $x-p$
• As any point on $L$ can be written as $x=x_0+t.u$ for a scalar $t$, then we need to minize the $t^{\ast }$ (value of $t$ for $p^{\ast }$) $\to$ minimize $f(t):=||tu-(p-x_0)|{|}_2^2$
• A lot more proofs, then we have, $t^{\ast }=u^T(p-x_0)$ and $|u^{T}p|\le ||p|{|}_2$

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