Let $W$ be a vector subspace of $V$, a space with a dot product; $v\in V$. Let $p_W(v)$ be the orthogonal projection of $v$ onto $W$ and $w\in W, w\neq p_W(v)$.
How can i prove that $||v-w|| > ||v-p_W(v)||$?
Let $W$ be a vector subspace of $V$, a space with a dot product; $v\in V$. Let $p_W(v)$ be the orthogonal projection of $v$ onto $W$ and $w\in W, w\neq p_W(v)$.
How can i prove that $||v-w|| > ||v-p_W(v)||$?
On
Since $W \subset V$ is a fixed subspace, I am going to drop the subscript "$W"$ from $P$; in this answer, $P = P_W$.
Recall that an orthogonal projection $P$ satisfies
$P^2 = P = P^T; \tag 1$
thus for any
$x, y \in V \tag 2$
we have
$\langle x, Py \rangle = \langle P^Tx, y \rangle = \langle Px, y \rangle; \tag 3$
now consider the three vectors $v$, $Pv$, and $w$;
$Pv, w \in W \Longrightarrow Pv - w \in W, \tag 4$
$\langle v - Pv, w \rangle = \langle v, w \rangle - \langle Pv, w \rangle = \langle v, w \rangle - \langle v, Pw \rangle = \langle v, w \rangle - \langle v, w \rangle = 0, \tag 5$
where we have used (3) and the fact that $Pw = w$ for $w \in W$; since this holds for every $w \in W$, we have established that
$v - Pv \in W^\bot; \tag 6$
we write
$v - w = (v - Pv) + (Pv - w), \tag 7$
and by virtue of (4) and (6)
$\langle v - Pv, Pv - w \rangle = 0, \tag 8$
whence
$\Vert v - w \Vert^2 = \langle v - w, v - w \rangle = \langle (v - Pv) + (Pv - w), (v - Pv) + (Pv - w) \rangle$ $= \langle v - Pv, v - Pv \rangle -2\langle v - Pv, Pv - w \rangle + \langle Pv - w, Pv - w \rangle$ $= \Vert v - Pv \Vert^2 + \Vert Pv - w \Vert^2;\tag 9$
since each term on the right is non-negative, we find that
$\Vert v - w \Vert^2 \ge \Vert v - Pv \Vert^2, \tag{10}$
whence
$\Vert v - w \Vert \ge \Vert v - Pv \Vert, \; \forall w \in W, \tag{11}$
with equality holding precisely when
$\Vert w - Pv \Vert = 0 \Longleftrightarrow w = Pv; \tag{12}$
thus
$w \ne Pv \Longrightarrow \Vert v - w \Vert > \Vert v - Pv \Vert. \tag{13}$
$OE\Delta$.
Let $v = v_\| + v_\bot $, where $v_\| = p_W(v)$, $v_\bot \in W^\bot$.
$ \|v-w\|^2 = (v-w)^2 = v^2 + w^2 - 2vw = v_\bot^2 + v_\|^2 + w^2 - 2v_\|w = v_\bot^2 + (v_\|-w)^2$
$ \|v-v_\|\|^2 = v_\bot^2 < v_\bot^2 + (v_\|-w)^2 $, QED.