Minimum distance to a subspace is attained by an orthogonal vector

Question

Given a subspace $S$ in some inner product space $X$, we know that $\forall p \in X$, $\exists s_o \in S$, s.t. $\forall s \in S$, $\|p - s_o\| \leq \|p - s\| \Leftrightarrow p - s_o \perp S$

We want to show that given a translated (affine) space $M = d + S$, $\forall p \in X$, $\exists m_o \in M$, s.t. $\|p - m_o\| \leq \|p - m\| \Leftrightarrow p - m_o \perp M$

Attempt:

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Let $m_o = s_o+d, m = s+d$

$\langle p - s_o, s \rangle = 0$

$\Rightarrow \langle p - (m_o - d), m - d \rangle = 0$

Is there a way to argue from here that the above implies:

$\langle p - m_o, m \rangle = 0$

Answer 1

It is just the Pythagorean theorem. Given $p\in X$, let $p_0$ be the projection of $p$ on $S$ and assume that $$ \min_{s\in S}\|p-s\| $$ is attained by some $q\in S$ with $q\neq p_0$. Then $(p-p_0)\perp (p_0-q)$, hence: $$ \|p-q\|^2 = \|(p-p_0)+(p_0-q)\|^2 = \|p-p_0\|^2 + \|p_0-q\|^2 > \|p-p_0\|^2 $$ contradicting the minimality of $q$.