Let $Ax$ and $Ay$ minimize distance to $b \in \mathbb{R^m}.$ Show $x-y \in \ker(A).$ Let $A$ be an $m \times n$ real matrix and $b \in \mathbb{R^m}.$  Suppose $Ax$ and $Ay$ both minimize distance to $b,$ i.e. $||Ax-b|| = ||Ay-b||.$  Prove that $x-y \in \ker(A).$
Seems like there should be a straightforward proof, but I have not been able to provide one.  Any suggestions would be appreciated, thanks in advance.

My thoughts:
\begin{align}
||Ax - b||^2 &= ||Ay - b||^2\\
\langle Ax - b, Ax - b \rangle &= \langle Ay - b, Ay - b  \rangle\\
||Ax||^2 - ||Ay||^2 &= 2(\langle Ax,b \rangle - \langle Ay,b \rangle).
\end{align}
If the LHS equals zero in the last expression, then $0 = \langle A(x-y),b \rangle$ by linearity and then we are done.  Not sure why the LHS would be zero though.
 A: This is a classic Basic Exam question, and one common approach is to use the parallelogram identity. Fix $b \in \mathbb{R}^{m}$, and let $u, v \in \mathbb{R}^{m}$ simultaneously minimize $D(x) = ||A(x)-b||$. Put $\alpha = A(u)-b, \beta = A(v)-b$, and let $k = ||\alpha|| = ||\beta||$. Then by the parallelogram identity,
$$||\alpha-\beta||^{2} = 2||\alpha||^{2}+2||\beta||^{2} - ||\alpha+\beta||^{2} = 4k^{2} - ||\alpha+\beta||^{2}$$
Note that
$$\alpha+\beta = A(u)+A(v)-2b = 2\left(A\left(\frac{u+v}{2}\right)-b\right)$$
so
$$||\alpha+\beta||^{2} = 4||A\left(\frac{u+v}{2}\right)-b||^{2} = 4D\left(\frac{u+v}{2}\right)^{2} \geqslant 4k^{2}$$
so
$$||\alpha-\beta||^{2} \leqslant 4k^{2} - 4k^{2} = 0$$
whence $||\alpha-\beta||^{2} = 0$, so 
$$\alpha-\beta = A(u)-b - (A(v)-b) = A(u)-A(v) = A(u-v) = 0$$
so $u-v \in \ker(A)$. 
A: First of all, I think your "i.e." should be followed by "$= \min_{z \in \mathbb R^n} \|Az -b\|$".
That is, you should use the fact that $x$ and $y$ minimize the distance from $\operatorname{ran} A$ to $b$.
Then a hint: observe that $\|Ax - Ay\| \leq \|Ax - \lambda b\| + \|Ay - \lambda b\|$ for any $\lambda$.
