If $A$ is skew-symmetric, then how $\langle y-x, Ax + b\rangle = \langle y-x, Ay + b\rangle $? I am sorry to ask and wondering, how come the following is true?
$$\langle y-x, Ax + b\rangle = \langle y-x, Ay + b\rangle $$
if $A$ is skew-symmetric, that is, $A^T = -A$. 
 A: Observe that
$\langle y - x, Ax + b \rangle = \langle y - x, Ax \rangle + \langle y - x, b \rangle, \tag 1$
and that
$\langle y - x, Ay + b \rangle = \langle y - x, Ay \rangle + \langle y - x, b \rangle; \tag 2$
thus
$\langle y - x, Ax + b \rangle = \langle y - x, Ay + b \rangle \tag 3$
if and only if
$\langle y - x, Ax \rangle = \langle y - x, Ay \rangle; \tag 4$
we have
$\langle y - x, Ax \rangle = \langle y, Ax \rangle - \langle x, Ax \rangle \tag 5$
and also
$\langle y - x, Ay \rangle = \langle y, Ay \rangle - \langle x, Ay \rangle; \tag 6$
since 
$A^T = -A, \tag 7$
we have
$\langle x, Ax \rangle = \langle A^Tx, x \rangle = -\langle Ax, x \rangle = -\langle x, Ax \rangle, \tag 8$
which implies
$\langle x, Ax \rangle = 0; \tag 9$
likewise, 
$\langle y, Ay \rangle = 0; \tag{10}$
therefore,
$\langle y - x, Ax \rangle = \langle y, Ax \rangle = \langle A^Ty, x \rangle =$
$-\langle Ay, x \rangle = -\langle x, Ay \rangle = \langle y - x, Ay \rangle. \tag{11}$
The desired result (3) now follows from this in light of (1), (2).
A: By definition and assumption $\langle A^tx,y\rangle = \langle x,Ay\rangle = -\langle Ax,y\rangle$. So in particular $\langle x,Ax\rangle = -\langle Ax,x\rangle = -\langle x,Ax\rangle = 0$. This holds for $y$ as well. Therefore
\begin{align*}
\langle y-x,Ax+b\rangle & = \langle y,Ax\rangle + \langle y,b\rangle - \langle x,Ax \rangle -\langle x,b\rangle \\
& = -\langle x,Ay\rangle + \langle y,b\rangle + \langle y,Ay \rangle - \langle x,b\rangle \\
& = \langle y-x,Ay+b\rangle.
\end{align*}
