matrix algebra and idempotent matrix

Question

I'm having a little trouble understanding a few derivations in my book for least squares regression.

$\textbf{Question 1}$:

If $\textbf{M}^0 \textbf{i} = [\textbf{I} - \frac{1}{n}\textbf{ii'}]\textbf{i} = \textbf{i}-\frac{1}{n}\textbf{i(i'i)}=\textbf{0}$

Then how does this imply that $\textbf{i'M}^0 = \textbf{0'}$? Shouldn't it be $(\textbf{M}^0\textbf{i})' = \textbf{0}'$?

and moreover how do we get that

$$\sum_{i=1}^n(x_i -\bar x)=\textbf{i}'[\textbf{M}^0\textbf{x}]$$

It's not explicitly stated, but i'm pretty sure $\textbf{i}$ is a row vector of 1's.

And also, evidently $\textbf{M}^0$ is being defined as $[\textbf{I}-\frac{1}{n}\textbf{i}\textbf{i}']$

$\textbf{Question 2}$:

Why is it the case that for the expansion of $\textbf{e}'_0\textbf{e}_0 = (\textbf{y-Xb})'(\textbf{y - xb})= $ $$\textbf{y'y - b'X'y - y'Xb + b'X'Xb} = \textbf{y'y - 2y'Xb + b'X'Xb}$$

In particular, how can we group the $\textbf{b'X'y}$ and $\textbf{y'Xb}$ as the same?

Q1 is from the appendix of William H. Greene - Econometric analysis 7th ed.

Answer 1

For Question 1,

Then how does this imply that $\textbf{i'M}^0 = \textbf{0'}$? Shouldn't it be $(\textbf{M}^0\textbf{i})' = \textbf{0}'$?

$i'M^0 = (M^0 i)^{'}$, since $M^0$ is symmetric ($(M^0)^T=M^0$).

For Question 2, both $b'X'y$ and $y'Xb$ are scalars and taking the transpose of a scalar doesn't change its value, hence $b'X'y = y'Xb$.