Exercise 4.3.12

(a) $a=(1,\dots ,1)$ has $a^{T}a=m,a^{T}b=b_1+\cdots +b_m$. Therefore $\hat{x}=a^{T}b∕m$ is the mean of the $b$ ’s (their average value)

(b) $e=b-\hat{x}a$ and $\parallel e{\text{∥}}^2={\left(b_1-\text{ mean }\right)}^2+\cdots +{\left(b_m-\text{ mean }\right)}^2=$ variance (denoted by ${\sigma }^2$ ).

(c) $p=(3,3,3)$ and $e=(-2,-1,3)p^{T}e=0$. Projection matrix $P=\frac{1}{3}\left[\begin{array}{ccc}1\hfill & 1\hfill & 1\hfill \\ 1\hfill & 1\hfill & 1\hfill \\ 1\hfill & 1\hfill & 1\hfill \end{array}\right]$.