This problem projects [tex]b = (b_1,b_2....,b_m)[/tex] onto the line through [tex]a = (1, 1, 1, ....1)[/tex]. We solve m equations ax = b in 1 unknown (by least squares).

(a) Solve [tex]a^T~a~\hat{x} = a^T~b[/tex] to show that [tex]\hat{x}[/tex]is the mean (the average) of the b’s.

(b) Find [tex]e = b - a \hat{x}[/tex] and the variance [tex]||e||^2[/tex] and the standard deviation [tex]||e||[/tex].

(c) The horizontal line [tex]\hat{b} = 3[/tex] is closest to b = (1, 2, 6). Check that p = (3, 3 3) is perpendicular to e and find the 3 by 3 projection matrix P.

Ans(a): Becauase a = (1,1,1,....1), therefore [tex]a^T a = 1 + 1 + 1 +....+ 1 = 1~*~m = m[/tex]

And [tex]a^T b = b_1 + b_2 + .... + b_m[/tex]

So [tex]\hat{x} = \frac{b_1 + b_2 + b_3 + .... + b_m}{m} = b_{avg}[/tex]

Ans(b): Need help..

Ans(c): Need help..