S 5.1 (P 2.24)
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a = [-1 7 2 4].'
b = [3 0 -1 -5].'

(i)
||a||^2 = a'*a = 1+49+4+16 = 70.  Therefore ||a||=sqrt(70).

||b||^2 = b'*b = 9+0+1+25 = 35.  Therefore ||b|| = sqrt(35)

b-a = [4 -7 -3 -9].'  ||b-a||^2 = 16+49+9+81 = 155

||b-a|| = sqrt(155)

(ii)
cos(theta) = ( a'*b ) / ( ||a||*||b|| ) 

Here a'*b = -3+0-2-20 = -25, so

cos(theta) = -0.5051  and 

theta = 2.1003 radians (or 118.8 degrees)

(iii)
f = q*a where q = (a'*b)/(a'*a) = -25/70 = -5/14 = -.03571

g = m*b where m = (a'*b)/(b'*b) = -25/35 = -5/7 = -0.7143

(iv)
b-f = b+(5/14)*a = (1/14)*[37 35 -4 -50].'

(b-f)'*a = (1/14)*(-37+245-8-200) = 0

a-g = a+(5/7)*b = (1/7)*[8 56 9 3].'

(a-g)'*a = (1/7)*(24-0-9-15) = 0


S 5.2 (P 2.34)
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(i) Let the columns of V be v1, v2, v3 and v4.

	||v1||^2 = ||V2||^2 = 4
	
	||v3||^2 = ||v4||^2 = 2

The remaining inner products are all zero:

	<v1,v2> = <v1,v3> = <v1,v4> = <v2,v3> = <v2,v4> = <v3,v4>

	= 0

Therefore

	V'*V = diag([4 4 2 2])

(ii) One can certainly solve V*c = x by Gaussian elimination, but
it's much faster to exploit the orthogonality of V and project:

	V*c = x  is equivalent to  V'*V*c = V'*x

and since V'*V is diagonal, we have:

	c1 = <v1,x>/4 = 1/2
	
	c2 = <v2,x>/4 = 0

	c3 = <v3,x>/2 = -1/2

	c4 = <v4,x>/2 = -5/2

Thus c = [1/2 0 -1/2 -5/2].'


S 5.3
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A'*A = [ 100  0    0    0 
	 0    100  0    0
         0    0    100  0
         0    0    0    100 ]

(Since all columns are circular shifts of the first column,
there are only 3 distinct dot products to compute.)

The columns of A are mutually orthogonal, with norm = 10.

(ii) A*c = b can be solved by projection, i.e., 

	(A'*A)*c = A'*b    or  100*c = A'*b

For b = [2 -1  8  -9].', we have

	c = (1/100)*[0 50 -100 50].'  = [0 1/2 -1 1/2].'

Thus b is a linear combination of the second, third and fourth
columns of A.


S 5.4
_____

i) V = [ v1 v2 v3 ]

	v1'*v2 = 3(a+bj) - 18j + 6j(a-jb)

v1'*v2 = 0 for orthogonality, so 

	Re( v1'*v2 ) = 3a + 6b = 0 -> a = -2b
	Im( v1'*v2 ) = 3b - 18 + 6a = 0 -> b = -2, a = 4

The same terms are obtained in v2'*v3 and v3'*v1, so these inner
products are zero also.  The square norm of each column then equals 65,
and
	V'*V = diag([65 65 65])

ii) 
	v1'*s =  65
	v2'*s = -130
	v3'*s = -65

Therefore s = v1 - 2*v2 - v3

(iii)  The projection of y onto x equals lambda*x, where

		x'*(y - lambda*x) = 0

i.e.,
		lambda = (x'*y)/||x||^2 

Since x and y are given in terms of the orthogonal vectors v1, v2 and v3,
inner products involving x and y can be computed directly using those 
coefficients (i.e., in the new coordinate system).  All the coefficients 
involved are real, therefore the same is true for the inner products:

	x'*y = (1 - 3/4 - 3)*65

	||x||^2 = (1/4 + 1/4 + 9)*65
Thus
	lambda =  (1 - 3/4 - 3)/(1/4 + 1/4 + 9) = -11/38