_______________
v1 = [1 1 1].'
v2 = [2 1 3].'
s = [1 2 3].'
(i)
V = [v1 v2]
v1^2 = 1+1+1 = 3
v2^2 = 4+1+9 = 13
v1'*v2 = 2 1 +3 = 0
(v1 and v2 are orthogonal)
V'*V = [3
0
0
14]
V'*V is diagonal, so its inverse is given by
[1/3 0
0
1/14]
Now
s_hat = c(1)*v1 + c(2)*v2,
where (V'*V)*c = V'*s, or c = (V'*V)\(V'*s)
Here
v1'*s = 1+2+3 = 4
v2'*s = 22+9 = 9
and thus
c1 = 4/3
and
c2 = 9/14
(Since v1 and v2 are orthogonal, the projection of s onto R(V) is the
sum of the projections onto v1 and v2.)
s_hat = V*c = (1/42)*[2 29 137].'
= [0.0476
0.6905
3.2619]'
(ii)
Projection of s onto v1 is q*v1, where
q = (v1'*s)/(v1'*v1) = 4/3
(this was computed above).
Projection of s_hat onto v1 is m*v1, where
m = (v1'*s_hat)/(v1'*v1)
Since
v1'*s_hat = (1/42)*[2 29 137]*[1 1 1].' = 4
v1'*v1 = 3
the same answer is obtained.
In three dimensions, we know that the projection of a point
which lies above a plane onto a line lying on the plane can be
accomplished in two stages:
first project onto the plane, then
project that point onto the line.
The equation V'*s_hat = V'*s
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 Spring '08
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