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Unformatted text preview: and x ∈ Cm , then a vector y ∈ V0 satisﬁes
y − x ⊥ V0 if and only if
y − x ≤ inf v − x .
v ∈V 0 Moreover, there is only one such point y in V0 with these properties.
Proof. First we assume v ∈ V0 , y − x ⊥ x. Choose any u ∈ V0 , u = y , then u = y + w with
w = 0. Consider for t ∈ R
f (t) = y + tw − x2 = (y − x) + tw2
then by the orthogonality of y − x and w ∈ V0 ,
f (t) = y − x2 + t2 w2
which has a global minimum at t = 0. Since w = 0, f (0) = P x − x2 < f (1) = u − x + w2 =
u − x2 , so among all u ∈ V0 the unique minimizer for the distance u − x is y . Conversely,
assume that there is y ∈ V0 which satisﬁes that y − x ≤ inf v∈V0 v − x. Choosing any
w ∈ V0 \ {0} and α ∈ C then gives that
f (t) = y + tαw − x2
is minimal at t = 0, so 0 = f (0) = 2Reαw, y − x. Taking the supremum of the real part over
all complex α gives that
w, y − x = 0
which me...
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 Fall '12
 BernhardBodmann
 Math, Matrices

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