ECE 275A Homework # 1 Solutions
1. To solve this problem, we need to start with some known relationship between
∥
x
∥
,
∥
y
∥
, and
⟨
x, y
⟩
, together with one or more free parameters (if possible) to give us some
ﬂexibility in deriving the CS inequality. Perhaps the simplest such relationship we
can envision (having only one free parameter) is
0
≤ ∥
x
−
αy
∥
2
=
⟨
x
−
αy, x
−
αy
⟩
=
∥
x
∥
2
−
2 Re
α
⟨
x, y
⟩
+

α

2
∥
y
∥
2
.
(1)
Note that this relationship involves the squares of the norms of
x
and
y
, whereas the
CS inequaliy involves the norms themselves. Thus, in order to understand better
where we are heading, let us rewrite the CS inequality as
⟨
x, y
⟩
2
∥
y
∥
2
≤ ∥
x
∥
2
.
(2)
We can now try to ﬁnd a value for
α
which will place equation (1) into the form (2).
Without too much work, one can show that the value,
α
=
⟨
x, y
⟩
∥
y
∥
2
,
will do the job. Note the combination of educated guessing and luck involved in this
proof, which is a standard one encountered in many textbooks.
Comment:
Don’t feel at a loss when studying the optimizationbased proof used in the
Moon text. Although the proof he gives is valid
qua
proof, it is not one that is natural
to use at this stage of knowledge. It would be pedagogically useful to be given an
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 Fall '08
 Chandrasekara
 Linear Algebra, #, xH Ax

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