Linear Algebra II (pdf, 500 kB)

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Since t = 0, we have 〈t, t〉 = 0, therefore we must have λt = 0. Since this works
for any t ∈ T , the linear combination must have been trivial.
9.10. Theorem (Bessel’s Inequality). Let V be an inner product space, and
let {e1, . . . , en} ⊂ V be an orthonormal set. Then for all x ∈ V , we have the
inequality
n
〈x, ej〉 2 ≤ x 2 .
j=1
Let U = L(e1, . . . , en) be the subspace spanned by e1, . . . , en. Then for x ∈ V , the
following statements are equivalent.
(1) x ∈ U;
n
(2) 〈x, ej〉 2 = x 2 ;
j=1
(3) x =
n
〈x, ej〉ej;
j=1
(4) for all y ∈ V , 〈x, y〉 =
n
〈x, ej〉〈ej, y〉.
j=1
In particular, statements (2) to (4) hold for all x ∈ V when e1, . . . , en is an ONB
of V .
When e1, . . . , en is an ONB, then (4) (and also (2)) is called Parseval’s Identity.
The relation in (3) is sometimes called the Fourier expansion of x relative to the
given ONB.
Proof. Let z = x −
n
〈x, ej〉ej. Then
j=1
0 ≤ 〈z, z〉 = 〈x, x〉 −
n
〈x, ej〉〈ej, x〉 = x 2 −
j=1
n
〈x, ej〉 2 .
This implies the inequality and also gives the implication (2) ⇒ (3). The implication
(3) ⇒ (4) is a simple calculation, and (4) ⇒ (2) follows by taking y = x.
(3) ⇒ (1) is trivial. Finally, to show (1) ⇒ (3), let
Then
〈x, ek〉 =
x =
n
j=1
λjej .
n
λj〈ej, ek〉 = λk ,
j=1
which gives the relation in (3).
Next, we want to discuss linear maps on inner product spaces.
j=1