Material de Estudio Humanidades Uni 2024
Material de Estudio Humanidades Uni 2024
Material de Estudio Humanidades Uni 2024
(17, 20) + 𝑈
Suppose 𝑈 is a subspace of 𝑉. Then the quotient space 𝑉/𝑈 is the set of all
translates of 𝑈. Thus
𝑉/𝑈 = {𝑣 + 𝑈 ∶ 𝑣 ∈ 𝑉}.
100 Chapter 3 Linear Maps
• If 𝑈 = {(𝑥, 2𝑥) ∈ 𝐑2 ∶ 𝑥 ∈ 𝐑}, then 𝐑2/𝑈 is the set of all lines in 𝐑2 that have
slope 2.
• If 𝑈 is a line in 𝐑3 containing the origin, then 𝐑3/𝑈 is the set of all lines in 𝐑3
parallel to 𝑈.
• If 𝑈 is a plane in 𝐑3 containing the origin, then 𝐑3/𝑈 is the set of all planes in
𝐑3 parallel to 𝑈.
Our next goal is to make 𝑉/𝑈 into a vector space. To do this, we will need the
next result.
𝑣 − 𝑤 ∈ 𝑈 ⟺ 𝑣 + 𝑈 = 𝑤 + 𝑈 ⟺ (𝑣 + 𝑈) ∩ (𝑤 + 𝑈) ≠ ∅.
𝑣 + 𝑢 = 𝑤 + ((𝑣 − 𝑤) + 𝑢) ∈ 𝑤 + 𝑈.
𝑣 + 𝑢1 = 𝑤 + 𝑢2 .
(𝑣 + 𝑈) + (𝑤 + 𝑈) = (𝑣 + 𝑤) + 𝑈
𝜆(𝑣 + 𝑈) = (𝜆𝑣) + 𝑈
As part of the proof of the next result, we will show that the definitions above
make sense.
Section 6B Orthonormal Bases 205
If 𝑣 ∈ 𝑉, then the map that sends 𝑢 The next result is named in honor
to ⟨𝑢, 𝑣⟩ is a linear functional on 𝑉. The of Frigyes Riesz (1880–1956), who
next result states that every linear func- proved several theorems early in the
tional on 𝑉 is of this form. For example, twentieth century that look very much
we can take 𝑣 = (2, −5, 1) in Example like the result below.
6.40.
Suppose we make the vector space 𝒫5 (𝐑) into an inner product space by
defining ⟨𝑝, 𝑞⟩ = ∫−11
𝑝𝑞. Let 𝜑 be as in Example 6.41. It is not obvious that there
exists 𝑞 ∈ 𝒫5 (𝐑) such that
1
∫ 𝑝(𝑡)(cos(𝜋𝑡)) 𝑑𝑡 = ⟨𝑝, 𝑞⟩
−1
for every 𝑝 ∈ 𝒫5 (𝐑) [we cannot take 𝑞(𝑡) = cos(𝜋𝑡) because that choice of 𝑞 is
not an element of 𝒫5 (𝐑)]. The next result tells us the somewhat surprising result
that there indeed exists a polynomial 𝑞 ∈ 𝒫5 (𝐑) such that the equation above
holds for all 𝑝 ∈ 𝒫5 (𝐑).
𝜑(𝑢) = ⟨𝑢, 𝑣⟩
for every 𝑢 ∈ 𝑉.
Proof First we show that there exists a vector 𝑣 ∈ 𝑉 such that 𝜑(𝑢) = ⟨𝑢, 𝑣⟩ for
every 𝑢 ∈ 𝑉. Let 𝑒1 , …, 𝑒𝑛 be an orthonormal basis of 𝑉. Then
𝜑(𝑢) = 𝜑(⟨𝑢, 𝑒1 ⟩𝑒1 + ⋯ + ⟨𝑢, 𝑒𝑛 ⟩𝑒𝑛 )
= ⟨𝑢, 𝑒1 ⟩𝜑(𝑒1 ) + ⋯ + ⟨𝑢, 𝑒𝑛 ⟩𝜑(𝑒𝑛 )
for every polynomial 𝑝 ∈ 𝒫2 (𝐑). To do this, we make 𝒫2 (𝐑) into an inner product
space by defining ⟨𝑝, 𝑞⟩ to be the right side of the equation above for 𝑝, 𝑞 ∈ 𝒫2 (𝐑).
Note that the left side of the equation above does not equal the inner product
in 𝒫2 (𝐑) of 𝑝 and the function 𝑡 ↦ cos(𝜋𝑡) because this last function is not a
polynomial.
Define a linear functional 𝜑 on 𝒫2 (𝐑) by letting
1
𝜑(𝑝) = ∫ 𝑝(𝑡)(cos(𝜋𝑡)) 𝑑𝑡
−1
for each 𝑝 ∈ 𝒫2 (𝐑). Now use the orthonormal basis from Example 6.34 and
apply formula 6.43 from the proof of the Riesz representation theorem to see that
if 𝑝 ∈ 𝒫2 (𝐑), then 𝜑(𝑝) = ⟨𝑝, 𝑞⟩, where
1 1
𝑞(𝑥) = (∫ √ 12 cos(𝜋𝑡) 𝑑𝑡)√ 12 + (∫ √ 32 𝑡 cos(𝜋𝑡) 𝑑𝑡)√ 32 𝑥
−1 −1
1
+ (∫ √ 45
8
(𝑡2 − 13 ) cos(𝜋𝑡) 𝑑𝑡)√ 45
8
(𝑥2 − 13 ).
−1