Solution: be Now Thus Let Let G any Subspace of the vector space R 24 vector 2₂ A V = 2 04-1₂3-40 y= 012 V = 0 x₂ { [2] R² | +--} x-x2 +213 - xy =0 1 0 x2-x3 + xy 04 243 r + [1₂-13. Mr may EV Satisfies [1. The above computation shows that any vector xin V can be written as a linean combination of the vectores V₁, V₂ & V₂ Pexy J.E xy D Hence the set V₁, V2, } is a Now we show that { v₁₂, 13] is a Consider ; av₁ ta₂k₂ + az v z 20 9₁-92 +93 G₁ 92 az 10 Spanning set for the subspace V linearly independent. а, газ газ го The set {V₁, V₂, ^3} is Lineaaly independent spanning set for V, thus it is a basis for the subspace V. Since the basis consists of 3 vectors the dimension d of the subspace V is 3. } (0.00) of the vecton space R4 and the dimension d'is 3. is a basis • for a subspace
Solution: be Now Thus Let Let G any Subspace of the vector space R 24 vector 2₂ A V = 2 04-1₂3-40 y= 012 V = 0 x₂ { [2] R² | +--} x-x2 +213 - xy =0 1 0 x2-x3 + xy 04 243 r + [1₂-13. Mr may EV Satisfies [1. The above computation shows that any vector xin V can be written as a linean combination of the vectores V₁, V₂ & V₂ Pexy J.E xy D Hence the set V₁, V2, } is a Now we show that { v₁₂, 13] is a Consider ; av₁ ta₂k₂ + az v z 20 9₁-92 +93 G₁ 92 az 10 Spanning set for the subspace V linearly independent. а, газ газ го The set {V₁, V₂, ^3} is Lineaaly independent spanning set for V, thus it is a basis for the subspace V. Since the basis consists of 3 vectors the dimension d of the subspace V is 3. } (0.00) of the vecton space R4 and the dimension d'is 3. is a basis • for a subspace
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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transcribe the following text in digital format
![Solution:
be
Now
Thus
x2
ER
(3²)
213
ху
a subspace of the vector space R
vector
any
[3]
Let
Let V =
04-1₂3-450
7 24 =
2
2
Y=
V₁
047
012
22-23+x4
23
y=
my
·x₂
S
из
O
+ M13
23
O
24
212
1/₂ ²
23
[1₂-13+14
x
M2
+ my
1
[1].
The above computation shows that any vector x in V can be
written as
a linear combination of the vectores V₁, V₂ & 1₂
my
O
20-214-23-²4}
x-x2 +213 x 20
EV Satisfies
ky
2
H
Hence the set {V₁, V₂, V}
Now
we show that { V₁, V₂, 13} is linearly independent.
; avita₂v₂ + az V₂ 20
Consider
=)
is a spanning set for the subspace V
→ [9₁-92 +93]
ay
a
O
аз
гаг газо
spanning
The set {V₁, V₂,^3} is Linearly independent
set for V, thus it is a basis for the subspace V.
Since the basis consists of 3 vectors, the dimension of
of the subspace V is 3.
is a
basis
• for a subspace
O
of the vecton space R4 and the dimension d'is 3.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F45fe7bab-247d-4d9b-8e43-be55c82fe47a%2Fc59f67a3-a044-408c-9f74-d2b9d3ae8ef4%2F4z3tsp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Solution:
be
Now
Thus
x2
ER
(3²)
213
ху
a subspace of the vector space R
vector
any
[3]
Let
Let V =
04-1₂3-450
7 24 =
2
2
Y=
V₁
047
012
22-23+x4
23
y=
my
·x₂
S
из
O
+ M13
23
O
24
212
1/₂ ²
23
[1₂-13+14
x
M2
+ my
1
[1].
The above computation shows that any vector x in V can be
written as
a linear combination of the vectores V₁, V₂ & 1₂
my
O
20-214-23-²4}
x-x2 +213 x 20
EV Satisfies
ky
2
H
Hence the set {V₁, V₂, V}
Now
we show that { V₁, V₂, 13} is linearly independent.
; avita₂v₂ + az V₂ 20
Consider
=)
is a spanning set for the subspace V
→ [9₁-92 +93]
ay
a
O
аз
гаг газо
spanning
The set {V₁, V₂,^3} is Linearly independent
set for V, thus it is a basis for the subspace V.
Since the basis consists of 3 vectors, the dimension of
of the subspace V is 3.
is a
basis
• for a subspace
O
of the vecton space R4 and the dimension d'is 3.
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