Linear Algebra 5th Edition Chapter 1 Vector Spaces
Page 54 Problem 1 Answer
Here, the zero vector space has some basis. So the statement is false.
A vector is a quantity that has a magnitude and direction.
There are different types of vectors.
A vector that has a magnitude equal to zero is called a zero vector.
Considering the definitions of zero vector and vector.
The vector is a property or quantity that has both magnitude and direction.
So even if the magnitude is zero, a vector still holds some significance because of its direction.
Therefore a zero vector space has some basis.
A vector is defined by two parameters.
So, even if one is zero the other still holds some significance. Therefore the statement is can not true.
Therefore given statement “the zero vector space has no basis” is false.
Linear Algebra 5th Edition Chapter 1 Page 54 Problem 2 Answer
Read and Learn More Stephen Friedberg Linear Algebra 5th Edition Solutions
Page 54 Problem 3 Answer
Every vector has an infinite base. So the statement is false.
A vector is a quantity that has a magnitude and direction.
There are different types of vectors.
The vector is a property or quantity that has both magnitude and direction.
So even if the direction is fixed, a vector also holds some significance because of its magnitude that can have any value.
Therefore a vector space has an infinite basis.
A vector is defined by two parameters.
So, even if one is fixed the other still can change.
Therefore the statement is can not true.
Therefore the given statement “every vector space has a finite basis” is false.
Chapter 1 Exercise 1.6 Vector Spaces Solved Examples Page 54 Problem 4 Answer
The zero vector space has some basis. So the statement is false.
A vector is a quantity that has a magnitude and direction.
There are different types of vectors.
The vector is a property or quantity that has both magnitude and direction.
So even if one property is fixed and can have only one definition, the other can have multiple definitions.
Therefore a zero vector space has some basis.
A vector is defined by two parameters.
The vector’s basis depends on both these parameters.
So, the statement is not true.
Therefore the given statement “a vector space cannot have more than one basis” is false.
Page 54 Problem 5 Answer
For a vector space of finite basis, the number of vectors in every basis is the same.
So the statement is true. A vector is a quantity that has a magnitude and direction.
There are different types of vectors.
For a vector space that has a finite basis, all parameters defining the space are fixed.
So, for every basis in it, the number of vectors will be the same.
The entire space is finite.
Therefore, if a vector space has a finite basis, then the number of vectors in every basis is the same.
The vector space is finite.
So the statement is not false.
Therefore the given statement “if a vector space has a finite basis, then the number of vectors in every basis is the same” is true.
Linear Algebra 5th Edition Chapter 1 Page 54 Problem 6 Answer
The dimension of Pn(F) is n+1. So the given statement is false.
A vector is a quantity that has a magnitude and direction.
There are different types of vectors.
From the replacement theorem we know that, for a finite-dimensional vector space, there is no linearly independent vector subset of it that can contain more vectors than its own dimension.
Following this, we can say that as we do not know the value of n the given vector space becomes an infinite-dimensional vector space.
So, the dimension of Pn(F) is n+1 as the dimension of a vector space depends on its field of scalars.
The dimension of a vector space depends on its field of scalars. So the statement is not true.
Therefore the given statement “the dimension of P(n) is n” is false.
Page 54 Problem 7 Answer
The dimensions of Mm×n(F) is mn.
So the statement is false.
A vector is a quantity that has a magnitude and direction.
There are different types of vectors.
From the replacement theorem we know that, for a finite-dimensional vector space, there is no linearly independent vector subset of it that can contain more vectors than its own dimension.
The replacement theorem also gives a heavy wealth of information about dimensions of vector spaces.
So, from the replacement theorem, the dimension of Mm×n(F) is mn.
The dimension of Mm×n(F) is mn and not m+n. So the statement is not true.
Therefore the given statement “the dimension of Mm×n
(F) is m+n” is false.
Vector Spaces Exercise 1.6 Explained Friedberg Chapter 1 Page 54 Problem 8 Answer
A vector is a quantity that has a magnitude and direction.
There are different types of vectors.
From the replacement theorem we know that, for a finite-dimensional vector space, there is no linearly independent vector subset of it that can contain more vectors than its own dimension.
Hence, for any finite-dimensional vector space V, if S1 is a linearly independent subset of V and S2 is a subset of V that generates it, then the number of vectors in S1 is always less than number of vectors in S2.So the statement is true.
A linearly independent subset has less vector in comparison to the generating subset. So the statement is not false.
Therefore the given statement “suppose that V is a finite-dimensional vector space, that S1 is a linearly independent subset of V, and that S2 is a subset of V that generates V.
Then S1 cannot contain more vectors than S2 ” is true.
Linear Algebra 5th Edition Chapter 1 Page 54 Problem 9 Answer
If S is a linearly independent set that generates V, then the vectors in V can be written as combinations of vectors in S in only one way.
So the given statement is true.
A vector is a quantity that has a magnitude and direction.
There are different types of vectors.
Using the replacement theorem we know that, for a finite-dimensional vector space, no linearly independent vector subset of it can contain more vectors than its own dimension for a finite-dimensional vector space.
If S is a linearly independent vector space.
The for the vector space V that it generates, every vector in V
can be written as a linear combination of the vector in S but only in one way.
We know S is a linearly independent set. But, if S was not linearly independent then the statement would have been false.
So the statement is not false.
Therefore the given statement “if S generates the vector space V, then every vector in V can be written as a linear combination of vectors in S in only one way” is true.
Linear Algebra 5th Edition Chapter 1 Page 54 Problem 10 Answer
The subset of a finite-dimensional set is a finite-dimensional space.
So the statement is true.
A premise of a vector space is a most extreme straightly autonomous subset of it.
Each vector space has a premise. (It very well may be limited or boundless.) and each premise of a vector space has similar number of components.
The measurement a vector space is the quantity of components in any of its bases.
Each directly free subset of a vector space can be reached out to a premise.
Therefore, the quantity of components in a directly free subset of a vector space is not exactly or equivalent to its measurement.
It is practically impossible for a subset of a finite-dimensional set to be an infinite-dimensional set. Therefore the statement is not false.
Therefore the given statement “every subspace of a finite-dimensional space is finite-dimensional” is true.
Linear Algebra 5th Edition Chapter 1 Page 54 Problem 11 Answer
A vector is a quantity that has a magnitude and direction.
There are different types of vectors.
From the replacement theorem we know that, for a finite-dimensional vector space, there is no linearly independent vector subset of it that can contain more vectors than its own dimension.
The subspace of V with dimensions 0 is given as {0} and subspace with dimension n is V.
The statement is true.
The subspace with dimensions 0 is {0} and the subspace with dimension is V.
Therefore the statement is not false.
Therefore the given statement is true.
Linear Algebra 5th Edition Chapter 1 Page 54 Problem 12 Answer
A vector is a quantity that has a magnitude and direction. There are different types of vectors.
From the replacement theorem we know that, for a finite-dimensional vector space, there is no linearly independent vector subset of it that can contain more vectors than its own dimension.
Here,S can be a spanning set of V only if SpanS=V.So, If V is a vector space having dimension n, and if S is a subset of V with n vectors, then S is linearly independent if and only if S spans V.
The statement is true.
It could be false if SpanS≠V. Therefore the statement is not false.
Therefore the given statement “if V is a vector space having dimension n, and if S is a subset of V with n vectors, then S is linearly independent if and only if S spans V” is true.
Linear Algebra 5th Edition Chapter 1 Page 55 Problem 13 Answer
Given that {(1,0,−1),(2,5,1),(0,−4,3)} n the hypothesis of vector spaces, a bunch of vectors is supposed to be directly reliant in case there is a nontrivial straight mix of the vectors that approaches the zero vector.
On the off chance that no such direct mix exists, the vectors are supposed to be straightly autonomous.
These ideas are key to the meaning of measurement.
For bases of R3 the set must be linearly independent.
Use the relation for linearly independent to conclude the final answer.
Given that:
⇒ S={(1,0,−1),(2,5,1),(0,−4,3)}
For S to be the base of R3 it must be a linearly independent set.
Thus, for linearly independent set we have,
⇒ a1(1,0,−1)+a2(2,5,1)+a3(0,−4,3)=0
Solving further;
⇒ a1(1,0,−1)+a2(2,5,1)+a3(0,−4,3)=0
⇒ a1+2a2=0 ……(1)
⇒ 5a2−4a3=0 ……(2)
⇒ −a1+a2+3a3=0 ……(3)
Adding the equations (1),(2) and (3),
8a2−a3=0
This implies that,a2=0
As a result a1=a3
Thus,
⇒ a1=0
⇒ a3=0
Thus, the given set is a base of R3.
The given set is a bases of R3.
Friedberg 5th Edition Chapter 1 Exercise 1.6 Guide Page 55 Problem 14 Answer
Given that {(2,−4,1),(0,3,−1),(6,0,−1)}
In the hypothesis of vector spaces, a bunch of vectors is supposed to be directly reliant in case there is a nontrivial straight mix of the vectors that approaches the zero vector.
On the off chance that no such direct mix exists, the vectors are supposed to be straightly autonomous.
These ideas are key to the meaning of measurement.
For bases of R3 the set must be linearly independent.
Use the relation for linearly independent to conclude the final answer.
Given that: S={(2,−4,1),(0,3,−1),(6,0,−1)}
For S to be the base of R3 it must be a linearly independent set.
For linearly independent set we have,
⇒ a1(2,−4,1)+a2(0,3,−1)+a3(6,0,−1)=0
Solving further,
⇒ a1(2,−4,1)+a2(0,3,−1)+a3(6,0,−1)=0
⇒2a1+6a3=0 ……(1)
⇒−4a1+3a2=0 ……(2)
⇒a1−a2−a3=0 ……(3)
Adding the equations (1),(2) and (3)
we get,−a1+2a2+5a3=0
⇒ a1=2a2+5a3
The above equation is satisfied for the following values;
⇒ a1=3
⇒ a2=4
⇒ a3=−1
Thus, the given set is not a base of R3.
The given set is not a base of R3.
Linear Algebra 5th Edition Chapter 1 Page 55 Problem 15 Answer
Given that {(1,2,−1),(1,0,2),(2,1,1)}
In the hypothesis of vector spaces, a bunch of vectors is supposed to be directly reliant in case there is a nontrivial straight mix of the vectors that approaches the zero vector.
On the off chance that no such direct mix exists, the vectors are supposed to be straightly autonomous.
These ideas are key to the meaning of measurement.
For bases of R3 the set must be linearly independent.
Use the relation for linearly independent to conclude the final answer.
Given that:S={(1,2,−1),(1,0,2),(2,1,1)}
For S to be a base of R3
we have,
⇒ a1(1,2,−1)+a2(1,0,2)+a3(2,1,1)=0
Solving further;
⇒ a1(1,2,−1)+a2(1,0,2)+a3(2,1,1)=0
⇒a1+a2+2a3=0 ……(1)
⇒2a1+a3=0 ……(2)
⇒−a1−2a2+a3=0 ……(3)
Adding the equations (1),(2) and (3)
we get,
⇒ 2a1−a2+4a3=0
⇒ a2=2a1+4a3
The above equation is true only if all the scalars are zero.
Thus, the given set is a base of R3 for the above condition.
The given set is a bases of R3.
Linear Algebra 5th Edition Chapter 1 Page 55 Problem 16 Answer
Given that {(−1,3,1),(2,−4,−3),(−3,8,2)}
In the hypothesis of vector spaces, a bunch of vectors is supposed to be directly reliant in case there is a nontrivial straight mix of the vectors that approaches the zero vector.
On the off chance that no such direct mix exists, the vectors are supposed to be straightly autonomous.
These ideas are key to the meaning of measurement.
For bases of R3 the set must be linearly independent.
Use the relation for linearly independent to conclude the final answer.
Given that:
⇒ S={(−1,3,1),(2,−4,−3),(−3,8,2)}
For the given set to be the base of R3
we have,
⇒ a1(−1,3,1)+a2(2,−4,−3)+a3(−3,8,2)=0
Solving further;
⇒ a1(−1,3,1)+a2(2,−4,−3)+a3(−3,8,2)=0
⇒−a1+2a2−3a3=0 ……(1)
⇒3a1−4a2+8a3=0 ……(2)
⇒a1−6a2+2a3=0 ……(3)
Adding equations (1),(2) and (3) we get,3a1−8a2+7a3=0
The above equation is only possible for the scalars to be zero.
Thus, the given set is a base of R3 only for the above condition.
The given set is a bases of R3.
Linear Algebra 5th Edition Chapter 1 Page 55 Problem 17 Answer
Given that {(1,−3,−2),(−3,1,3),(−2,−10,−2)}
In the hypothesis of vector spaces, a bunch of vectors is supposed to be directly reliant in case there is a nontrivial straight mix of the vectors that approaches the zero vector.
On the off chance that no such direct mix exists, the vectors are supposed to be straightly autonomous.
These ideas are key to the meaning of measurement.
For bases of R3 the set must be linearly independent.
Use the relation for linearly independent to conclude the final answer.
Given that:
⇒ S={(1,−3,−2),(−3,1,3),(−2,−10,−2)}
For the given set to be a base of R3
we have,
⇒ a1(1,−3,−2)+a2(−3,1,3)+a3(−2,−10,−2)=0
Solving further,
⇒ a1(1,−3,−2)+a2(−3,1,3)+a3(−2,−10,−2)=0
⇒a1−3a2−2a3=0 …(1)
⇒−3a1+a2−10a3=0 …(2)
⇒−2a1+3a2−2a3=0 …(3)
Adding equations (1),(2) and (3)
we get,−4a1+a2−14a3=0
⇒ a2=4a1+14a3
The above equation is true only if all scalars are zero.
The given set is a base of R3 for the above condition.
The given set is a base of R3.
Exercise 1.6 Notes From Friedberg Linear Algebra 5th Edition Chapter 1 Page 55 Problem 18 Answer
Given, {−1−x+2×2,2x+x−2×2,1−2x+4×2}.Determine which of the following sets are basis for P2(R).
Using dimensions and results will get.
The dim of (P2(R))=3.
∴, The set is a basis for P2(R) if it contains three linearly independent vectors.
The sets are α(−1−x+2×2)+β(2+x−2×2)+γ(1−2x+4×2)=0
⇒ (−α+2β+γ)+(−α+β−2γ)x+(2α−2β+4γ)x2=0
⇒ −α+2β+γ=0……..(1)
⇒ −α+β−2γ=0……..(2)
⇒ 2α−2β+4γ=0…….(3)
Subtract (1) from (2) equation also multiply (1) equation by 2 and add (1) and (3) equation.
⇒ −α+2β+γ=0……..(4)
⇒ −β−3γ=0……….(5)
⇒ 2β+5γ=0……….(6)
Multiply (5) equation by 2 and add (5) and (6) equation.
⇒ −α+2β+γ=0
⇒ −β−3γ=0
⇒ 0=0
⇒ α=7t
⇒ β=−3t
⇒ γ=t∀t∈R
{−1−x+2×2,2+x−2×2,1−2x+4×2} is linearly independent.
The set is not a basis for P2(R).
Hence, the set is not a basis for P2(R).
Linear Algebra 5th Edition Chapter 1 Page 55 Problem 19 Answer
Given, {1+2x+x2,3+x2,x+x2}.Determine which of the following sets are basis for P2(R).Using dimensions and results will get.
The dim (P2(R))=3.
∴, The set is a basis for P2(R) if it contains three linearly independent vectors.
The set is α(1+2x+x2)+β(3+x2)+γ(x+x2)=0
(α+3β)+(2α+γ)x+(α+β+γ)x2=0 α+3β=0……….(1)
⇒ 2α+γ=0……….(2)
⇒ α+β+γ=0…….(3)
Multiply (1) equation by 2 and subtract (1) from (2) and then subtract (1) from (3).
⇒ α+3β=0………..(4)
⇒ −6β+γ=0……….(5)
⇒ −2β+γ=0……….(6)
Subtract (5) from (6).
⇒ α+3β=0
⇒ −6β+γ=0
⇒ 4β=0
⇒ α=0
⇒ β=0
⇒ γ=0
{(1+2x+x2)+(3+x2)+(x+x2)} is a linearly independent set of three vectors.
The set is not a basis for P2(R).
Hence, the set is not a basis for P2(R).
Linear Algebra 5th Edition Chapter 1 Page 55 Problem 20 Answer
Given, {1−2x−2×2,−2+3x−x2,1−x+6×2}.Determine which of the following sets are basis for P2(R).
Using dimensions and results will get.
The dim (P2(R))=3.
∴, The set is a basis for P2(R) if it contains three linearly independent vectors.
The set is α(1−2x−2×2)+β(−2+3x−x2)+γ(1−x+6×2)=0
(α−2β+γ)+(−2α+3β−γ)x+(−2α−β+6γ)x2=0
⇒ α−2β+γ=0……….(1)
⇒ −2α+3β−γ=0……….(2)
⇒ −2α−β+6γ=0……….(3)
Multiply (1) equation by 2 and add (1) and (2)and then multiplying (1) equation by 2 and add (1) and (2).
⇒ α−2β+γ=0…………(4)
⇒ −β+γ=0…………(5)
⇒ −5β+8γ=0………..(6)
Multiplying (5) equation by 5 and subtract (5) from (6).
⇒ α−2β+γ=0
⇒ −β+γ=0
⇒ 3γ=0
⇒ α=0
⇒ β=0
⇒ γ=0
{1−2x−2×2,−2+3x−x2,1−x+6×2}is a linearly independent set of three vectors.
The set is a basis for P2(R).
Hence, the set is a basis for P2(R).
Study Resources For Chapter 1 Exercise 1.6 Vector Spaces Chapter 1 Page 55 Problem 21 Answer
Given, {1+2x+4×2,3−4x−10×2,−2−5x−6×2}.Determine which of the following sets are basis for P2(R).
Using dimensions and results will get.
The dim P2(R).
∴, The set is a basis for P2(R)if it contains three linearly independent vectors.
The set is α(−1+2x+4×2)+β(3−4x−10×2)+γ(−2−5x−6×2)=0
(−γ+3β−2γ)+(2α−4β−5γ)x+(4α−10β−64)x2=0
⇒ α+3β−2γ=0………(1)
⇒ 2α−4γ−5γ=0……….(2)
⇒ 4α−10β−6γ=0………(3)
Multiply (1) equation by 2 and add (1) and (2) equation multiplication (1) equation by 4 and add (1) and (3)equation.
⇒ −α+3β−2γ=0……….(4)
⇒ 2β−9γ=0………..(5)
⇒ 2β−14γ=0………(6)Subtract (5) from (6).
⇒ −α+3β−2γ=0
⇒ 2β−9γ=0
⇒ −5γ=0
{−1+2x+4×2,3−4x−10×2,−2−5x−6×2} is a linearly independent set of three vectors.
The set is a basis for P2(R).
Hence, the set is a basis for P2(R).
Linear Algebra 5th Edition Chapter 1 Page 55 Problem 22 Answer
Given, {1+2x−x2,4−2x+x2,−1+18−9×2}.Determine which of the following sets are basis for P2(R).
Using dimensions and results will get.
The dim (P2(R))=3.
∴, The set is a basis for P2(R) if it contains three linearly independent vectors.
he set isα(1+2x−x2)+β(4−2x+x2)+γ(−1+18x−9×2)=0
(α+4β−γ)+(2α−2β+18γ)x+(−α+β−9γ)x2=0
α+4β−γ=0……….(1)
2α−2β+18γ=0……….(2)
−α+β−9γ=0………..(3)
Multiply (1) equation by 2 and subtract (1) from (2) equation then add (1) and (3) equation.
α+4β−γ=0………..(4)
−10β+20γ=0……….(5)
5β−10γ=0………..(6)
Multiply (6) equation by (2) and add (5) and (6).
⇒ α+4β−γ=0
⇒ 0=0
⇒ 5β−10γ=0
⇒ α=−7t,
⇒ β=2t
⇒ γ=t∀t∈R
{1+2x−x2,4−2x+2×2,−1+18x−9×2} is linearly independent.
The set is not a basis for P2(R).
Hence, the set is not a basis for P2(R).
Page 55 Problem 23 Answer
Given, If {(1,4,−6),(1,5,8)(2,1,1)(0,1,0)} a linearly independent subset of R3 .Justify your answer.
Any linearly independent set which generates a vector space will have less number of vectors than the dimension of vector space.
R3 is a 3− dimensional vector space.
Hence, these cannot be a set of 4 linearly independent vectors in R3.
Hence, there cannot be a set of 4 linearly independent vectors in R3.
Linear Algebra 5th Edition Chapter 1 Page 55 Problem 24 Answer
We are given, u1=(2,−3,1)
⇒ u2=(1,4,−2)
⇒ u3=(−8,12,−4)
⇒ u4=(1,37,−17)
⇒ u5=(−3,−5,8) generate R3.
We have to find a subset of the set {u1,u2,u3,u4,u5} that is a basis for R3.
We check independence of u1 and u2.
αu1=u2
⇒ α(2,−3,1)=(1,4,−1)
⇒ 2α=1,
−3α=4
α=−2 (not possible)
⇒{u1,u2} is linearly dependent.
We check independence of u1,u2 and u3.
αu1+βu2=u3
⇒ α(2,−3,1)+β(1,4,−2)=(−8,12,−4)
⇒(2α+β,−3α+4β,α−2β)=(−8,12−4)
2α+β=−8
−3α+4β=12
α−2β=−4
solving these we get,
β=0
α=−4
⇒ the set {u1,u2,u3} is not linearly independent.
We check independence of u1,u2, and u4.
αu1+βu2=u4
⇒ α(2,−3,1)+β(1,4,−1)=(1,37,−17)
⇒ (2α+β,−3α+4β,α−2β)=(1,37,−17)
⇒ 2α+β=1,
−3α+4β=37,
α−2β=−17
solving these we get, β=7, α=−3
the set {u1,u2,u3}is not linearly independent.
We check independence of u1,u2 and u5.
αu1+βu2=u5
⇒α(2,−3,1)+β(1,4,−2)=(−3,−5,8)
⇒(2α+β,−3α+4β,α−2β)=(−3,−5,8)
⇒ 2α+β=−3,
−3α+4β=−5,
α−2β=8
From first two equations, we get
β=−7/5 and from first and third equation, we get
β=−19/5. This is not possible.
Therefore, the set {u1,u2,u5} is linearly independent, so, this forms our basis for R3
A subset of the set {u1,u2,u3,u4,u5} that forms a basis for R3 is {u1,u2,u5} or {(2,−3,1),(1,4,−2),(−3,−5,8)}.
Examples of Exercise 1.6 In Friedberg Linear Algebra Chapter 1 Page 55 Problem 25 Answer
We are given vectors, u1=(2,−3,4,−5,2),
⇒ u2=(−6,9,−12,15,−6),
⇒ u3=(3,−2,7,−9,1),
⇒ u4=(2,−8,2,−2,6),
⇒ u5=(−1,1,2,1,−3),
⇒ u6=(0,−3,−18,9,12),
⇒ u7=(1,0,−2,3,−2), and
⇒ u8=(2,−1,1,−9,7).
We have to find a subset of the set {u1,u2,u3,u4,u5} that is a basis for W, where W′ is a subspace for R5.
We check independence of u1 and u2.
αu1=u2
⇒α(2,−3,4,−5,2)=(−6,9,−12,15,−6)
⇒α=−3 [linearly dependent].
We check independence of u1 and u3.
αu1=u3
⇒ α(2,−3,4,−5,2)=(3,−2,7,−9,1)
There is no value of α that satisfies.
⇒ the set {u1,u3} is linearly independent.
We check independence of u1,u3 and u4.
αu1+βu3=u4
⇒ α(2,−3,4,−5,2)+β(3,−2,7,−9,1)=(2,−8,2,−2,6)
⇒(2α+3β,−3α−2β,4α+7β,−5α−9β,2α+β)=(2,−8,2,−2,6)
solving the equations,
⇒ 2α+3β=2,
⇒ −3α−2β=−8,
⇒ 4α+7β=2,
⇒ −5α−9β=−2,
⇒ 2α+β=6
we get,
⇒ α=4,
⇒ β=−2
{u1,u3,u4} is linearly independent.
We check independence of u1,u3 and u5.
αu1+βu3=u5
⇒ α(2,−3,4,−5,2)+β(3,−2,7,−9,1)=(−1,1,2,1,−3)
⇒ (2α+3β,−3α−2β,4α+7β,−5α−9β,2α+β)=(−1,1,2,1,−3)
⇒ 2α+3β=−1,
−3α−2β=1,
4α+7β=2,
−5α−9β=1,
2α+β)=−3
After solving the equations, we don’t have a unique value of α and β.
⇒ the set {u1,u3,u5} is linearly independent.
We check independence of u1,u3,u5 and u6.
αu1+βu2+γu5=u6
⇒α(2,−3,4,−5,2)+β(3,−2,7,−9,1)+γ(−1,1,2,1,−3)=(0,−3,−18,9,12)
⇒2α+3β−γ=0,
−3α−2β+γ=−3,
4α+7β+2γ=−18,
−5α−9β+γ=9,
2α+β−3γ=12
solving the above equations, we get
α=1,
β=−2,
γ=−4
⇒ the set {u1,u3,u5,u6} is linearly independent.
We check independence of u1,u3,u5,u7.
αu1+βu2+γu5=u7
⇒ α(2,−3,4,−5,2)+β(3,−2,7,−9,1)+γ(−1,1,2,1,−3)=(1,0,−2,3,−2)
⇒2α+3β−γ=1,
−3α−2β+γ=0,
4α+7β+2γ=−2,
−5α−9β+γ=3,
2α+β−3γ=−2
After solving the equations, we don’t have a unique value of α,β and γ.
⇒ the set {u1,u3,u5,u7} is linearly independent.
We check independence of u1,u3,u5,u7 and u8.
αu1+βu2+γu5+δu7=u8
⇒α(2,−3,4,−5,2)+β(3,−2,7,−9,1)+γ(−1,1,2,1,−3)+δ(1,0,−2,3,−2)=(2,−1,1,−9,7)
⇒2α+3β−γ+δ=2,
⇒ −3α−2β+γ=−1,
⇒ 4α+7β+2γ−2δ=1,
⇒ −5α−9β+γ+3δ=−9,
⇒ 2α+β−3γ−2δ=7
After solving the equations, we get
⇒ α=−1,
⇒ β=1,
⇒ γ=−2,
⇒ δ=−1
⇒ the set {u1,u3,u5,u7,u8} is linearly independent.
A subset of the set {u1,u2,u3,u4,u5,u6,u7,u8} that forms a basis for W is {u1,u3,u5,u7} or {(2,−3,4,−5,2),(3,−2,7,−9,1),(−1,1,2,1,−3),(1,0,−2,3,−2)}.
Linear Algebra 5th Edition Chapter 1 Page 56 Problem 26 Answer
We are given, u1=(1,1,1,1),
⇒ u2=(0,1,1,1),
⇒ u3=(0,0,1,1) and
u4=(0,0,0,1) from a basis for F4.
We have to find the unique representation of an arbitrary vector {a1,a2,a3,a4} in F4 as a linear combination of u1,u2,u3 and u4.
Let’s assume an arbitrary vector
a=(a1,a2,a3,a4)∈F4
⇒α1u1+α2u2+α3u3+α4u4=a
⇒α1(1,1,1,1)+α2(0,1,1,1)+α3(0,0,1,1)+α4(0,0,0,1)=(a1,a2,a3,a4)
⇒(α1,α1+α2,α1+α2+α3,α1+α2+α3+α4)=(a1,a2,a3,a4)
⇒α1=a1,
α1+α2=a2,α1+α2+α3=a3, α1+α2+α3+α4=a4.
After solving the above set of equations, we get
⇒ α1=a1,
⇒ α2=a2−a1,
⇒ α3=a3−a2,
⇒ α4=a4−a3.
Therefore, the unique representation of
a=(a1,a2,a3,a4) in terms of u1,u2,u3 and u4 is a=a1u1+(a2−a1)u2+(a3−a2)u3+(a4−a3)u4.