Linear Algebra 5th Edition Chapter 2 Linear Transformations and Matrices
Page 116 Problem 1 Answer
Given:
β={x1,x2,x3,x4,….,xn}
β′={x′1,x′2,x′3,x′4,….,x′n}
To find: The jth column of Q is [xj]β′.
To determine whether the statement is true or false.
Now,
xj′=i=1∑nQijxi,jxj′=1,2,3,….,n
Thus, the jth column of Q is [xj′]β
Since the jth column of Q is [xj′]β, and hence, the given statement is false.
Page 116 Problem 2 Answer
Given Statement: Every change of coordinate matrix is invertible.
To find: Whether the given statement is true or false.
Since the change of matrix can be written as,
⇒ Q=[IV]β′β
As IV is the identity transformation. So, it is invertible.
Since IV is invertible, then Q is also invertible.
Therefore, the given statement is true.
Since IV is invertible, then Q is also invertible and hence, the given statement is true.
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Stephen Friedberg Linear Algebra 5th Edition Solutions For Exercise 2.5 Chapter 2 Page 116 Problem 3 Answer
Given: T be a linear operator on a vector space with finite dimensions V.
Ordered bases for V are β and β′.
The change of coordinate matrix is Q.
To prove: [T]β=Q[T]β′
⇒ Q−1
Let, β and β′ be (x1,x2,x3,x4,….,xn) and (x′1,x′2,x′3,x′4,….,x′n) .
Since the change of matrix can be written as,
⇒ Q=[IV]β′β
As IV is the identity transformation, then use the identity property;
⇒ TI=IT
⇒ TI=T
Now, the matrix [T]β is
⇒ Q[T]β′=[I]−ββ
⇒ [T]−ββ
⇒ Q[T]β′
=Q[T]β
Hence,
⇒ Q[T]β′
=Q[T]β[T]β
=Q[T]β′Q−1
Therefore, the given statement is true.
Since, [T]β=Q[T]β′
Q−1 is proved and hence, the given statement is true.
Linear Algebra 5th Edition Chapter 2 Page 116 Problem 4 Answer
Linear Algebra 5th Edition Chapter 2 Page 116 Problem 5 Answer
Given: Let T be a linear operator on a vector space with finite dimensions V.
To prove: [T]β is similar to [T]γ.
According to similar matrices property,
If T is a linear operator on V then the matrices [T]β and [T]γ. So, it must follow the relation,
⇒ [T]γ=Q−1
⇒ [T]βQ
Therefore, the given statement is true.
Since, [T]γ=Q−1[T]βQ is proved and hence, the given statement is true.
Page 117 Problem 6 Answer
Given in the question are,
⇒ β={e1,e2} and β′={(a1,a2),(b1,b2)}
To find: [IR2]β′β=?
For finding the change of coordinate matrix, find (a1,a2) and (b1,b2), then put the values in [IR2]β′β.
Let, Q=[IR2]β′β
Now, find (a1,a2)
(a1,a2)=a1×(1,0)+a2×(0,1)(a1,a2)=a1×e1+a2×e2
Now, find (b1,b2)
(b1,b2)=b1×(1,0)+b2×(0,1)(b1,b2)=b1×e1+b2×e2
Therefore,
The change of coordinate matrix is,
Chapter 2 Exercise 2.5 Linear Transformations Solved Problems Linear Algebra 5th Edition Page 117 Problem 7 Answer
Given in the question are,
β={(−1,3),(2,−1)} and β′={(0,10),(5,0)}
To find: [IR2]β′β=?
For finding the change of coordinate matrix, find (0,10) and (5,0) , then put the values in [IR2]β′β.
Let, Q=[IR2]β′β
Now, find (0,10)
⇒ (0,10)=A×(−1,3)+B×(2,−1)
⇒ (0,10)=(−A,3A)+(2B,−B)
⇒ (0,10)=(−A+2B,2A−B)
On equating,
⇒ −A+2B=0 .. (1)
⇒ 2A−B=10 .. (2)
Solve the equation (1) and (2), get the results as
⇒ A=4
⇒ B=2
Therefore,
⇒ (0,10)=4×(−1,3)+2×(2,−1)
Now, find (0,10)
⇒ (5,0)=A×(−1,3)+B×(2,−1)
⇒ (5,0)=(−A,3A)+(2B,−B)
⇒ (5,0)=(−A+2B,2A−B)
On equating,
⇒ −A+2B=5 .. (3)
⇒ 2A−B=0 .. (4)
Solve the equation (3) and (4), get the results as
⇒ A=1
⇒ B=3
Therefore,
(5,0)=1×(−1,3)+3×(2,−1)
The change of coordinate matrix is,
Linear Algebra 5th Edition Chapter 2 Page 117 Problem 8 Answer
Given in the question are,
β={(2,5),(−1,−3)} and β′={e1,e2}
To find: [IR2]β′β=?
For finding the change of coordinate matrix, find (1,0) and (0,1), then put the values in [IR2]β′β.
Let, Q=[IR2]β′β
Now, find (1,0)
⇒ (1,0)=A×(2,5)+B×(−1,−3)
⇒ (1,0)=(2A,5A)+(−B,−3B)
⇒ (1,0)=(2A−B,5A−3B)
On equating,
⇒ 2A−B=1 .. (1)
⇒ 5A−3B=0 .. (2)
Solve the equation (1) and (2), get the results as
⇒ A=3
⇒ B=5
Therefore,
(1,0)=3×(2,5)+5×(−1,−3)
Now, find (0,1)
⇒ (0,1)=A×(2,5)+B×(−1,−3)
⇒ (0,1)=(2A,5A)+(−B,−3B)
⇒ (0,1)=(2A−B,5A−3B)
On equating,
⇒ 2A−B=0 .. (3)
⇒ 5A−3B=1 .. (4)
Solve the equation (3) and (4), get the results as
⇒ A=−1
⇒ B=−2
Therefore,
(0,1)=−1×(2,5)−2×(−1,−3)
The change of coordinate matrix is,
Linear Algebra 5th Edition Chapter 2 Page 117 Problem 9 Answer
Given in the question are, β={(−4,3),(2,−1)} and β′={(2,1),(−4,1)}
To find: [IR2]β′β=?
For finding the change of coordinate matrix, find (2,1) and (−4,1) , then put the values in [IR2]β′β.
Let, Q=[IR2]β′β
Now, find (2,1)
⇒ (2,1)=A×(−4,3)+B×(2,−1)
⇒ (2,1)=(−4A,3A)+(2B,−B)
⇒ (2,1)=(−4A+2B,3A−B)
On equating,−4A+2B=2 .. (1)
⇒ 3A−B=1 .. (2)
Solve the equation (1) and (2), get the results as
⇒ A=2
⇒ B=5
Therefore,
⇒ (2,1)=2×(−4,3)+5×(2,−1)
Now, find (−4,1)
⇒ (−4,1)=A×(−4,3)+B×(2,−1)
⇒ (−4,1)=(−4A,3A)+(2B,−B)
⇒ (−4,1)=(−4A+2B,3A−B)
On equating,−4A+2B=−4 .. (3)
⇒ 3A−B=1 .. (4)
Solve the equation (3) and (4), get the results as
⇒ A=−1
⇒ B=−4
Therefore,
⇒ (−4,1)=−1×(−4,3)−4×(2,−1)
The change of coordinate matrix is,
Linear Algebra Friedberg Exercise 2.5 Step-By-Step Solutions Chapter 2 Page 118 Problem 10 Answer
In the question, Given vector space over field Mn×n(F).
To prove “is similar to” is an equivalence relation on Mn×n(F) or not.
Prove that it is reflexive, symmetric and transitive to prove it equivalence.
Given vector space over field Mn×n(F).
Reflexive relation: Let A∈Mn×n(F)
Clearly, we can say that A is similar to A .
So, it is a reflexive relation.
Symmetric relation: Let A,B∈Mn×n(F) and A is similar to B .
Then, clearly, we can say that, B is similar to A .
So, it is a symmetric relation.
Transitive relation: Let A,B,C∈Mn×n(F) and A is similar to B and B is similar to C.
Then, clearly, we can say that A is similar to C .
So, it is a transitive relation.
Hence, “is similar to” is an equivalence relation on Mn×n(F)
Hence, “is similar to” is an equivalence relation on Mn×n(F) .
Page 118 Problem 11 Answer
Given in question that, A and B are similar n×n matrices.
In this we need to prove tr(A)=tr(B) .
As we know, if A and B are similar n×n matrices, then there exists an invertible matrix P such that B=P−1AP.
Solve by taking traces on both sides.
Given that, A and B are similar n×n matrices.
Consider B=P−1AP
Taking trace on both sides, we get,trB=tr(P−1AP)
Applying tr(A)=tr(B) , we get,
⇒ trB=tr(P−1(AP))
⇒ trB=tr((AP)P−1)
⇒ trB=tr(A(PP−1))
⇒ trB=tr(AI)
⇒ trB=trA
Hence, if A and B are similar n×n matrices, then tr(A)=tr(B).
Linear Algebra 5th Edition Chapter 2 Page 118 Problem 12 Answer
Given data: The given data is a finite-dimensional vector space.
To define: The trace of a linear operator on a finite-dimensional vector space and justify that the definition is well defined.
Observe that the trace of a linear operator T: V→V is tr([T]β) where β is the basis for V.
Now, consider α and β to be the 2 bases of V and consider T: V→V. Using exercise-8 it is clearly known that ∃ matrices P and P−1 as shown below,
Therefore,
⇒ [T]α=P[T]βP−1
And, henceforth by the definition of the similar matrices, it can be clearly obtained that [T]α and [T]β are similar matrices and using part (a), it implies,
⇒ tr([T]α)=tr([T]β)
Thus, it can be clearly the trace of a linear operator on a finite dimensional vector space can be clearly defined and justified that the definition is well defined.