To prove that | R ( A ) | = 2 r and conclude that A defines a permutation on S n only if A has full rank. Given information: r is the rank of matrix A. Preimage of y is defined by P ( A , y ) = { x : A x = y } , for given n × n matrix A and given a value y ∈ R ( A ) . Explanation: It can be assumed without generality loss, that the first r columns of A are linearly independent. Then for x 1 , x 3 ∈ S n , it can be concluded that A x 1 ≠ A x 2 based on following two conditions. In the first r entries x 1 and x 2 are not identical. And x 1 , x 2 have 0’s in the remaining entries. Now A x 1 ≠ A x 2 is true since the first r entries of each are a linear combination of the first r rows of A , and it is impossible to have two different linear combinations that are equal of them because they are independent. Now it is must that | R ( A ) | ≥ 2 r as there are at least 2 r non-equivalent vectors x ∈ S n . Alternatively, x is a vector that doesn’t consists of 0’s in the coordinates that larger than r . Therefore A x = ∑ x i a i where a i is the i t h column of A . Now as each of the last n − r columns of A is in fact a linear combination of the first r columns of A , this can be rewritten as a linear combination of the first r columns of A . Since all of these has been already, | R ( A ) | = 2 r . The range cannot include all 2 n elements of S n if A doesn’t have full rank so, permutation can be defined possible by A

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Answer 1

Question

Chapter D, Problem 2P

a.

Program Plan Intro

To prove that |R(A)|=2r and conclude that A defines a permutation on Sn only if A has full rank.

Given information:

r is the rank of matrix A.

Preimage of y is defined by P(A,y)={x:Ax=y} , for given n×n matrix A and given a value y∈R(A) .

Explanation:

It can be assumed without generality loss, that the first r columns of A are linearly independent.

Then for x1,x3∈Sn , it can be concluded that Ax1≠Ax2 based on following two conditions.

In the first r entries x1 and x2 are not identical. And x1 , x2
have 0’s in the remaining entries.

Now Ax1≠Ax2 is true since the first r entries of each are a linear combination of the first r rows of A , and it is impossible to have two different linear combinations that are equal of them because they are independent.

Now it is must that |R(A)|≥2r as there are at least 2r non-equivalent vectors x∈Sn . Alternatively, x is a vector that doesn’t consists of 0’s in the coordinates that larger than r . Therefore Ax=∑xiai where a_i is the ith column of A . Now as each of the last n−r columns of A is in fact a linear combination of the first r columns of A , this can be rewritten as a linear combination of the first r columns of A .

Since all of these has been already, |R(A)|=2r . The range cannot include all 2n elements of Sn if A doesn’t have full rank so, permutation can be defined possible by A

b.

Program Plan Intro

To prove that |P(A,y)|=2n−r if r is the rank of n×n matrix A and y∈R(A) .

c.

Program Plan Intro

To prove that |B(S',m)|=2r and that for each block in B(S',m) exactly 2m−r numbers in S map to that block.

d.

Program Plan Intro

To show that number of linear permutations of Sn is much less than number of permutations of S using counting argument.

e.

Program Plan Intro

To provide an example of a value n and permutation of Sn that cannot be achieved by any linear permutation.

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Answer 2

Question

Chapter D, Problem 2P

a.

Program Plan Intro

To prove that |R(A)|=2r and conclude that A defines a permutation on Sn only if A has full rank.

Given information:

r is the rank of matrix A.

Preimage of y is defined by P(A,y)={x:Ax=y} , for given n×n matrix A and given a value y∈R(A) .

Explanation:

It can be assumed without generality loss, that the first r columns of A are linearly independent.

Then for x1,x3∈Sn , it can be concluded that Ax1≠Ax2 based on following two conditions.

In the first r entries x1 and x2 are not identical. And x1 , x2
have 0’s in the remaining entries.

Now Ax1≠Ax2 is true since the first r entries of each are a linear combination of the first r rows of A , and it is impossible to have two different linear combinations that are equal of them because they are independent.

Now it is must that |R(A)|≥2r as there are at least 2r non-equivalent vectors x∈Sn . Alternatively, x is a vector that doesn’t consists of 0’s in the coordinates that larger than r . Therefore Ax=∑xiai where a_i is the ith column of A . Now as each of the last n−r columns of A is in fact a linear combination of the first r columns of A , this can be rewritten as a linear combination of the first r columns of A .

Since all of these has been already, |R(A)|=2r . The range cannot include all 2n elements of Sn if A doesn’t have full rank so, permutation can be defined possible by A

b.

Program Plan Intro

To prove that |P(A,y)|=2n−r if r is the rank of n×n matrix A and y∈R(A) .

c.

Program Plan Intro

To prove that |B(S',m)|=2r and that for each block in B(S',m) exactly 2m−r numbers in S map to that block.

d.

Program Plan Intro

To show that number of linear permutations of Sn is much less than number of permutations of S using counting argument.

e.

Program Plan Intro

To provide an example of a value n and permutation of Sn that cannot be achieved by any linear permutation.

Expert Solution & Answer

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See solution

Answer 3

Question

Chapter D, Problem 2P

a.

Program Plan Intro

To prove that |R(A)|=2r and conclude that A defines a permutation on Sn only if A has full rank.

Given information:

r is the rank of matrix A.

Preimage of y is defined by P(A,y)={x:Ax=y} , for given n×n matrix A and given a value y∈R(A) .

Explanation:

It can be assumed without generality loss, that the first r columns of A are linearly independent.

Then for x1,x3∈Sn , it can be concluded that Ax1≠Ax2 based on following two conditions.

In the first r entries x1 and x2 are not identical. And x1 , x2
have 0’s in the remaining entries.

Now Ax1≠Ax2 is true since the first r entries of each are a linear combination of the first r rows of A , and it is impossible to have two different linear combinations that are equal of them because they are independent.

Now it is must that |R(A)|≥2r as there are at least 2r non-equivalent vectors x∈Sn . Alternatively, x is a vector that doesn’t consists of 0’s in the coordinates that larger than r . Therefore Ax=∑xiai where a_i is the ith column of A . Now as each of the last n−r columns of A is in fact a linear combination of the first r columns of A , this can be rewritten as a linear combination of the first r columns of A .

Since all of these has been already, |R(A)|=2r . The range cannot include all 2n elements of Sn if A doesn’t have full rank so, permutation can be defined possible by A

b.

Program Plan Intro

To prove that |P(A,y)|=2n−r if r is the rank of n×n matrix A and y∈R(A) .

c.

Program Plan Intro

To prove that |B(S',m)|=2r and that for each block in B(S',m) exactly 2m−r numbers in S map to that block.

d.

Program Plan Intro

To show that number of linear permutations of Sn is much less than number of permutations of S using counting argument.

e.

Program Plan Intro

To provide an example of a value n and permutation of Sn that cannot be achieved by any linear permutation.

Expert Solution & Answer

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See solution

Answer 4

Question

Chapter D, Problem 2P

a.

Program Plan Intro

To prove that |R(A)|=2r and conclude that A defines a permutation on Sn only if A has full rank.

Given information:

r is the rank of matrix A.

Preimage of y is defined by P(A,y)={x:Ax=y} , for given n×n matrix A and given a value y∈R(A) .

Explanation:

It can be assumed without generality loss, that the first r columns of A are linearly independent.

Then for x1,x3∈Sn , it can be concluded that Ax1≠Ax2 based on following two conditions.

In the first r entries x1 and x2 are not identical. And x1 , x2
have 0’s in the remaining entries.

Now Ax1≠Ax2 is true since the first r entries of each are a linear combination of the first r rows of A , and it is impossible to have two different linear combinations that are equal of them because they are independent.

Now it is must that |R(A)|≥2r as there are at least 2r non-equivalent vectors x∈Sn . Alternatively, x is a vector that doesn’t consists of 0’s in the coordinates that larger than r . Therefore Ax=∑xiai where a_i is the ith column of A . Now as each of the last n−r columns of A is in fact a linear combination of the first r columns of A , this can be rewritten as a linear combination of the first r columns of A .

Since all of these has been already, |R(A)|=2r . The range cannot include all 2n elements of Sn if A doesn’t have full rank so, permutation can be defined possible by A

b.

Program Plan Intro

To prove that |P(A,y)|=2n−r if r is the rank of n×n matrix A and y∈R(A) .

c.

Program Plan Intro

To prove that |B(S',m)|=2r and that for each block in B(S',m) exactly 2m−r numbers in S map to that block.

d.

Program Plan Intro

To show that number of linear permutations of Sn is much less than number of permutations of S using counting argument.

e.

Program Plan Intro

To provide an example of a value n and permutation of Sn that cannot be achieved by any linear permutation.

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Chapter D, Problem 2P

a.

Program Plan Intro

To prove that |R(A)|=2r and conclude that A defines a permutation on Sn only if A has full rank.

Given information:

r is the rank of matrix A.

Preimage of y is defined by P(A,y)={x:Ax=y} , for given n×n matrix A and given a value y∈R(A) .

Explanation:

It can be assumed without generality loss, that the first r columns of A are linearly independent.

Then for x1,x3∈Sn , it can be concluded that Ax1≠Ax2 based on following two conditions.

In the first r entries x1 and x2 are not identical. And x1 , x2
have 0’s in the remaining entries.

Now Ax1≠Ax2 is true since the first r entries of each are a linear combination of the first r rows of A , and it is impossible to have two different linear combinations that are equal of them because they are independent.

Now it is must that |R(A)|≥2r as there are at least 2r non-equivalent vectors x∈Sn . Alternatively, x is a vector that doesn’t consists of 0’s in the coordinates that larger than r . Therefore Ax=∑xiai where a_i is the ith column of A . Now as each of the last n−r columns of A is in fact a linear combination of the first r columns of A , this can be rewritten as a linear combination of the first r columns of A .

Since all of these has been already, |R(A)|=2r . The range cannot include all 2n elements of Sn if A doesn’t have full rank so, permutation can be defined possible by A

b.

Program Plan Intro

To prove that |P(A,y)|=2n−r if r is the rank of n×n matrix A and y∈R(A) .

c.

Program Plan Intro

To prove that |B(S',m)|=2r and that for each block in B(S',m) exactly 2m−r numbers in S map to that block.

d.

Program Plan Intro

To show that number of linear permutations of Sn is much less than number of permutations of S using counting argument.

e.

Program Plan Intro

To provide an example of a value n and permutation of Sn that cannot be achieved by any linear permutation.

Answer 6

Chapter D, Problem 2P

a.

Program Plan Intro

To prove that |R(A)|=2r and conclude that A defines a permutation on Sn only if A has full rank.

Given information:

r is the rank of matrix A.

Preimage of y is defined by P(A,y)={x:Ax=y} , for given n×n matrix A and given a value y∈R(A) .

Explanation:

It can be assumed without generality loss, that the first r columns of A are linearly independent.

Then for x1,x3∈Sn , it can be concluded that Ax1≠Ax2 based on following two conditions.

In the first r entries x1 and x2 are not identical. And x1 , x2
have 0’s in the remaining entries.

Now Ax1≠Ax2 is true since the first r entries of each are a linear combination of the first r rows of A , and it is impossible to have two different linear combinations that are equal of them because they are independent.

Now it is must that |R(A)|≥2r as there are at least 2r non-equivalent vectors x∈Sn . Alternatively, x is a vector that doesn’t consists of 0’s in the coordinates that larger than r . Therefore Ax=∑xiai where a_i is the ith column of A . Now as each of the last n−r columns of A is in fact a linear combination of the first r columns of A , this can be rewritten as a linear combination of the first r columns of A .

Since all of these has been already, |R(A)|=2r . The range cannot include all 2n elements of Sn if A doesn’t have full rank so, permutation can be defined possible by A

Answer 7

Chapter D, Problem 2P

a.

Program Plan Intro

To prove that |R(A)|=2r and conclude that A defines a permutation on Sn only if A has full rank.

Given information:

r is the rank of matrix A.

Preimage of y is defined by P(A,y)={x:Ax=y} , for given n×n matrix A and given a value y∈R(A) .

Explanation:

It can be assumed without generality loss, that the first r columns of A are linearly independent.

Then for x1,x3∈Sn , it can be concluded that Ax1≠Ax2 based on following two conditions.

In the first r entries x1 and x2 are not identical. And x1 , x2
have 0’s in the remaining entries.

Now Ax1≠Ax2 is true since the first r entries of each are a linear combination of the first r rows of A , and it is impossible to have two different linear combinations that are equal of them because they are independent.

Now it is must that |R(A)|≥2r as there are at least 2r non-equivalent vectors x∈Sn . Alternatively, x is a vector that doesn’t consists of 0’s in the coordinates that larger than r . Therefore Ax=∑xiai where a_i is the ith column of A . Now as each of the last n−r columns of A is in fact a linear combination of the first r columns of A , this can be rewritten as a linear combination of the first r columns of A .

Since all of these has been already, |R(A)|=2r . The range cannot include all 2n elements of Sn if A doesn’t have full rank so, permutation can be defined possible by A

Answer 8

b.

Program Plan Intro

To prove that |P(A,y)|=2n−r if r is the rank of n×n matrix A and y∈R(A) .

Answer 9

b.

Program Plan Intro

To prove that |P(A,y)|=2n−r if r is the rank of n×n matrix A and y∈R(A) .

Answer 10

c.

Program Plan Intro

To prove that |B(S',m)|=2r and that for each block in B(S',m) exactly 2m−r numbers in S map to that block.

Answer 11

c.

Program Plan Intro

To prove that |B(S',m)|=2r and that for each block in B(S',m) exactly 2m−r numbers in S map to that block.

Answer 12

d.

Program Plan Intro

To show that number of linear permutations of Sn is much less than number of permutations of S using counting argument.

Answer 13

d.

Program Plan Intro

To show that number of linear permutations of Sn is much less than number of permutations of S using counting argument.

Answer 14

e.

Program Plan Intro

To provide an example of a value n and permutation of Sn that cannot be achieved by any linear permutation.

Answer 15

e.

Program Plan Intro

To provide an example of a value n and permutation of Sn that cannot be achieved by any linear permutation.

Answer 16

Expert Solution & Answer

Answer 17

Expert Solution & Answer

Answer 18

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Answer 19

Ch. D.1 - Prob. 1E Ch. D.1 - Prob. 2E Ch. D.1 - Prob. 3E Ch. D.1 - Prob. 4E Ch. D.2 - Prob. 1E Ch. D.2 - Prob. 2E Ch. D.2 - Prob. 3E Ch. D.2 - Prob. 4E Ch. D.2 - Prob. 5E Ch. D.2 - Prob. 6E

Solution Summary: The author concludes that A defines a permutation on S_n only if A has full rank.

Concept explainers

To prove that |P(A,y)|=2n−r if r is the rank of n×n matrix A and y∈R(A) .

To prove that |B(S',m)|=2r and that for each block in B(S',m) exactly 2m−r numbers in S map to that block.

To show that number of linear permutations of Sn is much less than number of permutations of S using counting argument.

To provide an example of a value n and permutation of Sn that cannot be achieved by any linear permutation.

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Chapter D Solutions