To prove that the determinant of the Vandermonde matrix V is det ( V ( x 0 , x 1 , ... , x n ) ) = ∏ 0 ≤ j < k ≤ n − 1 ( x k − x j )

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter D, Problem 1P

Program Plan Intro

To prove that the determinant of the Vandermonde matrix V is $det(V(x0,x1,...,xn))=∏0≤j<k≤n−1(xk−xj)$

Expert Solution & Answer

Explanation of Solution

Given information:

Vandermonde matrix V is given by $V(x0,x1,...xn−1)=(1x0x02x0n−11x1x12x1n−1............1xn−1xn−12xn−1n−1)$

Explanation:

Start with induction on n .

For n = 1, the Vandermonde matrix is of size $1×1$ with only one entry i.e. 1.It is easy to note that this matrix’s determinant is 1. Also product is one as no distinct i and j can be picked in the product on the right hand side.

Now, assume that the given condition is true for $n×n$ matrix. Using this let’s show that it is true for $(n+1)×(n+1)$ matrices.

Start from second last column on right side that is $n−1$ and go towards left adding $(−x0)$ times that column to the one to its right.

The determinant will not be changed from this operation. All entries in the top row, except the leftmost one, have become zero as $x0j−1−x0j−2⋅x0=0$ . This means that expansion by minors along the top row can be done in order to compute the determinant. The top left column will be the only non-zero term in this operation. Thus determinant remains the same as the determinant of that one minor. Also, the factor of every item in i^t^h of that minor is $(xi−x0)$ . This implies that taking all those factors out will give expression $∏i=1n−1(xi−x0)$ i.e. all the factors in the formula that include $x0$ .

After taking factors out,a $n×n$ Vandermonde matrix on the variables $x1,...,xn−1$ is left.This will only contain all the other factors that are needed to get the right hand side.

Thus, this completes the induction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

.NET Interactive Solving Sudoku using Grover's Algorithm We will now solve a simple problem using Grover's algorithm, for which we do not necessarily know the solution beforehand. Our problem is a 2x2 binary sudoku, which in our case has two simple rules: •No column may contain the same value twice •No row may contain the same value twice If we assign each square in our sudoku to a variable like so: 1 V V₁ V3 V2 we want our circuit to output a solution to this sudoku. Note that, while this approach of using Grover's algorithm to solve this problem is not practical (you can probably find the solution in your head!), the purpose of this example is to demonstrate the conversion of classical decision problems into oracles for Grover's algorithm. Turning the Problem into a Circuit We want to create an oracle that will help us solve this problem, and we will start by creating a circuit that identifies a correct solution, we simply need to create a classical function on a quantum circuit that…

using r language

8. Cash RegisterThis exercise assumes you have created the RetailItem class for Programming Exercise 5. Create a CashRegister class that can be used with the RetailItem class. The CashRegister class should be able to internally keep a list of RetailItem objects. The class should have the following methods: A method named purchase_item that accepts a RetailItem object as an argument. Each time the purchase_item method is called, the RetailItem object that is passed as an argument should be added to the list. A method named get_total that returns the total price of all the RetailItem objects stored in the CashRegister object’s internal list. A method named show_items that displays data about the RetailItem objects stored in the CashRegister object’s internal list. A method named clear that should clear the CashRegister object’s internal list. Demonstrate the CashRegister class in a program that allows the user to select several items for purchase. When the user is ready to check out, the…

Answer 2

Question

Chapter D, Problem 1P

Program Plan Intro

To prove that the determinant of the Vandermonde matrix V is $det(V(x0,x1,...,xn))=∏0≤j<k≤n−1(xk−xj)$

Expert Solution & Answer

Explanation of Solution

Given information:

Vandermonde matrix V is given by $V(x0,x1,...xn−1)=(1x0x02x0n−11x1x12x1n−1............1xn−1xn−12xn−1n−1)$

Explanation:

Start with induction on n .

For n = 1, the Vandermonde matrix is of size $1×1$ with only one entry i.e. 1.It is easy to note that this matrix’s determinant is 1. Also product is one as no distinct i and j can be picked in the product on the right hand side.

Now, assume that the given condition is true for $n×n$ matrix. Using this let’s show that it is true for $(n+1)×(n+1)$ matrices.

Start from second last column on right side that is $n−1$ and go towards left adding $(−x0)$ times that column to the one to its right.

The determinant will not be changed from this operation. All entries in the top row, except the leftmost one, have become zero as $x0j−1−x0j−2⋅x0=0$ . This means that expansion by minors along the top row can be done in order to compute the determinant. The top left column will be the only non-zero term in this operation. Thus determinant remains the same as the determinant of that one minor. Also, the factor of every item in i^t^h of that minor is $(xi−x0)$ . This implies that taking all those factors out will give expression $∏i=1n−1(xi−x0)$ i.e. all the factors in the formula that include $x0$ .

After taking factors out,a $n×n$ Vandermonde matrix on the variables $x1,...,xn−1$ is left.This will only contain all the other factors that are needed to get the right hand side.

Thus, this completes the induction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter D, Problem 1P

Program Plan Intro

To prove that the determinant of the Vandermonde matrix V is $det(V(x0,x1,...,xn))=∏0≤j<k≤n−1(xk−xj)$

Expert Solution & Answer

Explanation of Solution

Given information:

Vandermonde matrix V is given by $V(x0,x1,...xn−1)=(1x0x02x0n−11x1x12x1n−1............1xn−1xn−12xn−1n−1)$

Explanation:

Start with induction on n .

For n = 1, the Vandermonde matrix is of size $1×1$ with only one entry i.e. 1.It is easy to note that this matrix’s determinant is 1. Also product is one as no distinct i and j can be picked in the product on the right hand side.

Now, assume that the given condition is true for $n×n$ matrix. Using this let’s show that it is true for $(n+1)×(n+1)$ matrices.

Start from second last column on right side that is $n−1$ and go towards left adding $(−x0)$ times that column to the one to its right.

The determinant will not be changed from this operation. All entries in the top row, except the leftmost one, have become zero as $x0j−1−x0j−2⋅x0=0$ . This means that expansion by minors along the top row can be done in order to compute the determinant. The top left column will be the only non-zero term in this operation. Thus determinant remains the same as the determinant of that one minor. Also, the factor of every item in i^t^h of that minor is $(xi−x0)$ . This implies that taking all those factors out will give expression $∏i=1n−1(xi−x0)$ i.e. all the factors in the formula that include $x0$ .

After taking factors out,a $n×n$ Vandermonde matrix on the variables $x1,...,xn−1$ is left.This will only contain all the other factors that are needed to get the right hand side.

Thus, this completes the induction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter D, Problem 1P

Program Plan Intro

To prove that the determinant of the Vandermonde matrix V is $det(V(x0,x1,...,xn))=∏0≤j<k≤n−1(xk−xj)$

Expert Solution & Answer

Explanation of Solution

Given information:

Vandermonde matrix V is given by $V(x0,x1,...xn−1)=(1x0x02x0n−11x1x12x1n−1............1xn−1xn−12xn−1n−1)$

Explanation:

Start with induction on n .

For n = 1, the Vandermonde matrix is of size $1×1$ with only one entry i.e. 1.It is easy to note that this matrix’s determinant is 1. Also product is one as no distinct i and j can be picked in the product on the right hand side.

Now, assume that the given condition is true for $n×n$ matrix. Using this let’s show that it is true for $(n+1)×(n+1)$ matrices.

Start from second last column on right side that is $n−1$ and go towards left adding $(−x0)$ times that column to the one to its right.

The determinant will not be changed from this operation. All entries in the top row, except the leftmost one, have become zero as $x0j−1−x0j−2⋅x0=0$ . This means that expansion by minors along the top row can be done in order to compute the determinant. The top left column will be the only non-zero term in this operation. Thus determinant remains the same as the determinant of that one minor. Also, the factor of every item in i^t^h of that minor is $(xi−x0)$ . This implies that taking all those factors out will give expression $∏i=1n−1(xi−x0)$ i.e. all the factors in the formula that include $x0$ .

After taking factors out,a $n×n$ Vandermonde matrix on the variables $x1,...,xn−1$ is left.This will only contain all the other factors that are needed to get the right hand side.

Thus, this completes the induction.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter D, Problem 1P

Program Plan Intro

To prove that the determinant of the Vandermonde matrix V is $det(V(x0,x1,...,xn))=∏0≤j<k≤n−1(xk−xj)$

Expert Solution & Answer

Explanation of Solution

Given information:

Vandermonde matrix V is given by $V(x0,x1,...xn−1)=(1x0x02x0n−11x1x12x1n−1............1xn−1xn−12xn−1n−1)$

Explanation:

Start with induction on n .

For n = 1, the Vandermonde matrix is of size $1×1$ with only one entry i.e. 1.It is easy to note that this matrix’s determinant is 1. Also product is one as no distinct i and j can be picked in the product on the right hand side.

Now, assume that the given condition is true for $n×n$ matrix. Using this let’s show that it is true for $(n+1)×(n+1)$ matrices.

Start from second last column on right side that is $n−1$ and go towards left adding $(−x0)$ times that column to the one to its right.

The determinant will not be changed from this operation. All entries in the top row, except the leftmost one, have become zero as $x0j−1−x0j−2⋅x0=0$ . This means that expansion by minors along the top row can be done in order to compute the determinant. The top left column will be the only non-zero term in this operation. Thus determinant remains the same as the determinant of that one minor. Also, the factor of every item in i^t^h of that minor is $(xi−x0)$ . This implies that taking all those factors out will give expression $∏i=1n−1(xi−x0)$ i.e. all the factors in the formula that include $x0$ .

After taking factors out,a $n×n$ Vandermonde matrix on the variables $x1,...,xn−1$ is left.This will only contain all the other factors that are needed to get the right hand side.

Thus, this completes the induction.

Answer 6

Chapter D, Problem 1P

Program Plan Intro

To prove that the determinant of the Vandermonde matrix V is $det(V(x0,x1,...,xn))=∏0≤j<k≤n−1(xk−xj)$

Expert Solution & Answer

Explanation of Solution

Given information:

Vandermonde matrix V is given by $V(x0,x1,...xn−1)=(1x0x02x0n−11x1x12x1n−1............1xn−1xn−12xn−1n−1)$

Explanation:

Start with induction on n .

For n = 1, the Vandermonde matrix is of size $1×1$ with only one entry i.e. 1.It is easy to note that this matrix’s determinant is 1. Also product is one as no distinct i and j can be picked in the product on the right hand side.

Now, assume that the given condition is true for $n×n$ matrix. Using this let’s show that it is true for $(n+1)×(n+1)$ matrices.

Start from second last column on right side that is $n−1$ and go towards left adding $(−x0)$ times that column to the one to its right.

The determinant will not be changed from this operation. All entries in the top row, except the leftmost one, have become zero as $x0j−1−x0j−2⋅x0=0$ . This means that expansion by minors along the top row can be done in order to compute the determinant. The top left column will be the only non-zero term in this operation. Thus determinant remains the same as the determinant of that one minor. Also, the factor of every item in i^t^h of that minor is $(xi−x0)$ . This implies that taking all those factors out will give expression $∏i=1n−1(xi−x0)$ i.e. all the factors in the formula that include $x0$ .

After taking factors out,a $n×n$ Vandermonde matrix on the variables $x1,...,xn−1$ is left.This will only contain all the other factors that are needed to get the right hand side.

Thus, this completes the induction.

Answer 7

Chapter D, Problem 1P

Program Plan Intro

To prove that the determinant of the Vandermonde matrix V is $det(V(x0,x1,...,xn))=∏0≤j<k≤n−1(xk−xj)$

Answer 8

Expert Solution & Answer

Explanation of Solution

Given information:

Vandermonde matrix V is given by $V(x0,x1,...xn−1)=(1x0x02x0n−11x1x12x1n−1............1xn−1xn−12xn−1n−1)$

Explanation:

Start with induction on n .

For n = 1, the Vandermonde matrix is of size $1×1$ with only one entry i.e. 1.It is easy to note that this matrix’s determinant is 1. Also product is one as no distinct i and j can be picked in the product on the right hand side.

Now, assume that the given condition is true for $n×n$ matrix. Using this let’s show that it is true for $(n+1)×(n+1)$ matrices.

Start from second last column on right side that is $n−1$ and go towards left adding $(−x0)$ times that column to the one to its right.

The determinant will not be changed from this operation. All entries in the top row, except the leftmost one, have become zero as $x0j−1−x0j−2⋅x0=0$ . This means that expansion by minors along the top row can be done in order to compute the determinant. The top left column will be the only non-zero term in this operation. Thus determinant remains the same as the determinant of that one minor. Also, the factor of every item in i^t^h of that minor is $(xi−x0)$ . This implies that taking all those factors out will give expression $∏i=1n−1(xi−x0)$ i.e. all the factors in the formula that include $x0$ .

After taking factors out,a $n×n$ Vandermonde matrix on the variables $x1,...,xn−1$ is left.This will only contain all the other factors that are needed to get the right hand side.

Thus, this completes the induction.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

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

To prove that the determinant of the Vandermonde matrix V is det ( V ( x 0 , x 1 , ... , x n ) ) = ∏ 0 ≤ j &lt; k ≤ n − 1 ( x k − x j )

Explanation of Solution

Want to see more full solutions like this?

Chapter D Solutions

To prove that the determinant of the Vandermonde matrix V is det ( V ( x 0 , x 1 , ... , x n ) ) = ∏ 0 ≤ j < k ≤ n − 1 ( x k − x j )