Vandermonde Determinant for NxN Matrices

In summary, the problem is to show that a 3x3 matrix has a determinant of (λ3 - λ2)(λ3 - λ1)(λ2 - λ1) and to generalize to the NxN case. While solving the 3x3 case is relatively easy by expanding the expression and using the rule of Sarrus, for the NxN case, one can use induction. Additionally, a 4x4 matrix with a 4th row being cubic will result in a 6th power polynomial form, with the number of combinations of 4 things taken 2 at a time being 6, making for the 6 factors. It is also observed that if two columns of a matrix are identical,
  • #1
grassstrip1
11
0
The problem I have is this: Show that

\begin{bmatrix} 1 & 1 & 1 \\ λ_{1} & λ_{2} & λ_{3} \\ λ_{1}^{2} & λ_{2}^{2} & λ_{3}^{2} \end{bmatrix}
Has determinant

$$ (λ_{3} - λ_{2}) (λ_{3} - λ_{1}) (λ_{2} - λ_{1}) $$

And generalize to the NxN case (proof not needed)Obviously solving the 3x3 was not hard, I simply expanded the expression for the determinate given and showed it to be the same as the one i calculated using the rule of Sarrus.

However for the nxn case I'm not sure how to proceed. I tried entering bigger and bigger matrices into wolfram but there was no clear pattern. Is there a non computational way to solve this?
 
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  • #3
grassstrip1 said:
The problem I have is this: Show that

\begin{bmatrix} 1 & 1 & 1 \\ λ_{1} & λ_{2} & λ_{3} \\ λ_{1}^{2} & λ_{2}^{2} & λ_{3}^{2} \end{bmatrix}
Has determinant

$$ (λ_{3} - λ_{2}) (λ_{3} - λ_{1}) (λ_{2} - λ_{1}) $$

And generalize to the NxN case (proof not needed)Obviously solving the 3x3 was not hard, I simply expanded the expression for the determinate given and showed it to be the same as the one i calculated using the rule of Sarrus.

However for the nxn case I'm not sure how to proceed. I tried entering bigger and bigger matrices into wolfram but there was no clear pattern. Is there a non computational way to solve this?
I don't have a complete answer, but a 4x4 with the 4th row being cubic will result in a 6th power polynomial form. The number of combinations of 4 things taken 2 at a time is 6, making for the 6 factors. (I don't know how to choose the minus or plus sign for the terms, but if it does generalize, this might be helpful.)...editing... One additional observation is if two columns of a matrix are identical, I think the determinant is zero. This is consistent with setting the product of the factors equal to zero.
 
Last edited:
  • #5
Thank you! The formula for the vandermonde determinant was exactly what I needed!
 

1. What is the definition of the determinant of an nxn matrix?

The determinant of an nxn matrix is a scalar value that can be calculated from the elements of the matrix. It represents the matrix's linear transformation of a vector space, and it is used to determine if the matrix is invertible or singular.

2. How is the determinant of an nxn matrix calculated?

The determinant of an nxn matrix is calculated by expanding the matrix using the Laplace expansion method or by using Gaussian elimination to reduce the matrix to upper triangular form and then multiplying the diagonal elements. There are also various online calculators that can be used to calculate the determinant of an nxn matrix.

3. What is the significance of the determinant of an nxn matrix?

The determinant of an nxn matrix has several important uses in mathematics, including determining if a matrix is invertible, finding the area or volume of a parallelogram or parallelepiped, and solving systems of linear equations. It is also used in many areas of physics and engineering.

4. Can the determinant of an nxn matrix be negative?

Yes, the determinant of an nxn matrix can be negative. The sign of the determinant depends on the order of the rows and columns in the matrix, and it can be positive, negative, or zero. A negative determinant indicates that the matrix has a negative scale factor and reflects the orientation of the vector space.

5. How does changing the elements of an nxn matrix affect its determinant?

Changing the elements of an nxn matrix can have a significant impact on its determinant. Adding a constant to a row or column will result in the determinant being multiplied by that constant, while swapping two rows or columns will result in the determinant being multiplied by -1. Multiplying a row or column by a constant will result in the determinant being multiplied by that constant. These operations can be used to simplify the calculation of the determinant or to solve systems of linear equations.

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