Relationship between determinant and eigenvalues?

In summary, the matrix has a trace of 7 and a determinant of 8. However, the product of the eigenvalues is not equal to 1.826.
  • #1
cookiesyum
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Homework Statement



Find the eigenvalues of B = [5 2 0 2], [3 2 1 0], [3 1 -2 4], [2 4 -1 2]. Compute the sum and product of eigenvalues and compare it with the trace and determinant of the matrix.

Homework Equations





The Attempt at a Solution



I get the characteristic polynomial x^4 -7x^3 - x^2 - 33x + 8. I used a computer program to solve it for 0 and got eigenvalues L1= 0.238 and L2= 7.673 roughly. Their sum is 7.911. Their product is 1.826. The trace of the matrix is 7. The determinant of the matrix is 8. The trace and the sum of the eigenvalues match up, approximately. The determinant and the product of the eigenvalues, however, don't. What am I doing wrong?
 
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  • #2
You have a 4x4 matrix, shouldn't you be getting 4 eigenvalues? (the other two are complex)
 
  • #3
gabbagabbahey said:
You have a 4x4 matrix, shouldn't you be getting 4 eigenvalues? (the other two are complex)

But when you add the real eigenvalues to the complex eigenvalues you'll get a complex answer, right? How can that end up equaling the trace of the matrix if the trace of the matrix is a real number?...Still confused about the relationships.
 
  • #4
The two eigenvalues you get will always be complex conjugates. This is easy to see in general for real polynomials:

If [tex]f(x) = \sum_{i=0}^{n} a_i x^i [/tex] then (using * for conjugation)

[tex]f(x)* = (\sum_{i=0}^{n} a_i x^i)* = \sum_{i=0}^{n} (a_i x^i)* = \sum_{i=0}^{n} a_i* x^i* = \sum_{i=0}^{n} a_i (x*)^i = f(x*)[/tex]

noting the coefficients are real. So if f(x) = 0, f(x*) = 0
 
  • #5
Office_Shredder said:
The two eigenvalues you get will always be complex conjugates. This is easy to see in general for real polynomials:

If [tex]f(x) = \sum_{i=0}^{n} a_i x^i [/tex] then (using * for conjugation)

[tex]f(x)* = (\sum_{i=0}^{n} a_i x^i)* = \sum_{i=0}^{n} (a_i x^i)* = \sum_{i=0}^{n} a_i* x^i* = \sum_{i=0}^{n} a_i (x*)^i = f(x*)[/tex]

noting the coefficients are real. So if f(x) = 0, f(x*) = 0

Oh! I see. Thanks for everyone's help.
 
Last edited:

1. What is the relationship between the determinant and eigenvalues?

The determinant and eigenvalues are two mathematical concepts that are closely related. The determinant of a square matrix is a scalar value that represents the scaling factor of the matrix. Eigenvalues, on the other hand, are the special values of a matrix that do not change the direction of any vector when multiplied by the matrix. In other words, the eigenvalues are the values that satisfy the equation Ax = λx, where A is the matrix, x is the vector, and λ is the eigenvalue. The determinant of a matrix is equal to the product of its eigenvalues.

2. How do the determinant and eigenvalues affect the invertibility of a matrix?

The invertibility of a matrix is determined by its determinant. If the determinant of a matrix is equal to zero, then the matrix is not invertible. This is because a zero determinant means that the matrix is singular, and there is no unique solution to the system of equations represented by the matrix. In terms of eigenvalues, a matrix is invertible if and only if all of its eigenvalues are non-zero. If even one eigenvalue is equal to zero, then the matrix is not invertible.

3. Can the determinant and eigenvalues be used to find the characteristic polynomial of a matrix?

Yes, the determinant and eigenvalues can be used to find the characteristic polynomial of a matrix. The characteristic polynomial is a polynomial function that is used to calculate the eigenvalues of a matrix. It is equal to the determinant of the matrix minus the identity matrix multiplied by the variable λ. The roots of the characteristic polynomial are the eigenvalues of the matrix.

4. How does the determinant and eigenvalues relate to the diagonalizability of a matrix?

A matrix is diagonalizable if it can be written as a product of three matrices: A = PDP^-1, where P is an invertible matrix and D is a diagonal matrix. The diagonal entries of D are the eigenvalues of A. A matrix is diagonalizable if and only if it has a full set of linearly independent eigenvectors. This is related to the determinant of the matrix because a matrix is diagonalizable if and only if its determinant is non-zero.

5. Can the determinant and eigenvalues be used to find the trace of a matrix?

Yes, the determinant and eigenvalues can be used to find the trace of a matrix. The trace of a matrix is defined as the sum of its diagonal elements. This is equal to the sum of its eigenvalues. Therefore, if we know the eigenvalues of a matrix, we can find its trace by simply adding them together. Additionally, the trace of a matrix is equal to the sum of the coefficients of its characteristic polynomial, which can also be calculated using the determinant and eigenvalues.

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