# Relationship between determinant and eigenvalues? (1 Viewer)

### Users Who Are Viewing This Thread (Users: 0, Guests: 1)

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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?

#### gabbagabbahey

Homework Helper
Gold Member
You have a 4x4 matrix, shouldn't you be getting 4 eigenvalues? (the other two are complex)

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.

#### Office_Shredder

Staff Emeritus
Gold Member
The two eigenvalues you get will always be complex conjugates. This is easy to see in general for real polynomials:

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

$$f(x)* = (\sum_{i=0}^{n} a_i x^i)* = \sum_{i=0}^{n} (a_i x^i)* = \sum_{i=0}^{n} a_i* x^i* = \sum_{i=0}^{n} a_i (x*)^i = f(x*)$$

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

The two eigenvalues you get will always be complex conjugates. This is easy to see in general for real polynomials:

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

$$f(x)* = (\sum_{i=0}^{n} a_i x^i)* = \sum_{i=0}^{n} (a_i x^i)* = \sum_{i=0}^{n} a_i* x^i* = \sum_{i=0}^{n} a_i (x*)^i = f(x*)$$

noting the coefficients are real. So if f(x) = 0, f(x*) = 0
Oh! I see. Thanks for everyone's help.

Last edited:

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving