Relationship between determinant and eigenvalues?

[cookiesyum]

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?

 

[gabbagabbahey]

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

 

[cookiesyum]

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]

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

 

[cookiesyum]

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.