Having difficulty working out a Char. Pol. for Eigenvalues

In summary, the conversation discusses the difficulty of finding the characteristic polynomial for a given matrix. The shortcut method for finding the polynomial is mentioned, but the speaker encounters trouble with it. They then mention the correct roots or eigenvalues for the polynomial, which are -2, 2, and 4. However, there seems to be a discrepancy with the matrix provided and the polynomial obtained.
  • #1
leo255
57
2
Given the following matrix:

2 3 0
3 2 4
0 4 2

I'm having a difficult time working out the characteristic polynomial. I used the shortcut that I saw on YT, where it is (I am using x instead of lambda) X^3 - (trace) X^2 + (A11+A22+A33) X - DET(A)

I got the following:

trace is just 2+2+2 = 6
A11+A22+A33 = (4-16) + (4-0) + (4-9) = -13
DET(A) = -42

This is giving me a characteristic polynomial of X^3 - 6X^2 - 13X + 42. Using synthetic division with the correct e-vals, it just doesn't come out right.

The correct roots/e-vals are -2, 2 and 4.

Thanks.
 
Physics news on Phys.org
  • #2
You could go through the process of working it out without the shortcut,
##det \left[\begin{pmatrix}2 -x&3 &0\\3&2-x&4\\0&4&2-x\end{pmatrix}\right] = 0##
##(2-x)^3 -25(2-x) =0##
I don't get -2, 2, and 4 though. Are you sure you copied the matrix down right?
 

FAQ: Having difficulty working out a Char. Pol. for Eigenvalues

1. What is a characteristic polynomial?

A characteristic polynomial is a special type of polynomial that is associated with a square matrix. It is used to find the eigenvalues of a matrix, which are the values that satisfy the equation Ax = λx, where A is the matrix and x is a non-zero vector.

2. Why is it important to find the characteristic polynomial for eigenvalues?

Finding the characteristic polynomial allows us to easily determine the eigenvalues of a matrix, which are useful in many applications in mathematics and science. Eigenvalues are also used in solving systems of differential equations and in understanding the behavior of dynamical systems.

3. How do you work out a characteristic polynomial?

To work out a characteristic polynomial, we first need to find the determinant of the matrix A-λI, where I is the identity matrix and λ is a variable. This will give us a polynomial in λ, which is the characteristic polynomial. We can then solve this polynomial to find the eigenvalues.

4. What do I do if I am having difficulty working out a characteristic polynomial?

If you are having difficulty working out a characteristic polynomial, it may be helpful to review the steps involved in finding the determinant of a matrix. You can also consult with a colleague or a textbook for additional guidance. Alternatively, there are many online resources and tutorials available that can help you understand the process.

5. Are there any tips for successfully working out a characteristic polynomial?

One tip for successfully working out a characteristic polynomial is to start with a small matrix and practice the steps involved before moving on to larger matrices. It is also helpful to understand the properties of determinants and how they relate to the characteristic polynomial. Additionally, double-checking your work and seeking assistance when needed can also lead to successful results.

Similar threads

A What physical meaning can the “determinant” of a divergency have?
Replies
3
Views
1K
MHB Is the Matrix A Diagonalizable with Real Eigenvalues?
Replies
10
Views
1K
MHB How Do I Find the Eigenvalues of \(F^2\) for the Derivative Mapping in \(X\)?
Replies
7
Views
2K
I Is a symmetric matrix with positive eigenvalues always real?
Replies
8
Views
2K
MHB Exploring the Spectrum of φ - Eigenspace in Finite Fields
Replies
15
Views
1K
MHB Technical problem with eigenvalues
Replies
2
Views
1K
MHB Almost all matrices have n distinct eigenvalues
Replies
1
Views
1K
Find the eigenvalues and eigenvectors
Replies
19
Views
3K
Eigenvalues and eigenvectors of a matrix
Replies
4
Views
3K
Why Are My Coupled Oscillator Eigenvalues Incorrect?
Replies
6
Views
1K

Hot Threads

Recent Insights

Back
Top