Technical problem with eigenvalues

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In summary, the conversation discusses the process of finding eigenvalues of a matrix. The characteristic polynomial is calculated and the eigenvalues are found to be 2, 2, and 4. The conversation then discusses techniques such as the Rational Root Theorem and transformations to factorize the polynomial and obtain the eigenvalues.
  • #1
Yankel
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Hello

I was trying to find eigenvalues of a matrix. I calculated the characteristic polynomial by calculating (A-lambdaI) and then calculating it's determinant. The results was:

[tex]-\lambda ^{3}+8\lambda ^{2}-20\lambda +16[/tex]

which is the correct calculation.

Now, the eigenvalues are 2,2,4, but I do not know, technically, how am I suppose to find it from:

[tex]-\lambda ^{3}+8\lambda ^{2}-20\lambda +16 = 0[/tex]

I mean, how do I expand this polynomial into:

[tex](\lambda -4)(\lambda -2)^{2}[/tex]

assuming that I do not see it in my eyes immediately, is there some hint to look for ?

Just to supply all information, the matrix is:

[tex]\begin{pmatrix} 3 &2 &3 \\ -1 &0 &-3 \\ 1 &2 &5 \end{pmatrix}[/tex]thanks !
 
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  • #2
The first place to look would be the Rational Root Theorem. Then, once you get one root, you can perform polynomial long division or synthetic division to obtain a quadratic times the linear factor you just discovered.
 
  • #3
Sometimes (this is the case) we can get factorized the characteristc polynomial with adequate transformations. For example, $R_3+R_2$ and $C_2-C_3$ provide:

$$\begin{aligned}\begin{vmatrix}{3-\lambda}&{2}&{3}\\{-1}&{-\lambda}&{-3}\\{1}&{2}&{5-\lambda}\end{vmatrix}&=\begin{vmatrix}{3-\lambda}&{2}&{3}\\{-1}&{-\lambda}&{-3}\\{0}&{2-\lambda}&{2-\lambda}\end{vmatrix}\\ &=\begin{vmatrix}{3-\lambda}&{-1}&{3}\\{-1}&{-\lambda+3}&{-3}\\{0}&{0}&{2-\lambda}\end{vmatrix}\\&=(2-\lambda)(\lambda^2-6\lambda+8)\\&=-(\lambda-2)^2(\lambda-4)\end{aligned}$$
 

FAQ: Technical problem with eigenvalues

1. What is a technical problem with eigenvalues?

The most common technical problem with eigenvalues is when a matrix has repeated eigenvalues, also known as eigenvalue degeneracy. This can make it difficult to accurately determine the eigenvectors and eigenvalues of the matrix.

2. How can I solve a technical problem with eigenvalues?

There are a few possible solutions to a technical problem with eigenvalues. One approach is to use specialized algorithms designed for matrices with repeated eigenvalues. Another option is to use perturbation techniques to slightly modify the matrix and avoid the degeneracy. Alternatively, you can also try using a different method for computing eigenvalues, such as the QR algorithm.

3. Can a technical problem with eigenvalues affect my results?

Yes, a technical problem with eigenvalues can definitely affect the accuracy of your results. If the eigenvectors and eigenvalues are not correctly determined due to the degeneracy, it can lead to incorrect interpretations or conclusions about the data.

4. How can I prevent technical problems with eigenvalues?

One way to prevent technical problems with eigenvalues is to carefully check for eigenvalue degeneracy before conducting any analysis. You can also use specialized software or programming libraries that have built-in functions for handling repeated eigenvalues.

5. Is there a way to simplify technical problems with eigenvalues?

While it is not always possible to completely eliminate technical problems with eigenvalues, there are some approaches that can help simplify them. For example, using a different coordinate system or normalizing the matrix can sometimes make the eigenvalues less degenerate and easier to work with.

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