What are the Eigenvalues and Eigenvectors of a 2x2 Matrix?

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Homework Help Overview

The discussion revolves around finding the eigenvalues and eigenvectors of a 2x2 matrix A, specifically given as A = [1 0; 2 3]. Participants are also considering the eigenvalues and eigenvectors of A^3.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the method for finding eigenvalues using the characteristic equation and the determinant. There are attempts to verify the eigenvalues found, with some participants expressing uncertainty about the eigenvector calculation. Questions arise regarding the process for calculating eigenvectors and the approach for A^3.

Discussion Status

Some participants have confirmed the eigenvalues as correct and provided guidance on the method to find them. However, there remains a lack of clarity on the eigenvector calculation, with requests for examples and further explanation. The discussion is ongoing, with multiple interpretations being explored.

Contextual Notes

Participants express confusion regarding the format of their responses and the steps involved in finding eigenvectors, indicating a need for clearer examples. There is also mention of the relationship between the calculations for A and A^3.

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Homework Statement



A=[1 0] Calculate
[2 3]
a) Eigenvalues of A
b) Eigenvectors of A
c) Eigenvalues and eigenvectors of A^3

The Attempt at a Solution


I had no idea what I was doing, but I saw someone attempt one somewhere and used the same method

Getting x=3 and 1 for part a)

However, I have no idea if this is correct, or even if it is in the correct format. Any help would be greatly appreciated.
 
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The eigenvalues of a matrix can be found as follows:

[tex]A\vec{x} = \lambda\vec{x}[/tex]

[tex](A - \lambda I) \vec{x} = 0[/tex]

Now we know that this equation will only have a nontrivial solution if:

[tex]det(A - \lambda I) = 0[/tex]

So to look at your question, we consider:

[tex]\left|\begin{array}{cc}1-\lambda&0\\2&{3-\lambda} \end{array}\right| = 0[/tex]

[tex](1 - \lambda)(3 - \lambda) - 0 = 0[/tex]

[tex]\lambda = 1, 3[/tex]

So you are right.

To find the eigenvectors, we go back and solve this equation:

[tex](A - \lambda I) \vec{x} = 0[/tex]

for each [itex]\lambda[/itex] in turn.
 
Yes, that's correct. The eigenvalues of a matrix A are those that satisfy the "characteristic equation"

[tex]|\lambda \textbf{I} - \textbf{A}| = 0.[/tex]

So for your A, we have

[tex](\lambda - 1)(\lambda - 3) - (0)(-2) = (\lambda - 1)(\lambda - 3) = 0.[/tex]

So the eigenvalues of A are [tex]\lambda_1 = 1[/tex] and [tex]\lambda_2 = 3.[/tex]

For part (b), the eigenvectors of A are all vectors in the nullspace of [tex]\lambda \textbf{I} - \textbf{A},[/tex] i.e., they satisfy the equationthe equation

[tex](\lambda \textbf{I} - \textbf{A})\vec{x} = \vec{0}.[/tex]

EDIT: I didn't see hgfalling's post until after I'd already posted...grrr...haha. Well here's mine for what it's worth anyways.
 
Ok, thanks for the help, but I still don't really understand the eigenvectors part of it. It would be useful if someone could write out an example. And for the A^3 bit is it the same as parts a) and b) but for AxAxA?
 

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