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

[eddysd]

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.

 

[hgfalling]

The eigenvalues of a matrix can be found as follows:

A\vec{x} = \lambda\vec{x}

(A - \lambda I) \vec{x} = 0

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

det(A - \lambda I) = 0

So to look at your question, we consider:

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

(1 - \lambda)(3 - \lambda) - 0 = 0

\lambda = 1, 3

So you are right.

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

(A - \lambda I) \vec{x} = 0

for each \lambda in turn.

 

[Raskolnikov]

Yes, that's correct. The eigenvalues of a matrix A are those that satisfy the "characteristic equation"

|\lambda \textbf{I} - \textbf{A}| = 0.

So for your A, we have

(\lambda - 1)(\lambda - 3) - (0)(-2) = (\lambda - 1)(\lambda - 3) = 0.

So the eigenvalues of A are \lambda_1 = 1 and \lambda_2 = 3.

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

(\lambda \textbf{I} - \textbf{A})\vec{x} = \vec{0}.

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.

 

[eddysd]

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?

 

[Raskolnikov]

http://tutorial.math.lamar.edu/Classes/DE/LA_Eigen.aspx

And yea, just do the same steps for A^3.