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

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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?