# Homework Help: Eigen Values of a 2x2 Matrix

1. Aug 4, 2010

### eddysd

1. The problem statement, all variables and given/known data

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

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.

2. Aug 4, 2010

### 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.

3. Aug 4, 2010

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.

4. Aug 4, 2010

### eddysd

Ok, thanks for the help, but I still dont 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?

5. Aug 4, 2010