# Homework Help: Zero eigenvector?

1. Sep 13, 2010

### Ryan007

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

I can calculate the proper eigenvalues, but when I plug them back into the matrix, I get x1=0 and x2=0. But this is not the answer Maple gives me! How do I solve for the eigenvector when it appears that a zero vector is the only solution?

2. Relevant equations

For example, for the matrix {1,1},{1,-1} (rows shown), Maple gives me: eigenvalue of sqrt(2) with eigenvector {1/((sqrt(2)-1),1} and eigenvalue of -sqrt(2) with eigenvector {1/(-(sqrt(2)-1),1}

3. The attempt at a solution

But I can't get these eigenvectors when I try to solve by hand! How do you solve in these situations?
What are these situations called?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 13, 2010

### jackmell

For sqrt(2), You end up with the eigenvector equation:

$$\left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right)\binom{x}{y}=\sqrt{2}\binom{x}{y}$$

or:

$$(1-\sqrt{2}) x+y=0$$

$$x-y(1+\sqrt{2})=0$$

which are redundant so any vector (x,y) that satisfies either equation is a suitable eigenvector like:

$$\binom{1+\sqrt{2}}{1}$$

and even:

$$\binom{\frac{1}{\sqrt{2}-1}}{1}$$

3. Sep 13, 2010

### Ryan007

Thanks! Oh duh! I didn't realize that the 2 equations were the same. That's why I was getting x=0 and y=0 as solutions. I had to multiply through by the appropriate constant to make the 2 equations look the same.