# eigenvectors

• Aug 27th 2010, 11:15 AM
eigenvectors
The matrix

4 0 1
-2 1 0
-2 0 1

I have got the eigenvalues 1,2,3

My first question is when you do 1 first you get

3 0 1
-2 0 0
-2 0 0

So you have the equations

3U1 + U3 = 0
-2U1 = 0
- 2U1 = 0

So U1 = 0 and U3 = 0, but what does U2 equal?

Then for the eigenvalue 2 you get

2 0 1
-2 -1 0
-2 0 -1

2U1 + U3 = 0
-2U1 - U2 = 0
-2U1 -U3 =0

so U1 = 1 U2 = -2 U3 = -2 or U1 = -1 U2 = 2 and U3 = 2. My lecturer got the second answer, but it doesnt matter does it? They're both right.
• Aug 27th 2010, 12:19 PM
Ackbeet
Quote:

So U1 = 0 and U3 = 0, but what does U2 equal?
Answer: Anything except zero. When finding eigenvectors, you should expect to find infinitely many solutions. After all, your eigenvalues are chosen so that the system you're solving, when trying to find eigenvectors, has infinitely many solutions! The reason you can't allow zero in this particular case is that if all three components are zero, you've got yourself a zero vector, which by definition is not an eigenvector.

Quote:

U1 = -1 U2 = 2 and U3 = 3.
For this eigenvalue of 2, this is not an eigenvector. It might be an eigenvector corresponding to 3, though. If you look at your system of

2U1 + U3 = 0
-2U1 - U2 = 0
-2U1 -U3 =0,

the first and third equation are the same. So you get

2U1 + U3 = 0
-2U1 - U2 = 0

Multiply the second row by -1:

2U1 + U3 = 0
2U1 + U2 = 0

From this, we can tell that U3 = U2, since they both equal -2U1. Therefore, the answer U1 = -1 U2 = 2 and U3 = 3 is impossible for this eigenvalue.
• Aug 28th 2010, 05:08 AM
Quote:

Originally Posted by Ackbeet
For this eigenvalue of 2, this is not an eigenvector. It might be an eigenvector corresponding to 3, though. If you look at your system of

2U1 + U3 = 0
-2U1 - U2 = 0
-2U1 -U3 =0,

the first and third equation are the same. So you get

2U1 + U3 = 0
-2U1 - U2 = 0

Multiply the second row by -1:

2U1 + U3 = 0
2U1 + U2 = 0

From this, we can tell that U3 = U2, since they both equal -2U1. Therefore, the answer U1 = -1 U2 = 2 and U3 = 3 is impossible for this eigenvalue.

Sorry i mistyped i meant U1 = 1 U2 = -2 U3 = -2 or U1 = -1 U2 = 2 and U3 = 2
• Aug 28th 2010, 08:47 AM
Ackbeet
Both are correct, since any multiple of an eigenvector of $A$ with eigenvalue $\lambda$ is again an eigenvector of $A$ with eigenvalue $\lambda$. Proof:

Suppose $x$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda.$ Then $x\not=0$. Suppose $c\not=0$ is a scalar. Then we have $cx\not=0,$ and

$A(cx)=c A x=c\lambda x=\lambda(cx).$

Therefore, $cx$ is an eigenvector of $A$ with corresponding eigenvalue $\lambda.$

So you really get a family of vectors called the eigenspace, not just one eigenvector. Does this clear things up?