How do I find the eigenvectors of this matrix?

Question

[2 3 0
3 2 0
0 0 1]

I can find the eigenvalues (5,-1,1) easily but I seem to be doing something wrong to find the vectors. (3-u)x1+2y1=0 and I should be able to find my vectors by plugging in the eigenvalues but! when I check my work (plug in the eigenvalues into another equation) I get an extra eigenvector.

Help!

Answer 1

subtract your matrix from xI and get

x-2 -3 0
-3 x-2 0
0 0 x-1

Now, find the determinant:

[(x-2)^2 - 9](x - 1) = (x - 5)((x + 1)(x - 1)

5, -1 and 1 are eigen values.

Now multiply your matrix by a vecorr (x1 x2 x3)
For the eigenvalue 1:

2x1+3x2 = x1
3x1+2x2 = x2
x3 = x3

or
x1 + 3x2 = 0
3x1 + x2 = 0

You don't have to work too hard to find that x1=x2=0, and x3 is any real number.

For the eigenvalue 5

2x1+3x2 = 5x1
3x1+2x2 = 5x2
x3 = 5x3

Well, x3 = 0
Let's find the dependency between x1 and x2.

-3x1+3x2=0
-2x1+2x2=0

Well, x1=x2
Take (1,1,0) as an eigenvector of 5

Answer 2

As you know, an eigenvector of a matrix A is a vector v such that Av = λv = λIv for some scalar value λ (where I is the identity matrix). These values of λ are called eigenvalues. In other words, an eigenvector is a vector v such that (A-λI)v = 0 where λ is one of the eigenvalues. This happens only when A-λI is singular. In other words, when determinant(A-λI)=0

You have already found the three eigenvalues, 5, -1, 1

So for each eigenvalue, let's find the nullspace (the vectors which map onto 0) for A-λI.

λ=1:
A-λI = [1 3 0; 3 1 0; 0 0 0]
It should be obvious that (0, 0, 1) is an eigenvector (since the bottom row is all 0s).

λ=-1:
A-λI = [3 3 0; 3 3 0; 0 0 2]
It should be pretty clear that (1, -1, 0) is an eigenvector (since the first two columns are identical).

λ=5:
A-λI = [-3 3 0; 3 -3 0; 0 0 -4]
We can eyeball this to see that (1, 1, 0) is an eigenvector (since the first two columns are negatives of each other).

The last two are not unit vectors, of course, but it's easy enough to divide by √2
This is perhaps not the most regular method, but for simple examples such as yours it can be effective.

Answer 3

Kindly find the explanation for your issue in these links. First link will have the answers and other links will equally explain you from the basics of finding eigenvalue and vectors. All the best.

http://mathforum.org/library/drmath/view/51975.html

http://www.riskglossary.com/link/eigenvalue.htm

http://ceee.rice.edu/Books/LA/eigen/

http://cepa.newschool.edu/het/essays/math/eigen.htm

http://www-math.mit.edu/18.013A/HTML/chapter32/section08.html

http://www.cs.ut.ee/~toomas_l/linalg/lin1/node16.html