Homework Help: Botched Eigen Vectors!

1. May 6, 2009

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

(a) Find the characteristic equation and
(b) the eigenvalues and corresponding eigen-vectors

[-5 0 0
3 7 0
4 -2 3]

2. Relevant equations

det($$\lambda$$*I - A)

3. The attempt at a solution
Finding the characteristic equation wasn't that challenging, I took lambda, $$\lambda$$ , and subtracted the given matrix values from it for every diagonal entry. The determiant of a 3x3 matrix is just the product of these diagonals.....I think so my characteristic equation is ($$\lambda$$ + 5)($$\lambda$$-7)($$\lambda$$ - 3)
Next the eigenvalues are the values for $$\lambda$$ that when plugged in would make the characteristic equation equal zero.....so the eigenvalues are -5, 7, 3
This is where I'm confused.
To find the eigen-vectors I'm supposed to multiply each eigenvalue by the identity matrix and subtract the original given matrix: so for the eigenvalue of -5....

[-5 0 0
0 -5 0
0 0 -5]
MINUS
[-5 0 0
3 7 0
4 -2 3]
which equals
[0 0 0
-3 -12 0
-4 2 -8]
I took the reduced row echelon form of this matrix and got...
[1 0 1.777
0 1 -.4444
0 0 0]
Is my eigen vector in the matrix above? I am so confused....I also tried to solve the system generated by multiplying a 3x1 vector of variable components x, y, z times
[0 0 0
-3 -12 0
-4 2 -8]
I appear to have failed miserably at this....because the vector I got did not work. Can someone help walk me through this?

2. May 6, 2009

HallsofIvy

Yes, in an "upper triangular" or "lower triangular" matrix like this, the eigenvalues are just the numbers on the main diagonal.