Botched Eigen Vectors

[SadSack]

Homework Statement

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

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

Homework Equations

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

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?

 

[HallsofIvy]

Homework Statement

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

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

Homework Equations

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

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

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

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]
Which is equivalent to the equations x+ 16/9 z= 0 and y- 4/9 z= 0 so x= -16/9 z and y= 4/9 z. Any eigenvector corresponding to eigenvalue -5 is of the form <-16/9 z, 4/9 z, z>. Since you only need one, you can take z= 9 to get rid of the fractions: <-16, 4, 9>.

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]

So your equations would be 0= 0, -3x- 12y= 0, and -4x+ 2y- 8z= 0. From -3x- 12y= 0, x= -4y. Putting that into -4x+ 2y- 8z= 0 gives 16y+ 2y- 8z= 0 so 18y= 8z and y= (4/9)z and x= -4y= (-16/9)z. That gives the same answer as before, <-16, 4, 9>.

[quoteI appear to have failed miserably at this...because the vector I got did not work. Can someone help walk me through this?