# Eigenvalues and Eigenlines

1. May 7, 2014

### Justin90

1.
Find the corresponding Eigenline

A= (3 -3 2 -4)

2. Relevant equations
A=(a b c d)

3. The attempt at a solution

k2-(3-(-4))k+(3(-4)-(-3)2)=0
k2+k-6=0
(k+2)(k-3)

So k=-2 and k=3

Eigenvector for k=3
(3 -3 2 -4)(x y) = 3(x y)
(3x -3y 2x -4y)= (3x 3y)

Hence 3y=0 and 2x-7y=0

Eigenvector for k=-2
(3 -3 2 -4)(x y) = -2(x y)
(3x -3y 2x -4y)=(-2x -2y)

Hence 5x-3y=0 and 2x-2y=0

This is where I get confused because my notes say that 3y=0 and 2x-7y=0 need to reduce to a single equation, and similarly 5x-3y=0 and 2x-2y=0 need to reduce to a single equation to give the eigenline for the given eigenvalue?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 7, 2014

### pasmith

Do you mean $$A = \begin{pmatrix} 3 & - 3 \\ 2 & -4 \end{pmatrix}$$
Here is your problem: $$(-2)^2 + (-2) - 6 = -4 \neq 0 \\ (3)^2 + 3 - 6 = 6 \neq 0$$ Therefore neither -2 nor 3 is an eigenvalue of $A$. Try solving $k^2 + k - 6 = 0$ again.

3. May 7, 2014

### Justin90

Thanks for that! Couldn't see the wood through the trees. Got it now.

K^2+k-6=0
(k-2)(k+3)=0
So k=2 and k=-3

So eventual both equations reduce to x=y/2

Can't believe I missed that.