Finding Eigenvalues and determine if there are invariant lines

[Luscinia]

Homework Statement

Find the eigen values of the following mapping and determine if there are invariant lines.

(2 -4)
(-3 3) is the mapping.

Homework Equations

det (L-λI)=0

The Attempt at a Solution

L-λI=
(2-λ -4)
(-3 3-λ)

det(L-λI)=0=ac-bd=(3-λ)(2-λ)-12
.: (3-λ)(2-λ)=12
6-3λ-2λ+λ² -12=0
λ²-5λ-6=0
λ=-6 and 1

but the answer is supposed to be -1 and 6.
Also, since there are 2 eigenvalues, I'm guessing that means that there are 2 invariant lines. How do we find these 2 invariant lines?

 

[Robert1986]

Homework Statement

Find the eigen values of the following mapping and determine if there are invariant lines.

(2 -4)
(-3 3) is the mapping.

Homework Equations

det (L-λI)=0

The Attempt at a Solution

L-λI=
(2-λ -4)
(-3 3-λ)

det(L-λI)=0=ac-bd=(3-λ)(2-λ)-12
.: (3-λ)(2-λ)=12
6-3λ-2λ+λ² -12=0
λ²-5λ-6=0
λ=-6 and 1

but the answer is supposed to be -1 and 6.
Also, since there are 2 eigenvalues, I'm guessing that means that there are 2 invariant lines. How do we find these 2 invariant lines?

Now, [itex]\lambda^2 -5\lambda -6 =0[/itex] looks like the right equation. Now, you need to solve for [itex]\lambda[/itex]. YOu can do this one of two (or both) ways: factor or use the quadratic formula. Using what you wrote would imply that [itex](\lambda - (-6))(\lambda - 1)=\lambda^2 - 5\lambda -6[/itex], but this isn't true. Remember, in factorisation, the factors are of the form [itex](\lambda - a)[/itex] where [itex]a[/itex] is the root.

 

[HallsofIvy]

Homework Statement

Find the eigen values of the following mapping and determine if there are invariant lines.

Well, first, eigenvalues are numbers not lines so this is not true. I expect that you mean that the lines in the direction of the eigenvectors are invariant.

(2 -4)
(-3 3) is the mapping.

Homework Equations

det (L-λI)=0

The Attempt at a Solution

L-λI=
(2-λ -4)
(-3 3-λ)

det(L-λI)=0=ac-bd=(3-λ)(2-λ)-12
.: (3-λ)(2-λ)=12
6-3λ-2λ+λ² -12=0
λ²-5λ-6=0
λ=-6 and 1

but the answer is supposed to be -1 and 6.

Yes, that's because -6 and 1 are incorrect. [itex](-6)^2- 5(-6)- 6= 36+ 30- 6= 60[/itex], not 0. And [itex](1)^2- 5(1)- 6= 1- 5- 6= -10[/itex], not 0. How did you get -6 and 1?

Also, since there are 2 eigenvalues, I'm guessing that means that there are 2 invariant lines. How do we find these 2 invariant lines?

No reason to guess. Distinct eigenvalues correspond to independent eigenvectors and those give the two lines. Do you understand the definition of "eigenvalue"?

A number [itex]\lambda[/itex] is an eigenvalue of an operator, T, if and only if there exist a non-zero vector, v, such that [itex]Tv= \lambda v[/itex]. Saying that -1 is an eigenvalue means there is a vector
[tex]\begin{bmatrix}x \\ y \end{bmatrix}[/tex]
so that
[tex]\begin{bmatrix}2 & -4 \\ -3 & 3\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}-x \\ -y\end{bmatrix}[/tex]

Multiplying on the left will give two equations for x and y. Of course x= y= 0 is a solution so there will be a non-zero solution only if those two equations are dependent and, in that case, there will be an infinite number of solutions- all vectors pointing in the same direction and so defining a line.