# Homework Help: Eigenvalues and eigenvectors

1. Dec 9, 2015

### says

1. The problem statement, all variables and given/known data
Given the linear transformation l : R 2 → R 2 defined below, find characteristic equation, real eigenvalues and corresponding eigenvectors. a) l(x, y) = (x + 5y, 2x + 4y)

2. Relevant equations
characteristic equation = det (A-λI) = 0

3. The attempt at a solution
l(x, y) = (x + 5y, 2x + 4y)

A =
[ 1 + 5 ]
[ 2 + 4 ]

det (A-λI) = 0 =
[ 1-λ + 5 ]
[ 2 + 4-λ ]

determinant works out to be

D =
2 - 5λ - 6) = 0

so eigenvalues are -1 and 6.

D =
[-1 0]
[ 0 6]

to get eigenvectors we multiply the original matrix A by D

[1 5 ] [ -1 0]
[ 2 4] [ 0 -1]

=
[ -1 -5 ]
[ -2 -4 ]

This is where I'm stuck

=
[ -1 -5 ] * [x] = 0
[ -2 -4 ] [y]

-x-5y = 0
-2x-4y = 0

I can't see how x in the first equation can = x in the second equation. The same with both y's. How can I find the eigenvector here? I'm going back through my steps now to see if I made an error somewhere. I'm having the same trouble finding eigenvectors with the second eigenvalue of 6 as well.

Last edited: Dec 9, 2015
2. Dec 9, 2015

### Staff: Mentor

No.
To get the eigenvector for $\lambda = -1$ solve this matrix equation:
$\begin{bmatrix} 1- (-1) & 5 \\ 2 & 4 - \lambda\end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0\end{bmatrix}$

Last edited: Dec 9, 2015
3. Dec 9, 2015

### says

Sorry, how did you get that matrix? Should the 6 be a 5?

4. Dec 9, 2015

### says

I got an eigenvector for λ=−1 of (-5,2)

5. Dec 9, 2015

### says

eigenvector for λ=6 of (1,1)

6. Dec 9, 2015

### Staff: Mentor

Yes, I must have hit the wrong key. I have fixed it in my earlier post.

Yes.

Yes.

You can check these yourself. If $\lambda$ is an eigenvalue associated with an eigenvector of x, it must be true that Ax = λx.