Question

eigenvalue/vector help needed!

X'(t)=(1 0 0 0
1 2 0 0
1 2 -1 1
1 1 0 4) X(t)

A) find the eigenvalues for the above system of equations

B) find the eigenvectors for the above system of equations

C) Show that answers A and B are correct.

Answer 1

I would take the determinant using the fact that the top row is almost all zero's. I assume you know how to apply the eigenvalue part to the procedure.

Answer 2

That's the thing... My book didn't explain it very well at all, so I have very little knowledge about how to do this... sorry. I'm hoping someone here could teach me better then my book

Answer 3

Do you know the method of cofactors?

Answer 4

no sorry

Answer 5

well taking this determinant will be a real pain without it, so I suggest you read up on that a bit http://www.mathwords.com/e/expansion_by_cofactors.htm
http://www.purplemath.com/modules/minors.htm

Answer 6

oh I guess I do know that just not for a 4x4 matrix. I'll look at it some more though

Answer 7

And eigenvalue \(\lambda\) is a scalar such that\[A=\lambda I\implies \det(A-\lambda I)=0\]so rewrite the matrix as \(A-\lambda I\), use the method of cofactors expanding along the top row to take the determinant, then solve for lambda to get the eigenvalues.

Answer 8

oh I dropped the vector from my definition, sorry

Answer 9

\[A\vec x=\lambda I\vec x\implies \det(A-\lambda I)=0\]

Answer 10

okay so I start by getting the determinant of my first vector, right?

Answer 11

well you start by taking the determinant of \(A-\lambda I\)

Answer 12

with 'I' being? sorry I'm a bit confused

Answer 13

that would be the identity matrix; a matrix with nothing but 1's on its diagonal

Answer 14

\[A-\lambda I=\left[\begin{matrix}1&0&0&0\\1&2&0&0\\2&2&-1&1\\1&1&0&4\end{matrix}\right]-\lambda\left[\begin{matrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{matrix}\right]\]

Answer 15

\[A-\lambda I=\left[\begin{matrix}1-\lambda&0&0&0\\1&2-\lambda&0&0\\2&2&-1-\lambda&1\\1&1&0&4-\lambda\end{matrix}\right]\]since \(A\vec x=\lambda\vec x\) the matrices must be linearly dependent, i.e. the determinant of the above is zero

Answer 16

\[\det(A-\lambda I)=\left|\begin{matrix}1-\lambda&0&0&0\\1&2-\lambda&0&0\\2&2&-1-\lambda&1\\1&1&0&4-\lambda\end{matrix}\right|=0\]expanding along the first row, this becomes...

Answer 17

\[\det(A-\lambda I)=(1-\lambda)\left|\begin{matrix}2-\lambda&0&0\\2&-1-\lambda&1\\1&0&4-\lambda\end{matrix}\right|=0\]since the rest of the expansion by cofactors have coefficient zero along row 1. Take this determinant and solve for the four values of lambda, i.e. the eigenvalues.

Answer 18

@TuringTest so then to find the eigenvector, I ttake the initial matrix minus one of the eigenvalues times the identity matrix, right?

Answer 19

well you don't subtract just *one* of the eigenvalues, they are all encompassed in the \(\lambda\). After taking the determinant you will get a 4th order equation in terms of lambda, from which you will get 4 solutions \(\lambda_1, \lambda_2. \lambda_3, \lambda_4\), i.e. your 4 eigenvalues.

Answer 20

right. I've got four eigenvalues but now I need the vectors. How do I get an answer as a vector?

Answer 21

use the definition of the eigenvalue and go back to the original equation. say one of your eigenvalues is 2, you then get\[A\vec b=6\vec b\] where\[\vec b=\left[\begin{matrix}x\\y\end{matrix}\right]\]you can then solve this system for x and y, which is one of your eigenvectors

Answer 22

oops, I meant "say one of your eigenvalues is 6..."

Answer 23

notice that we can subtract from both sides, in which case we wind up with the same thing we had before to find the polynomial for the eigenvalues\[(A-6I)\vec b=\vec 0\]so we would just solve\[\left[\begin{matrix}-5&0&0&0\\1&-4&0&0\\2&2&-7&1\\1&1&0&-2\end{matrix}\right]\vec b=\vec 0\]

Answer 24

@AccessDenied do you know anything about eigenvectors? Do you think you could help me with the eigenvector of 2 for this one? I think it's supposed to be (0,-2,-1,1) but I can't seem to get that for an answer

Answer 25

did you follow what I did above with the method of cofactors?

Answer 26

yep. I got it figured out but I posted another related question if you could check it out quick. Thanks for the help either way