Thread: Eigenvectors

Find the solution of the system

x (with a dot above it, dont know the Latex) = 2x - y
y (same, with dot) = 3x - 2y

so matrix A =;
[2 -1]
[3 -2]

satisfying x(0) = 2 and y(0) = -1.

Now i got the eigenvalues to be $\displaystyle \pm 1$

so how to i calculate the eigenvectors given that;

for $\displaystyle \lambda = 1$ we have A - I =;
[1 -1]
[3 -3]

and for $\displaystyle \lambda = -1$ we have A - I =;
[3 -1]
[3 -1]

I cant seem to find any examples that actually explain hows its done, every example just seems to state the eigenvector straight after stating the matrices.

Any help? Cheers.

Originally Posted by Deadstar

Find the solution of the system

x (with a dot above it, dont know the Latex) = 2x - y
y (same, with dot) = 3x - 2y

so matrix A =;
[2 -1]
[3 -2]

satisfying x(0) = 2 and y(0) = -1.

Now i got the eigenvalues to be $\displaystyle \pm 1$

so how to i calculate the eigenvectors given that;

for $\displaystyle \lambda = 1$ we have A - I =;
[1 -1]
[3 -3]

and for $\displaystyle \lambda = -1$ we have A - I =;
[3 -1]
[3 -1]

I cant seem to find any examples that actually explain hows its done, every example just seems to state the eigenvector straight after stating the matrices.

Any help? Cheers.

To find the eigen vectors you need a non-zero solution of (A-I)X=0.

By inspection [1,1]' is an eigen vector corresponding to lambda=1 and
[1,3]' is an eigen vector corresponding to lambda=-1 (assuming that you
have the A's right, also you may want to normalise them to unit vectors).

RonL

Check out mathway.com, allows you to enter a problem like this with step by step help and explanations. See the link for the complete step by step solution to this problem:

http://www.mathway.com/answer.aspx?p=line?p=matrix[2,-1:3,-2]?p=1319?p=?p=?p=?p=?p=?p=?p=0