For a 2 X 2 system, how can you have two eigenvectors for a repeated eigenvalue?
I'm trying to figure out if I am doing this correctly.
Given the matrix A:
[1 -.5]
[.5 2]
write down e^(tA) and calculate e^(tA)X, X = (1,1)
So far what I did was, I found the eigenvalues (1.5 repeated) and the eigenvectors ( [-1 1], [1 1]). (The eigenvectors are transposed)
Then I set e^(tA) = T^(-1)AT, where T = [eigenvector1, eigenvector 2]
Which leads to e^(tA):
[1.75 .75]
[0 1.5]
I Feel that I do not have the correct equation for e^(tA), can anyone help me?
First, it is very easy for a 2 by 2 matrix, having only one eigenvalue, to have two independent eigenvectors- the matrix
has the single eigenvalue 3 and has <1, 0> and <0, 1> as independent eigenvectors.
However, this matrix does not. It has only multiples of <1, -1> as eigenvectors. <1, 1> is a "generalized eigenvector".
But " ", even for a diagonalizable matrix, is NOT " ". Rather, where D is the diagonal matrix having the eigenvalues on the diagonal. Then where is the diagonal matrix having e to the eigenvalue powers on the diagonal.
That is, if then .
However, when we do not have a "complete set of eigenvectors" as here, we cannot diagonalize the matrix. What we can do is write it in "Jordan Normal Form" which, here, would be .
And if then
You mentioned that e^D will not be the diagonal matrix with the eigenvalues as the diagonal elements.
Am I supposed to solve e^(tA) = Te^(D)T^(-1)
where D:
(1.5 1)
(0 1.5)
So that e^D:
(e^(1.5t) te^1.5t)
(0 e^(1.5t))
or
e^D:
(e^(1.5) e^(1.5))
(0 e^(1.5))?
I did it the first way and solved it out to e^(tA)=
(-.5te^(1.5t)+e^(1.5t) -.5te^(1.5t))
(.5te^(1.5t) .5te^(1.5t)+e^(1.5t))