# 4x4 Complex repeated Eigenvalues

• Feb 27th 2011, 11:24 AM
Borkborkmath
4x4 Complex repeated Eigenvalues
I'm trying to find the general solution for the following 4x4 matrix:
[0 1 1 0]
[-1 0 0 1]
[0 0 0 1]
[0 0 -1 0]

So far, I have the eigenvalues as repeated i, i, - i ,-i
Eigenvector for i: [-i 1 0 0]^t (with multiplicity 2)
Eigenvector for -i: [1 i 0 0]^t (with multiplicity 2)

How do I get the general solution for this? And how do I find the adjoint eigenvectors for a 4x4 with complex repeated eigenvalues?
• Feb 27th 2011, 11:32 AM
FernandoRevilla
Quote:

Originally Posted by Borkborkmath
So far, I have the eigenvalues as repeated i, i, - i ,-i

Right.

Quote:

How do I get the general solution for this?

What do you mean by general solution?. General solution of ...
• Feb 27th 2011, 11:42 AM
Borkborkmath
Of the matrix,
Umm.. I do believe it should look something like x(t) = c_1e^(eigen)(eigenvector) + ... + c_ne^(eigen)(eigenvector)

This is from an ODE's class, but last time I posted a matrix question from ODE's a mod moved it and scolded me. So, that is why I put it in the linear algebra this time.
• Feb 27th 2011, 11:58 AM
FernandoRevilla
I guess you mean the general solution of the differential system $X'=AX$. You can express

$X(t)=e^{tA}K\quad (K=(k_1,k_2,k_3,k_4)^t)$

or in a vectorial form:

$X(t)=k_1C_1(t)+k_2C_2(t)+k_3C_3(t)+k_4C_4(t)$

where $C_i(t)$ are de columns of $e^{tA}$ .
• Feb 27th 2011, 12:02 PM
Borkborkmath
Yes, thats exactly what I mean.

How do I get the adjoint vectors of the repeated complex eigenvalues so that I can write the general solutions for X'= AX?
• Feb 27th 2011, 12:17 PM
FernandoRevilla
Quote:

Originally Posted by Borkborkmath
Yes, thats exactly what I mean. How do I get the adjoint vectors of the repeated complex eigenvalues so that I can write the general solutions for X'= AX?

Don't worry about it, all is included in the columns $C_i(t)$ of $e^{tA}$.
• Feb 27th 2011, 12:20 PM
Borkborkmath
So the general solution is:
X(t) = e^(tA)*(i 1 0 0) + e^(tA)*(1 i 0 0)?