Generalized Eigenvector for 4x4 matrix I'm working on Systems of Differential Equations and I'm looking to find the Generalized eigenvector for the following matrix:
$\left[\begin{array}{rrrr}
3 &-4 &1  &0 \\ 
 4& 3 &0  &1 \\ 
0 &0  &3  &-4 \\ 
0\ \ &0\ \ &4\ \
 &\>\>\>3 
\end{array}\right]$
I've found the eigenvalues, r = 3+4i, r = 3-4i, and one of the corresponding Eigenvectors
E1 = $\left[\begin{array}{rrrr}
i \\ 
 1 \\ 
0 \\ 
0\\  
\end{array}\right]$
So I'm looking for one generalized eigenvector. For some reason this question in particular has been causing me some problems. Any help is greatly appreciated
 A: Given:
$$A = \begin{pmatrix} 3 &-4 &1  &0 \\  4& 3 &0  &1 \\ 0 &0  &3  &-4 \\ 0 & 0 & 4 & 3 \end{pmatrix}$$
You have correctly found the eigenvalues:
$$ \lambda_1 = 3+4i, \lambda_2 = 3-4i$$
Each eigenvalue has an algebraic multiplicity of two.
For $\lambda_1 = 3 + 4i$, the row reduced echelon form (RREF) of $[A-\lambda_1 I]v_1 = 0$, is:
$$\begin{pmatrix} 1 &-i & 0  &0 \\ 0 & 0 & 1  & 0 \\ 0 & 0  & 1  & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}v_1 = 0$$
For this eigenvalue, you have correctly found one of the corresponding eigenvectors
$$v_1 = \begin{pmatrix} i \\  1 \\ 0 \\ 0\\  \end{pmatrix}$$
However, we only have a single eigenvector (geometric multplicity is one) and need to find a second generalized eigenvector. One approach is to use $[A - \lambda_1 I]v_2 = v_1$, yielding the augmented matrix:
$$
  \left[\begin{array}{rrrr|r}
    4i &-4 &1  &0 & i \\  4& 4i &0  &1 &1 \\ 0 &0  & 4i  &-4 &0\\ 0 & 0 & 4 & 4i&0
  \end{array}\right]
$$
The RREF of this matrix is:
$$
  \left[\begin{array}{rrrr|r}
    1 &-i & 0  & 0 & 0 \\  0& 0 &1  &0 &i \\ 0 &0  & 0  &1 &1 \\ 0 & 0 & 0 & 0&0
  \end{array}\right]
$$
From this RREF we have:
$$d = 1, c = i, a = i b$$
Choose $b = 0 \implies a = 0$, yielding a second generalized eigenvector:
$$v_2 = \begin{pmatrix} 0 \\ 0 \\ i \\  1 \end{pmatrix}$$
It is worth noting that the eigenvalues come in complex conjugate pairs and so do the eigenvectors. In other words, you can just write the other two eigenvectors for $\lambda_2$ from the work above by taking the complex conjugate of each eigenvector.
