Linear ODE repeated eigenvalues how to find more than 2 generalized eigenvectors So I've searched around the web for a few hours now, as
(i)
$\mathbf A = \begin{pmatrix}2&1\\0&2\end{pmatrix}$
The characteristic polynomial is $(\lambda-2)^2=0$, so $\lambda=2$, repeated.
A bit of algebra convinces us that $b=(1,0)^t$ is an eigenvector.
Setting $(\mathbf A-\lambda \mathbf I)r=b$, with $r$ and $b$ vectors. Maybe a bit unmotivated but lets proceed. We find that $r=(0,1)^t$ is a solution. Now the solution is 
$x(t)= c_1*e^{2t}+c_2(te^{2t}b+e^{2t}r)$
However lets say I have 3 or more repeated eigenvalues, or two eigenvalues that are both eigenvalues what then?
For instance: 
$\mathbf A = \begin{pmatrix}1&2&0\\0&1&3/4\\0&0&1\end{pmatrix}$
I find $\lambda=1$ and the vector $b_2=(1,0,0)^t$ is a solution. But I'm not sure how to proceed. 
Any help would be appreciated.

$\mathbf A = \begin{pmatrix}1&2&0\\0&1&3/4\\0&0&1\end{pmatrix}$
You should get an Eigensystem as follows:
$$\lambda_1 = 1, v_1 = (1,0,0)$$
$$\lambda_2 = 1, v_2 = (0, 1/2, 0)$$
$$\lambda_3 = 1, v_3 = (0, 0, 2/3)$$
Next, we can write the Jordan Normal Form as:
$$A = \begin{pmatrix}1&2&0\\0&1&3/4\\0&0&1\end{pmatrix} = P J P^{-1} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1/2 & 0\\ 0 & 0 & 2/3\end{bmatrix} \cdot \begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0 \\ 0 & 0 & 3/2\end{bmatrix}$$
What do you notice about $J$? What is it made from?
What do you notice about the columns of $P$? What is it made from?
If you would like to step through the manual process, just give a yell.
Here is a very nice example and procedure need help Jordan base (most time is overkill, but is also a general procedure).
