TITLE: Initial Value Problem with Repeated Eigenvalues
QUESTION [5 upvotes]: Given the matrix
$$
A=\left(\begin{array}{ccc}
1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)
$$
For $X'= AX.\quad$
$X\left(0\right)=\left(\begin{array}{r}1 \\ 0 \\ -2\end{array}\right)\,.\quad$
What is the solution ?.

REPLY [6 votes]: We find the eigenvalues and eigenvectors as:

$\lambda_{1,2,3} = 1$, which gives one eigenvector and two generalized eigenvectors as:
$v_1 = (0,1,0), v_2 = (1,0,0), v_3 = (0,0,1)$

We can write the general solution as:
$$X(t) = e^t\left(c_1v_1 + c_2(t v_1 + v_2) + c_3\left(\frac{t^2}{2!}v_1 + t v_2 + v_3\right)\right)$$
Using the initial condition, $X(0)$, we find:
$c_1 = 0, c_2 = 1, c_3 = -2$
Our final solution is then:

$x(t) = e^t(1 - 2t)$
$y(t) = e^t(t - t^2)$
$z(t) = e^t(-2)$

Note, there are many ways to do these types of problems from the matrix exponential, fundamental matrix, set of linear equations...