Solving x'=Ax for 4x4 matrix with repeated eigenvalues

Question

$$\begin{bmatrix} 2 & 1 & 4 & 0 \\ 0 & 2 & 1 & -1 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix}$$

I have the following matrix, with eigenvalue 2 (multiplicity 4). I am trying to solve x' = Ax, where A is the matrix shown above.

I know that the equation can be solved as follows:

$x(t) = e^{At}x_0 = Pe^{Jt}P^{-1}x_0 $

J is the Jordan canonical form of the matrix A:

$$\begin{bmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix}$$

How do I go about finding the eigenvectors in order to construct P and $P^{-1}$ in order to solve for x(t)? In addition, what is $e^{Jt}$ and how do I find it?

Answer 1

you do not need Jordan form this time. Your matrix is of the form $A = 2I + N,$ where $N$ is the strict upper diagonal part as a square matrix, otherwise put $0.$ The two parts also commute, so $$ e^A = e^{2I} e^N $$ and $$ e^{At} = e^{2tI} e^{tN} $$ As $N$ is nilpotent, one can find the exponential from its (finite) Taylor series. Find $N, N^2, N^3.$ By my count $N^4 = 0.$

Note that the process with Jordan form is the same, you have a matrix $J = D + N$ with $N$ nilpotent, $D$ diagonal, and $DN=ND.$ The additional work is that it really is necessary to find a matrix $P$ such that $PJP^{-1} = A$ is the original matrix. Then $e^{At} = P e^{Dt} e^{Nt} P^{-1}.$ You need to find the actual matrix $P$ and its inverse.

Answer 2

Eigenvectors are not going to cut it, your matrix is not diagonalizable. You'll need generalized eigenvectors.

Find an eigenvector by solving $(A-2I)x=0$ (or by observing your matrix). Then look for generalized eigenvectors which satisfy $$(A-2I)^2x=0, (A-2I)^3x=0, (A-2I)^4x=0$$ You want them to be linearly independent. They form the matrix $P$.

$e^{Jt}$ is the exponential of $Jt$ that is $$e^{Jt}=\sum_{n=0}^{\infty}\dfrac{t^nJ^n}{n!}$$ Since $J$ is not diagonal, $J^n$ is not so easily computed. You may write $J=2I+N$ where $N$ is nilpotent to help with practical calculations.