Find the solution of the system

Question

Find the solution of the system $ \ \ y'(t)=\begin{pmatrix} 1 & 2 & 0 \\ 1 & 1 & 2 \\ 0 & -1 & 1 \end{pmatrix}\vec{y(t)} $ , the repeated eigen values are 1,1,1 . $$ $$ Corresponding to the repeated eigen vector 1 , the eigen vector is $ \ \vec{ \alpha}= \begin{pmatrix} -2 \\ 0 \\ 1 \end{pmatrix} $. But I can not not find the other eigen vector in order to get the solution. Please help me

Answer 1

You can’t find any other eigenvectors because there aren’t any other eigenvectors of this matrix that are linearly independent of the one you’ve found. Calling the coefficient matrix $A$, row-reducing $A-I$ produces $$\pmatrix{1&0&2\\0&1&0\\0&0&0}$$ so the kernel of $A-I$ and so also the eigenspace of $1$ is clearly only one-dimensional. $A$ is therefore not diagonalizable.

In theory, this means that you have to compute the Jordan decomposition of $A$, finding generalized eigenvectors &c., but in practice, you don’t really need the Jordan basis. Observe that if we set $N=A-I$, the Cayley-Hamilton theorem tells us that $N$ is nilpotent. You can thus write $A$ as the sum of a multiple of the identity matrix and a nilpotent matrix. These matrices commute, so $e^{t(I+N)}=e^{tI}e^{tN}$. The exponential of a diagonal matrix is trivial to compute, and you can use the power series definition of the exponential of a matrix to compute $e^{tN}$, taking advantage of the fact that the series will be truncated after a few terms because $N$ is nilpotent.