finding a basis to the solutions of the following differential equation

Question

I am supposed to find the solutions to the differential equation:

$x'(t)= Ax(t)$, where $$A= \left(\begin{array}{rrr}-1 &1& 0\\0& -1& 0\\0& 0& 2\end{array}\right).$$ We haven't really developed many methods yet to solving differential equations. I first tried to calculate the exponential of the matrix to plug it into the solution formula but ran into trouble since the Matrix wasn't diagonal. I then tried to diagonalize the matrix which isn't possible(unless I made a mistake). (By not possible I mean with my very modest methods)

Answer 1

The matrix is already in Jordan normal form, and is not diagonalizable. To find its exponential, write it as the sum of a diagonal matrix $D$ and nilpotent matrix $N$: $$\pmatrix{-1&0&0\\0&-1&0\\0&0&2}+\pmatrix{0&1&0\\0&0&0\\0&0&0}.$$ These two matrices commute, so $e^{tA}=e^{t(D+N)}=e^{tD}e^{tN}$. I trust that you can find the exponential of a diagonal matrix. To find $e^{tN}$, expand it as a power series and note that $N^2=0$, so the series is truncated after two terms. Thus, $$e^{tA}=e^{tD}(I+tN)=\pmatrix{e^{-t}&0&0\\0&e^{-t}&0\\0&0&e^{2t}}\pmatrix{1&t&0\\0&1&0\\0&0&1}=\pmatrix{e^{-t}&te^{-t}&0\\0&e^{-t}&0\\0&0&e^{2t}}.$$

Answer 2

The exponential of $A$ is $$ \mathrm{e}^{tA}=\left(\begin{array}{ccc}\mathrm{e}^{-t}&t\mathrm{e}^{-t}& 0\\ 0&\mathrm{e}^{-t}&0 \\ 0&0& \mathrm{e}^{2t}\end{array}\right) $$ and a basis of the space of the solutions of $x'=Ax$ consists of the columns of $\mathrm{e}^{tA}$.