# Finding a solution basis

Find a real solution basis of $$y'=\left( \begin{matrix}-1&-2&0\\0&2&0\\-1&-3&2\\ \end{matrix} \right)y.$$

The characteristic equation of this matrix is $$P(t) = (1-t)(2-t)^2.$$ So next I calculated eigenvectors for the eigenvalues $1$ and $2$, which are $$u\overset{def}=(1,0,1) \text{ and }v\overset{def}=(0,0,1) \text{ respectively}.$$ The eigenvalue $2$ has algebraic multiplicity $2$ but it only has one eigenvector. So if I'm correct we need a principal vector. I computed this and it is $$v_p\overset{def}=(2,-1,0).$$

Now the solution basis is $$B=\Big\{t\mapsto u e^t, t\mapsto ve^{2t}, ?? \Big\}.$$

My question is, what is the third function? What solution does the principal vector I have computed correspond to?

Thank you.

• Shouldn't your characteristic polynomial be $(1+t)(2-t)^2$? – paw88789 Aug 19 '14 at 10:29
• You're right, sorry :) – rehband Aug 19 '14 at 11:02

If the constant matrix $A$ of one such homogeneous system is defective, for each eigenvalue $\lambda$ whose geometric multiplicity $m$ isn't big enough (i.e., it doesn't coincide with the algebraic multiplicity), the following functions are linearly independent solutions to the differential equation:
\begin{align} &t\mapsto e^{\lambda t}v_1\\ &t\mapsto e^{\lambda t}(tv_1+v_2)\\ &t\mapsto e^{\lambda t}\left(\dfrac{t^2}{2!}v_1+tv_2+v_3\right)\\ &t\mapsto e^{\lambda tv}\left(\dfrac{t^{m-1}}{(m-1)!}v_1+\ldots +tv_{m-1}+v_m\right), \end{align} where for each $i\in \{1, \ldots m\}$, $v_i$ is a generalized eigenvector which can be found iteratively by solving the system $(A-\lambda I)v_i=v_{i-1}$ for $i\ge 2$ and $v_1$ being an eigenvector.
Assuming what you did is correct, you get a third solution (which makes the set of solutions linearly independent) by considering $t\mapsto e^{2t}(tv+w)$, where $w$ is a solution to $(A-2I)x=v$. I found one such solution to be $\begin{bmatrix} \frac 2 7 \\ -\frac 3 7\\ 0\end{bmatrix}$.