# Generalized eigenvector question (ODE system)

This is an example in Boyce-Diprima. I have the following system$$x'=\begin{bmatrix} 1 & -1 \\ 1 & 3 \end{bmatrix}x$$ I solve for the eigenvalues, which is just $$\lambda=2$$ in this case, and it has has eigenvector $$\xi = (1,-1)$$. I set up the equation to solve for the generalized eigenvector which gives me $$-\eta_1 + \eta_2=1$$. Now, here is my question. If I set $$\eta_1=k$$ and the solve the equation for the generalized eigenvector, I get $$\eta= \begin{bmatrix} 0 \\ -1\end{bmatrix} + k\begin{bmatrix} 1 \\ -1\end{bmatrix}$$ But if I instead set $$\eta_2=-k$$ and solve the equation for the generalized eigenvector, I get

$$\eta= \begin{bmatrix} -1 \\ 0\end{bmatrix} + k\begin{bmatrix} 1 \\ -1\end{bmatrix}$$ They are not multiplicative constants of each other, which seems strange. Are both correct, and if so, why?

Thanks!

• Your $\lambda$ should be $2$. In generalized eigenvector, in many times, the generalized eigenvector won't be multiple of each other(why?) since it is a solution to nonhomogenous equation rather than to homogenous one. – Azlif Oct 20 '19 at 13:30
• I have changed it to 2. I'm afraid I don't understand what you mean. Could you explain it some other way? – user5744148 Oct 20 '19 at 14:22

If $$A=\begin{bmatrix} 1 & -1 \\ 1 & 3 \end{bmatrix}$$, then $$A-2I=\begin{bmatrix} -1& -1 \\ 1 & 1 \end{bmatrix}$$.

Therefore $$(A-2I)^2=\begin{bmatrix} -1& -1 \\ 1 & 1 \end{bmatrix}^2=\begin{bmatrix} 0&0 \\ 0&0\end{bmatrix}$$.

We can therefore see that $$(A-2I)^2$$ will reduce every vector to zero.

Equivalently, we can say that $$(A-2I)$$ will map any vector to a multiple of the eigenvector $$\begin{bmatrix} 1 \\ -1\end{bmatrix}$$.

• Okey, so any vector will be reduced to zero. I take that as meaning that either of the above eta vectors are correct as representing the generalized vectors. Is that a correct interpretation? – user5744148 Oct 20 '19 at 15:26
• Yes, that's correct. – S. Dolan Oct 20 '19 at 15:27