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?


  • $\begingroup$ 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. $\endgroup$ – Azlif Oct 20 '19 at 13:30
  • $\begingroup$ I have changed it to 2. I'm afraid I don't understand what you mean. Could you explain it some other way? $\endgroup$ – 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 $\begin{bmatrix} 1 \\ -1\end{bmatrix}$.

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  • $\begingroup$ 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? $\endgroup$ – user5744148 Oct 20 '19 at 15:26
  • $\begingroup$ Yes, that's correct. $\endgroup$ – S. Dolan Oct 20 '19 at 15:27

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