Finding generalised eigenvector - $(A-\lambda*I)^2$ is all zeros.

Question

So I'm solving some systems of differential equations, and while doing that I have to find the generalised eigenvector. But when I calculate $(A-\lambda I)^2$ all I get is a matrix full of zeros.

This is the matrix A i start with.

$ \begin{pmatrix} 3 & 5 \\ -5 & -7 \end{pmatrix} $

I find $\lambda = -2 $ with algebraic multiplicity 2.

Giving me $(A+2I)$ = $ \begin{pmatrix} 5 & 5 \\ -5 & -5 \end{pmatrix} $

and eigenvector $ \begin{pmatrix} -1 \\ 1 \end{pmatrix} $

So now, to find the generalised eigenvector, i need to find $\ker(A+2I)^2$

But when finding $(A+2I)^2$ i get : $ \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} $

What do i do in this situation?

Answer 1

If $(A+2I)^{2}=0$, then $\ker(A+2I)^{2}=\mathbb{R}^{2}$ (if you're working on $\mathbb{R}$), so the only possible Jordan form is $\left[\begin{matrix}-2 & 1 \\ 0 & -2\end{matrix}\right]$.