I have a matrix $$A = \begin{bmatrix}1 & 1 \\ -1 & 3\end{bmatrix}$$ I want to find out the generalised Eigenvectors. The Eigen values corresponding to the characteristic equation is $\lambda = 2$ and the Eigenvector correspondig to the eigenvalue is found to be $\begin{bmatrix}1 \\ 1\end{bmatrix}$.
So how to calculate the generalised Eigen vector for this matrix.
What I did is ,
I took $(A-\lambda\cdot I)^2 \nu = 0$.
then solving the $(A-\lambda I)^2 = \begin{bmatrix}1-\lambda & 1 \\ -1 & 3-\lambda \end{bmatrix}^2 = \begin{bmatrix}\lambda^2-2\lambda & 4 – 2\lambda \\ 2\lambda-4 & \lambda^2 – 6\lambda+8 \end{bmatrix}$
At this point I don't know whether I am doing the things correct . as finding the determinant will take this to $\lambda^4$.
Best Answer
Since $\lambda=2$ has algebraic multiplicity equal to 2, the Jordan form is
$$J = \begin{bmatrix}2 & 1 \\ 0 & 2\end{bmatrix}$$
then by $P=[v_1\,v_2]$
$$P^{-1}AP=J \implies AP=PJ$$
that is