# Eigenvectors and generalized Eigenvectors not same!!

Let $A = \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 4 & 0 & -3\\ \end{bmatrix}$. One can see that the corresponding eigenvalues are $\{-2,-2,1 \}$. Finding the eigenvectors we see that we get one one eigen vector corr to $-2$ which is $\begin{bmatrix} 1\\ -2\\ 4\\ \end{bmatrix}$ and the eigenvector corr to $1$ is $\begin{bmatrix} 1\\ 1\\ 1\\ \end{bmatrix}$. We have eigenvalue $-2$ of multiplicity $2$ but got only one independent eigen vector so the matrix $A$ is defective. So we use method to fing generalized eigen vector.

But when I calculate $(A+2I)^2(x)=0$ to find eigen vectors I get a completely different set of eigen vectors $\{\begin{bmatrix} 1\\ 0\\ -4\\ \end{bmatrix} , \begin{bmatrix} 0\\ 1\\ -4\\ \end{bmatrix} \}.$

But how is this this possible??!! • The determinant is $-4$.and the product of your eigenvelues is $4$ Mar 24, 2016 at 14:45
• The vector $(1,-2,4)$ is in the span of the vectors $(1,0,-4)$ and $(0,1,-4)$. Indeed, it is $(1,-2,4)=(1,0,-4) - 2\cdot (0,1,-4)$. How did you find that there was only one eigenvector originally? There is likely another that you missed and the eigenspaces should be equal (although perhaps different representations of the same space). Mar 24, 2016 at 14:49
• @Upstart see that det(A) is 4. Mar 24, 2016 at 14:53

There is no problem. So when you solve $(A+2I)^2X=0$, you find two generalized eigenvectors $(1,0,-4)$ and $(0,1,-4)$. Now you don't know that one of these has to be an eigenvector, but you do know that there is an eigenvector in the space spanned by these two vectors. And indeed, $(1,-2,4)=(1,0,-4)-2(0,1,-4)$.
If you solve $(A+2I)^2(x)=0$ the solutions are no longer Eigenvectors (but I don't know the english word for it). By taking the solution of $(A+2I)^2(x)=0$ you can get the Jordan normal form, instead of diagonalize it.