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??!!
Having a great confusion. Please Help!
$\endgroup$ 32 Answers
$\begingroup$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, $(1,-2,4)=(1,0,-4)-2(0,1,-4)$.
Generalized eigenvectors generalize eigenvectors in the sence that the space spanned by eigenvectors belonging to some eigenvalue is contained in the space spannen by the corresponding generalized eigenvectors.
$\endgroup$ $\begingroup$The point is. The algebraic multiplicity of Eigenvalue -2 is 2. But the geometric multiplicity is 1. So there is just 1 Eigenvector to this Eigenvalue.
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.
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