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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??!!

BY Mathematica

Having a great confusion. Please Help!

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  • $\begingroup$ The determinant is $-4$.and the product of your eigenvelues is $4$ $\endgroup$ – Upstart Mar 24 '16 at 14:45
  • $\begingroup$ 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). $\endgroup$ – JMoravitz Mar 24 '16 at 14:49
  • $\begingroup$ @Upstart see that det(A) is 4. $\endgroup$ – user8795 Mar 24 '16 at 14:53
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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.

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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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  • $\begingroup$ but generalized eigen vectors gives an extension of eigen vectors set!! $\endgroup$ – user8795 Mar 24 '16 at 14:58
  • $\begingroup$ I've learned it in another context, but you might be right, I guess $\endgroup$ – MorphhproM Mar 24 '16 at 14:59

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