I have the matrix

$$A=\begin{pmatrix} 5 & 1 & 0\\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}$$

and I should determine generalised eigenvectors, if they exist.

I found one eigenvalue with algebraic multiplicity $3$.


I calculated two eigenvectors:

$$\vec{v_{1}} =\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \qquad{} \vec{v_{2}} =\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$$

Also, I know this formula for generalized vector

$$\left(A-\lambda I\right)\vec{x} =\vec{v}$$

Finally, my question is:

How do I know how many generalised eigenvectors I should calculate?

For every eigenvector one generalised eigenvector or?

My university book is really confusing, and I saw there that they calculated generalised eigenvector only for some eigenvectors, and for some not. But I don't understand how to know that.

  • 2
    $\begingroup$ How can eigenvectors of a 2×2 matrix have 3 coordinates? $\endgroup$ – Bernard May 20 '16 at 21:41
  • $\begingroup$ Sorry. My mistake in writing matrix. I will edit it now. $\endgroup$ – Ana Matijanovic May 20 '16 at 21:42
  • $\begingroup$ by "generalised vector" do you mean generalised eigenvector? $\endgroup$ – Josh R May 20 '16 at 21:44
  • $\begingroup$ Yes I mean on that. $\endgroup$ – Ana Matijanovic May 20 '16 at 21:45
  • $\begingroup$ In your example all nonzero vectors are generalised eigenvectors. $\endgroup$ – Marc van Leeuwen May 20 '16 at 21:52

Your matrix is in Jordan normal form. You can read on the matrix that $e_1$ and $e_3$ are eigenvectors for the eigenvalue $2$, and $e_2$ is a generalised eigenvector.

  • $\begingroup$ So I dont have to calculate anything, I have just to see in matrix vector that is not eigenvector I got? $\endgroup$ – Ana Matijanovic May 20 '16 at 21:57
  • $\begingroup$ In the present case, yes. $\endgroup$ – Bernard May 20 '16 at 22:03
  • $\begingroup$ But what would mean present case, because I have to solve more than this matrix. They all are in Jordan normal form. So can I use this on them too? $\endgroup$ – Ana Matijanovic May 20 '16 at 22:15
  • $\begingroup$ I think so. However if all of them are in Jordan normal form, I don't see the point of the exercise. $\endgroup$ – Bernard May 20 '16 at 22:34

The generalised eigenspace of $A$ for an eigenvalue $\lambda$ is the kernel of $(A-\lambda I)^k$ for sufficiently large $k$ (meaning that the kernel won't get bigger by further increasing $k$). The multiplicity of $\lambda$ as root of the characteristic polynomial is always sufficiently large. In the example $(A-\lambda)^2=0$ so $k=2$ suffices and the generalised eigenspace is the whole space.

It is common to find a basis for the kernel with exponent $1$ first (the ordinary eigenspace) then extend to a basis for exponent$~2$, and so forth until$~k$. This basis is somewhat better than just any basis for the generalised eigenspace, but it remains non unique in general. Though there are infinitely many generalised eigenvectors, it is not useful to list linearly dependent ones among them, so one stops having found a basis for the generalised eigenspace (here after $3$ independent vectors).


The dimension of the nullspace of A minus lamda*I will give you the number of 'generalizable' eigenvectors for any particular eigenvalue. The sum of this for all different eigenvalues is the dimension of the eigenspace. Your matrix does not have 3 generalizable eigenvectors so it is not diagonizable.

Sorry for the lack of formulas. Doing this on my phone. Will come back later to edit.

  • $\begingroup$ Nice bamboozle with formulas. $\endgroup$ – nikoliazekter Apr 13 '17 at 22:18

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