# What is a generalised eigenvector and what they have to do with the dimension they span?

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Hi guys, I was doing some past papers for my linear algebra course and came across this question. Part i) is where I am stuck on and would appreciate some guidance as to how to go about it.

As far as my understanding goes regarding this question, the eigenvalue of -2 has algebraic multiplicity of 3 and geometric multiplicity of 2. This means that the eigenspace is 2D. However I am unsure of how to relate the span of this eigenspace to that of 3D space as well as how generalised eigenvectors come into play.

Also, as far as my knowledge of generalised eigenvectors go, I know that they are vectors not in the kernel of (A - (lambda)*I) (and for higher power of this). That's about as much as I know about generalised eigenvectors.

Any help on this question would be greatly appreciated,

Thanks.

The definition of a generalised eigenvector for a (square) matrix $$B$$ corresponding to eigenvalue $$\lambda$$, is a non-zero vector $$v$$ such that $$v \in \operatorname{ker}(B - \lambda I)^m$$ for some $$m \in \mathbb{N}$$. When $$m = 1$$, you get the eigenvectors (which are still generalised eigenvectors!).
One of the main results along the way of building the Jordan Normal Form is that, given an $$n \times n$$ matrix (or equivalently, a linear operator on a dimension $$n$$ space $$V$$), the generalised eigenvectors direct sum to $$\mathbb{R}^n$$ (or $$V$$, respectively). Therefore, the only way you will have every vector being a generalised eigenvector is if the matrix/operator has only one eigenvalue.
In this case, we must have $$\lambda = -2$$ is the only eigenvalue, given the information about $$\operatorname{ker}(B + 2I)$$. Note that the eigenspace is two-dimensional, so if the question is right, then we'd need to have $$\operatorname{ker}(B + 2I)^2 = \mathbb{R}^3.$$ This fact should be easily verifiable. Compute $$(B + 2I)^2$$, and you'll find that it comes to the $$0$$ matrix (which contains every vector in its kernel.)