# Subspaces of generalised eigenspace

If $$A$$ is an endomorphism of a vector space $$V$$ and $$\lambda$$ an eigenvalue then $$Ker (A-\lambda I)$$ is the eigenspace corresponding to $$\lambda$$ and $$V_\lambda=Ker (A-\lambda I)^{dim \ V}$$ is the generalised eigenspace corresponding to $$\lambda$$.

The space $$V_\lambda$$ has a direct sum decomposition into a number of subspaces $$V_i$$ equal to the geometric multiplicity of $$\lambda$$ and there exists a basis of $$V_i$$ such that $$A|_{V_i}$$ is represented by a matrix consisting of a single Jordan block.

Do the spaces $$V_i \subset V_\lambda$$ have a standard name? What is a convenient way to refer to them without having to repeat all of the above?

I can't say for sure that nobody has named them, but I suspect not, simply because the decomposition into these subspaces is not unique in general. For example, if we consider the operator $$T(x, y, z) = (y, 0, 0)$$, then the standard matrix for $$T$$ is in JNF: $$\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$ From this, we get one possible decomposition: $$\operatorname{span}(i, j), \operatorname{span}(k)$$.

But, on the other hand, consider the basis $$B = (i, j, i + k)$$. The matrix for this basis is exactly the same as above: $$i$$ and $$j$$ form a chain of generalised eigenvectors, and $$i + k$$ is a linearly independent eigenvector. So, we could just as easily decompose it like so: $$\operatorname{span}(i, j), \operatorname{span}(i + k)$$.

The point is, this decomposition is a property of the specific Jordan basis, not of the operator itself, which suggests to me that it likely has no name for itself.

Given a Jordan basis $$B$$, you could refer to these subspaces as "spans of chains of generalised eigenvectors". A bit of a mouthful though.