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?


1 Answer 1


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.


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