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Let $T$ be a linear operator on a complex vector space $V$, where $n<\infty$, and let $A_1,\dots,A_m$ be the Jordan blocks of the matrix of $T$ with respect to some Jordan basis. For each $A_i$ (of size $d\times d$), there is a natural associated subspace $U_i\subseteq V$ (of dimension $d$) for which $\mathcal M(T|_{U_i})=A_i$.

We can decompose $V$ as

$$V=U_1\oplus\dots \oplus U_m,$$

which is an immediate consequence of the existence (and definition) of a Jordan basis. What is less clear is whether the subspaces $U_i$ are unique, i.e., independent of choice of Jordan basis.

Are the subspaces $U_i$ (defined above) of dimension at least two unique, i.e., independent of choice of Jordan basis?


Although it is a standard result that the Jordan matrix is unique up to reordering of blocks, this does not guarantee that these subspaces are unique. In fact, as a good warning, they definitely are not if we drop the requirement of considering only subspaces of dimension at least two. (As a counterexample, consider $T=I$ on $\mathbb C^2$ and compare the subspaces for the Jordan bases $\{(0,1),(1,0)\}$ and $\{(1,1),(1,-1)\}$.)

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The subspaces are not unique. For instance, consider the transformation with matrix $$ A = \pmatrix{\lambda&1&0&0\\0&\lambda&0&0\\0&0&\lambda&1\\0&0&0&\lambda}. $$ Note that the standard basis and the basis $$ \{(1,0,-1,0),(0,1,0,-1),(1,0,1,0),(0,1,0,1)\} $$ are two Jordan bases that lead to distinct subspaces.

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  • $\begingroup$ On the other hand, I believe that a similar idea with Weyr canonical form yields a unique collection of subspaces $\endgroup$ – Omnomnomnom May 19 at 1:25
  • $\begingroup$ Thanks. The answer makes sense, and the comment was especially helpful. It seems like the Weyr canonical form provides more clear insight into the structure of the operator's generalized eigenspaces. $\endgroup$ – WillG May 19 at 16:05
  • $\begingroup$ It seems that the subspace associated with the first $n$ Weyr blocks corresponding to $\lambda$ is precisely the subspace $\operatorname{null}(T-\lambda I)^n$, which is certainly independent of a basis. $\endgroup$ – WillG May 19 at 16:07
  • $\begingroup$ @WillG That makes sense. I haven't thought about WCF for a while so I wasn't confident. $\endgroup$ – Omnomnomnom May 19 at 16:43

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