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 $\{(1,1),(1,-1)\}$.)
Best Answer
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.