# Are the subspaces corresponding to Jordan blocks unique?

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

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.
• 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. – WillG May 19 at 16:07