Do linear operators that map one space into a different space have a Jordan canonical form?

Question

I know that this answer is most likely "yes", and that, in the setting of matrices, all matrices are similar to its Jordan form, which is unique (up to the ordering of the Jordan blocks.)

But what about linear operators that map one vector space into a different space, e.g., a mapping from $V \to W$?

Does it make sense for this operator to have a Jordan canonical form?

I feel it doesn't, since in the problems that I've been working on this past weekend, I am computing generalized eigenspaces (which may or may not factor into smaller, cyclic subspaces), but this means that the spaces are invariant under the action of the operator - but this only makes sense if the mapping is from $V \to V$.

Where am I going wrong with my thinking?

When I re-read the theorems, the only assumption is that the operator should have a characteristic polynomial that splits.

Also, all the matrices that I computed (before computing its Jordan form) turned out to be nilpotent, square matrices, but I don't see any definitions that require a nilpotent operator to have to map one space onto itself - but only that it has some (finite) nilpotence degree, for some positive integer k.

EDIT: the spaces are invariant under the action of $A-\lambda I$, not $A$, but I still essentially have the same question. Thanks.

Answer 1

No, it does not make sense to talk about a Jordan canonical form for a linear map $\varphi:V\to W$. Alternatively, one could say it only makes sense once you choose a particular isomorphism $\psi:W\xrightarrow{\;\cong\;}V$, at which point what you're really talking about is the Jordan canonical form of the linear map $(\psi\circ \varphi):V\to V$. (In particular, this is only possible if $V$ and $W$ are isomorphic.)

When you ask "where am I going wrong with my thinking?", you haven't made any mistake at all, generalized eigenspaces, invariant subspaces, etc. only make sense for a map from a space to itself, and that is an accurate way of realizing why it doesn't make sense to think of a Jordan canonical form for a map between two different spaces.

When you say "I don't see any definitions that require a nilpotent operator to have to map one space onto itself", that's incorrect, the very idea of nilpotency does require that; a linear map $\varphi:V\to V$ is nilpotent if it has the property that if you compose it with itself enough times you get the zero map, but you can't compose a map $\varphi:V\to W$ with itself any times if $V$ and $W$ are different spaces.

Another comment: the Jordan canonical form of a linear map $\varphi:V\to V$ corresponds to the decomposition of $V$ as a $k[x]$-module, where $x$ acts on $V$ as the linear map $\varphi$. A linear map $\varphi:V\to V$ creats a $k[x]$-module structure because we can talk about ring homomorphisms $k[x]\to\mathrm{Hom}_k(V,V)$, whereas $\mathrm{Hom}_k(V,W)$ does not have a built-in ring structure if $V\neq W$ so that there is no $k[x]$-module in sight to decompose if we're only given a linear map $\varphi:V\to W$.

Answer 2

I just wanted to point out that linear transformations between different spaces do have a very nice form:

If $T:V\to W$ then we can let $w_1, \dots, w_r$ be a basis for the range of $T$, and we can extend with vectors $w_{r+1}, \dots , w_m$ to a basis for $W$. Then we can let $v_1, \dots, v_r$ be such that $Tv_i=w_i$. Note that this implies that $v_1, \dots , v_r$ are linearly independent so that they can be extended with vectors $v_{r+1}, \dots , v_n$ to a basis of $V$. Moreover, by the rank-nullity theorem, we may choose $v_{r+1}, \dots , v_n$ to be a basis for the nullspace of $T$, so that we have $Tv_i=w_i$ for $1\leq i \leq r$, and $Tv_i=0$ for $i>r$. Hence, the matrix of $T$ with respect to these bases for $V$ and $W$ is extremely nice - it is diagonal, and its first $r=$ rank$(T)$ diagonal entries are $1$, and the rest are zero.

Therefore, we actually get a form of $T$ which is much nicer than Jordan canonical form. The reason we get such a nice matrix is that we're free to choose separate bases for the domain and target spaces, whereas the significance of the Jordan canonical form is that given an endomorphism of a finite-dimensional complex vector space, one can find a single basis for the vector space so that the matrix of the operator with respect to that basis is in Jordan canonical form.

By the way, this matrix form has a nice application: you can use it to provide a simple proof that the rank of a matrix is equal to the rank of its transpose.