Give the chains of rangespaces and nullspaces for the zero and identity transformations.
For the zero transformation, no matter what the space, the chain of rangespaces is and the chain of nullspaces is . For the identity transformation the chains are and .
For each map, give the chain of rangespaces and the chain of nullspaces, and the generalized rangespace and the generalized nullspace.
Iterating twice gives
and any higher power is the same map. Thus, while is the space of quadratic polynomials with no linear term , and is the space of purely-quadratic polynomials , this is where the chain stabilizes . As for nullspaces, is the space of purely-linear quadratic polynomials , and is the space of quadratic polynomials with no term , and this is the end .
The second power
is the zero map. Consequently, the chain of rangespaces
and the chain of nullspaces
each has length two. The generalized rangespace is the trivial subspace and the generalized nullspace is the entire space.
Iterates of this map cycle around
and the chains of rangespaces and nullspaces are trivial.
Thus, obviously, generalized spaces are and .
and so the chain of rangespaces
and the chain of nullspaces
each has length two. The generalized spaces are the final ones shown above in each chain.
Prove that function composition is associative and so we can write without specifying a grouping.
Each maps .
Check that a subspace must be of dimension less than or equal to the dimension of its superspace. Check that if the subspace is proper (the subspace does not equal the superspace) then the dimension is strictly less. (This is used in the proof ofLemma 1.3.)
Recall that if is a subspace of then any basis for can be enlarged to make a basis for . From this the first sentence is immediate. The second sentence is also not hard: is the span of and if is a proper subspace then is not the span of , and so must have at least one vector more than does .
Prove that the generalized rangespace is the entire space, and the generalized nullspace is trivial, if the transformation is nonsingular. Is this "only if" also?
It is both "if" and "only if". We have seen earlier that a linear map is nonsingular if and only if it preserves dimension, that is, if the dimension of its range equals the dimension of its domain. With a transformation that means that the map is nonsingular if and only if it is onto: (and thus , etc).
The nullspaces form chains because because if then and and so .
Now, the "further" property for nullspaces follows from that fact that it holds for rangespaces, along with the prior exercise. Because the dimension of plus the dimension of equals the dimension of the starting space , when the dimensions of the rangespaces stop decreasing, so do the dimensions of the nullspaces. The prior exercise shows that from this point on, the containments in the chain are not proper— the nullspaces are equal.
Give an example of a transformation on a three dimensional space whose range has dimension two. What is its nullspace? Iterate your example until the rangespace and nullspace stabilize.
(Of course, many examples are correct, but here is one.) An example is the shift operator on triples of reals . The nullspace is all triples that start with two zeros. The map stabilizes after three iterations.
Show that the rangespace and nullspace of a linear transformation need not be disjoint. Are they ever disjoint?
The differentiation operator has the same rangespace as nullspace. For an example of where they are disjoint— except for the zero vector— consider an identity map (or any nonsingular map).