Linear Algebra/Self-Composition/Solutions

Solutions edit

Problem 1

Give the chains of rangespaces and nullspaces for the zero and identity transformations.

Answer

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  .

Problem 2

For each map, give the chain of rangespaces and the chain of nullspaces, and the generalized rangespace and the generalized nullspace.

  1.  ,  
  2.  ,
     
  3.  ,  
  4.  ,
     
Answer
  1. 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  .
  2. 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.
  3. Iterates of this map cycle around
     
    and the chains of rangespaces and nullspaces are trivial.
     
    Thus, obviously, generalized spaces are   and  .
  4. We have
     
    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.
Problem 3

Prove that function composition is associative   and so we can write   without specifying a grouping.

Answer

Each maps  .

Problem 4

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 of Lemma 1.3.)

Answer

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  .

Problem 5

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?

Answer

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).

Problem 6

Verify the nullspace half of Lemma 1.3.

Answer

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.

Problem 7

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.

Answer

(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.

Problem 8

Show that the rangespace and nullspace of a linear transformation need not be disjoint. Are they ever disjoint?

Answer

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).