403/503- Homework 2

This homework is due February 28.

1. From the textbook: Solve exercises 2.14, 3.3, 3.4, 3.9, 3.10, 3.16, 3.25.

2. a.Suppose  that T:{\mathbb R}^n\to {\mathbb R}^m satisfies linearity (i.e., what the book calls additivity). Suppose also that T is continuous. Show that T is linear (i.e., it also satisfies homogeneity).
b. Give an example of a T that is additive but not homogeneous.

3. The goal of this exercise is to state and prove the rank-nullity theorem (Theorem 3.4 from the book) without the assumption that V is finite dimensional. What we want to show is that if V,W are vector spaces and T:V\to W is linear, then

V/{\rm null}(T)\cong {\rm ran}(T).

First, we need to make sense of V/{\rm null}(T). Recall that if A is a set, an equivalence relation \sim on A is a relation \sim\subseteq A\times A such that:

  • a\sim a for any a\in A (reflexivity),
  • Whenever a\sim b, then also b\sim a (symmetry),
  • If a\sim b and b\sim c, then also a\sim c (transitivity).

Given such an equivalence relation, the equivalence class of an element a\in A is the subset {}[a]\subseteq A consisting of all those b\in A such that a\sim b. The quotient A/\sim is the collection of all equivalence classes, so if {\bf x}\in A/\sim then there is some a\in A such that {\bf x}=[a].

The point is that the equivalence classes form a partition of A into pairwise disjoint, non-empty sets: Each {}[a] is nonempty, since a\in[a]. Clearly, the union of all the classes is A (again, because any a is in the class {}[a]), and if {}[a]\cap[b]\ne\emptyset, then in fact {}[a]=[b] (check this).

Ok. Back to T:\to W. Define, in V, an equivalence relation \sim by: v_1\sim v_2 iff Tv_1=Tv_2 (Check that this is an equivalence relation). Then, as a set, we define V/{\rm null}(T) to be V/\sim. The reason why the null space is even mentioned here is because of the following (check this): v_1\sim v_2 iff v_1-v_2\in{\rm null}(T).

We want to define addition in V/{\rm null}(T) and scalar multiplication so that V/\sim is actually a vector space.

  • Given {\bf x} and {\bf y} in V/{\rm null}(T), set {\bf x}+{\bf y}={\bf z}, where if {\bf x}=[a] and {\bf y}=[b], then {\bf z}=[a+b]. The problem with this definition is that in general there may be infinitely many c such that {}[c]=[a] and infinitely many d such that {}[d]=[b]. In order for this definition to make sense, we need to prove that for any such c,d, we {}[a+b]=[c+d]. Show this.
  • Given {\bf x}\in V/{\rm null}(T), and a scalar \alpha, define \alpha{\bf x}={\bf y}, where if {\bf x}=[a], then {\bf y}=[\alpha a]. As before, we need to check that this is well-defined, i.e., that if {}[a]=[b], then also {}[\alpha a]=[\alpha b].
  • Check that V/{\rm null}(T) is indeed a vector space with the operations we just defined.

Now we want to define a linear transformation from V/{\rm null}(T) to {\rm ran}(T), and argue that it is an isomorphism. Define {\bf T}:V/{\rm null}(T)\to W by {\bf T}({\bf x})=Tv where {\bf x}=[v]. Once again, check that this is well-defined. Also, check that this is indeed linear, and a bijection.

Finally, to see that this is the “right” version of Theorem 3.4, we want to verify that {\rm dim}(V/{\rm null}(T))={\rm dim}(V)-{\rm dim}({\rm null}(T)) if V is finite dimensional. Prove this directly (i.e., without using the statement of Theorem 3.4).


2 Responses to 403/503- Homework 2

  1. Thanks, Tommy. I think it is fixed now.

  2. Tommy Chartier says:

    Hi Dr. Caicedo,

    I just want to point out a possible typo. I believe A \setminus sim is supposed to be A \setminus \sim.

    May the Math Be With You!

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