- #1
zenterix
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- Homework Statement
- Theorem 3.107 (Linear Algebra Done Right)
Suppose ##V## and ##W## are finite-dimensional and ##T\in L(V,W)##. Then
(a) null T' = (range ##T)^0##
- Relevant Equations
- I would like to understand item (a) in an intuitive, descriptive way.
There are multiple concepts involved and all are quite abstract. I would like to go over them in my own words before speaking of the theorem in question.
In what follows I do so and end with my understanding of the presented theorem.
Consider the concepts of dual space, dual basis, dual map, and annihilator.
Given a linear map ##T\in L(V,W)##, the dual space of ##T## is the vector space ##V'=L(V,\mathbb{F})## where ##\mathbb{F}## is a field.
Note that given any basis ##v_1, ..., v_n## of ##V##, each distinct linear functional in ##V'## maps the basis vectors ##v_i## in a unique way to scalars in ##\mathbb{F}##.
Consider the ##n## linear functionals defined as
$$\phi_j(v_i)=\begin{cases} 1, \text{ if } i=j \\ 0 \text{ otherwise } \end{cases}$$
for ##j=1,...,n##.
It isn't difficult to see that each linear functional in ##V'## is a linear combination of these ##n## ##\phi_j## linear functionals.
In addition, it can be shown that the ##\phi_j##'s are linearly independent and so are a basis for ##V'##.
We call them the dual basis of ##v_1, ..., v_n## and they are a basis of ##V'##.
Next let's consider what a dual map is. I find this concept to be the most difficult to grasp so far in linear algebra.
Given a linear map ##T\in L(V,W)##, the dual map of ##T## is a specific linear map in ##L(W',V')##, namely ##T'\in L(W',V')## defined by
$$T'(\phi)=\phi\circ T, \text{ for } \phi \in W'\tag{1}$$
Let me try to understand this.
The vectors in the vector space ##L(W', V')## are linear maps that map a linear functional in ##L(W,\mathbb{F})## to a linear functional in ##L(V,\mathbb{F})##.
One specific of these linear maps in ##L(W',V')## is the dual map of ##T\in L(V,W)##.
It is the linear map that maps a ##\phi\in L(W,\mathbb{F})## to ##\phi\circ T##.
##\phi\circ T## is a linear map that maps ##V## to ##W## and then ##W## to ##\mathbb{F}## and thus is in ##V'##.
Next, consider the concept of annihilator.
Given a vector space ##V## and a subset ##U\subset V##, the annihilator of ##U## is the set of all vectors in ##V'## (ie, linear functionals in ##L(V,\mathbb{F})##) that map ##U## (ie every vector in ##U##) to 0.
The annihilator of ##U## is denoted ##U^0## and is a subspace of ##V'##.
Now we get to an extra layer of abstraction.
Consider the null space of the dual map ##T'##.
Since ##T'\in L(W',V')## then its null space is a subspace of ##W'##.
There is a theorem that says that this null space is the same as the annihilator of the range of ##T##.
##T\in L(V,W)## so the range of ##T## is a subspace of ##W##.
The annihilator of range ##T## is then a subspace of ##W'## (the linear functionals in ##W'## that map range ##T## to 0).
The claim in bold above is that the linear functionals mapped by ##T'## to 0 (that is, the linear functionals ##\phi## such that ##\phi\circ T=0##) are the exact same linear functionals that annihilate the range of ##T##.
Now that I wrote it and thought about it more it finally made clicked and made sense:
Since the dual map is formed by functionals that take a ##v\in V## pass it through ##T## to get a ##w\in \text{range} T## and then pass that through a functional ##\phi## in ##W'##, this can only be zero always if the functional ##\phi## is in the annihilator subspace. Thus, any such ##\phi## in the null space of ##T'## is in the annihilator of range ##T##.
Given a linear map ##T\in L(V,W)##, the dual space of ##T## is the vector space ##V'=L(V,\mathbb{F})## where ##\mathbb{F}## is a field.
Note that given any basis ##v_1, ..., v_n## of ##V##, each distinct linear functional in ##V'## maps the basis vectors ##v_i## in a unique way to scalars in ##\mathbb{F}##.
Consider the ##n## linear functionals defined as
$$\phi_j(v_i)=\begin{cases} 1, \text{ if } i=j \\ 0 \text{ otherwise } \end{cases}$$
for ##j=1,...,n##.
It isn't difficult to see that each linear functional in ##V'## is a linear combination of these ##n## ##\phi_j## linear functionals.
In addition, it can be shown that the ##\phi_j##'s are linearly independent and so are a basis for ##V'##.
We call them the dual basis of ##v_1, ..., v_n## and they are a basis of ##V'##.
Next let's consider what a dual map is. I find this concept to be the most difficult to grasp so far in linear algebra.
Given a linear map ##T\in L(V,W)##, the dual map of ##T## is a specific linear map in ##L(W',V')##, namely ##T'\in L(W',V')## defined by
$$T'(\phi)=\phi\circ T, \text{ for } \phi \in W'\tag{1}$$
Let me try to understand this.
The vectors in the vector space ##L(W', V')## are linear maps that map a linear functional in ##L(W,\mathbb{F})## to a linear functional in ##L(V,\mathbb{F})##.
One specific of these linear maps in ##L(W',V')## is the dual map of ##T\in L(V,W)##.
It is the linear map that maps a ##\phi\in L(W,\mathbb{F})## to ##\phi\circ T##.
##\phi\circ T## is a linear map that maps ##V## to ##W## and then ##W## to ##\mathbb{F}## and thus is in ##V'##.
Next, consider the concept of annihilator.
Given a vector space ##V## and a subset ##U\subset V##, the annihilator of ##U## is the set of all vectors in ##V'## (ie, linear functionals in ##L(V,\mathbb{F})##) that map ##U## (ie every vector in ##U##) to 0.
The annihilator of ##U## is denoted ##U^0## and is a subspace of ##V'##.
Now we get to an extra layer of abstraction.
Consider the null space of the dual map ##T'##.
Since ##T'\in L(W',V')## then its null space is a subspace of ##W'##.
There is a theorem that says that this null space is the same as the annihilator of the range of ##T##.
##T\in L(V,W)## so the range of ##T## is a subspace of ##W##.
The annihilator of range ##T## is then a subspace of ##W'## (the linear functionals in ##W'## that map range ##T## to 0).
The claim in bold above is that the linear functionals mapped by ##T'## to 0 (that is, the linear functionals ##\phi## such that ##\phi\circ T=0##) are the exact same linear functionals that annihilate the range of ##T##.
Now that I wrote it and thought about it more it finally made clicked and made sense:
Since the dual map is formed by functionals that take a ##v\in V## pass it through ##T## to get a ##w\in \text{range} T## and then pass that through a functional ##\phi## in ##W'##, this can only be zero always if the functional ##\phi## is in the annihilator subspace. Thus, any such ##\phi## in the null space of ##T'## is in the annihilator of range ##T##.
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