Null space of dual map of T = annihilator of range T

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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##.
 
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You may want to break this into chunks. Given your vector spaces are finite -dimensional, you can represent a linear map , once you've chosen a basis, as a matrix M. Then the dual map is given by its transpose ##M^{T}##. Find the kernel and see if the result is independent of the choice of basis.
 

FAQ: Null space of dual map of T = annihilator of range T

What is the null space of a linear map?

The null space of a linear map \( T: V \to W \) is the set of all vectors in \( V \) that are mapped to the zero vector in \( W \). Formally, it is defined as \( \text{null}(T) = \{ v \in V \mid T(v) = 0 \} \).

What is the dual map of a linear transformation?

The dual map \( T^*: W^* \to V^* \) of a linear transformation \( T: V \to W \) is defined by \( T^*(\phi) = \phi \circ T \) for every linear functional \( \phi \in W^* \). Here, \( W^* \) and \( V^* \) are the dual spaces of \( W \) and \( V \), respectively.

What is the annihilator of a subspace?

The annihilator of a subspace \( U \subseteq V \) is the set of all linear functionals in \( V^* \) that map every vector in \( U \) to zero. It is denoted as \( U^\perp \) and defined as \( U^\perp = \{ \phi \in V^* \mid \phi(u) = 0 \text{ for all } u \in U \} \).

How are the null space of the dual map and the annihilator of the range related?

The null space of the dual map \( T^* \) of a linear transformation \( T \) is the annihilator of the range of \( T \). This means \( \text{null}(T^*) = (\text{range}(T))^\perp \). This relationship arises because a functional in \( W^* \) is in the null space of \( T^* \) if and only if it annihilates every vector in the range of \( T \).

Can you provide an example illustrating this relationship?

Consider a linear map \( T: \mathbb{R}^3 \to \mathbb{R}^2 \) defined by \( T(x, y, z) = (x + y, y + z) \). The range of \( T \) is spanned by the vectors \( (1, 1) \) and \( (0, 1) \) in \( \mathbb{R}^2 \). The annihilator of this range in \( (\mathbb{R}^2)^* \) consists of all functionals that map both \( (1, 1) \) and \( (0, 1)

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