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

[zenterix]

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 $T$.

 

[WWGD]

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.