# Prove that range ##T'## = ##(\text{null}\ T)^0##

• zenterix
zenterix
Homework Statement
Prove the following theorem (Axler, Linear Algebra Done Right, Theorem 3.109)

Suppose ##V## and ##W## are finite-dimensional and ##T\in L(V,W)##. Then

(a) ##\text{dim range}\ T' = \text{dim range}\ T##

(b) ##\text{range}\ T'=(\text{null}\ T)^0##
Relevant Equations
(a)

##\text{dim null}\ T' = \text{dim null}\ T + \text{dim}\ W-\text{dim}\ V\tag{1}##

##=\text{dim} W'-\text{dim range}\ T'\tag{2}##

##=\text{dim}\ W-\text{dim range}\ T'\tag{3}##

From (1) and (3) we have

##\text{dim range}\ T'=\text{dim}\ V-\text{dim null}\ T=\text{dim range}\ T\tag{4}##
My question is about item (b).

(b)

Here is what I drew up to try to visualize the result to be proved

The general idea, I think, is that

1) ##(\text{null}\ T)^0## and ##\text{range}\ T'## are both subspaces of ##V'=L(V,\mathbb{F})##.

2) We can show that they have the same dimension.

3)We can show that ##\text{range}\ T' \subseteq (\text{null}\ T)^0##

4) From (2) and (3) we can prove that the two subspaces are in fact the same subspace.

Honestly, I did steps (1)-(3) myself and was looking for a way to infer (4) but didn't realize that I could use (2) and (3) to do so, so I looked at the proof in the book.

I'd like to know if there is another way to infer (4).

Here is what I have

##(\text{null}\ T)^0## is by definition all the linear functionals in ##V'## that map ##\text{null}\ T## to 0 in ##\mathbb{F}##.

For every ##\varphi\in W'##, the linear functional ##\varphi\circ T## maps ##\text{null}\ T## to 0 in ##\mathbb{F}##, so ##\varphi\circ T## is in ##(\text{null}\ T)^0##.

Now, at this point, it could be that there are other elements in ##(\text{null}\ T)^0## that are not one of the ##\varphi\circ T##.

How can I prove that this is not possible (in an alternative manner to (4))?

Last edited:
I think I found the answer to my question. It uses the dimensions as well.

Let's use a proof by contradiction.

Suppose there were an ##\alpha\in (\text{null }\ T)^0## that is not in ##\text{range}\ T'##.

From (a) we know that range ##T'## and range ##T## have the same dimension and so are isomorphic.

In addition, we have dim ##V## = dim null ##T## + dim range ##T##.

However, the dimension of the complement of null ##T## in ##V## also has dimension dim range ##T##.

Thus this complement is isomorphic with range ##T##.

Let ##v_1,...,v_n## be a basis for this complement subspace. ##\alpha\in(\text{null}\ T)^0## maps these to the scalars ##f_1, ..., f_n## in ##\mathbb{F}##.

##T## maps ##v_1,...,v_n## to ##w_1, ...,w_n## in range ##T##.

Let's define a ##\varphi\in W'## by ##\varphi(w_i)=f_i## for ##i=1,...,n##.

Then, ##\varphi\circ T## maps ##v_1,...,v_n## to ##f_1,...,f_n## and so ##\varphi\circ T=\alpha##.

Thus, ##\alpha\in \text{range}\ T'##, contradicting our initial assumption.

We can then infer, by proof by contradiction, that all vectors in ##(\text{null }\ T)^0## are in range ##T'##.

And since we have proved that range ##T'\subseteq (\text{null}\ T)^0## we can conclude that

$$\text{range}\ T'=(\text{null}\ T)^0$$

## What does "range T'" refer to in linear algebra?

In linear algebra, "range T'" refers to the range (or image) of the transpose of a linear transformation T. If T is a linear transformation from vector space V to vector space W, then T' (or T^T) is a transformation from the dual space W' to the dual space V'. The range T' is the set of all possible outputs of T'.

## What is the null space of T?

The null space (or kernel) of a linear transformation T is the set of all vectors in the domain that map to the zero vector in the codomain. Formally, if T: V → W, then null T = {v ∈ V | T(v) = 0}. It represents the solutions to the homogeneous equation T(v) = 0.

## What does the notation ##(\text{null}\ T)^0## mean?

The notation ##(\text{null}\ T)^0## typically denotes the orthogonal complement of the null space of T. In other words, it is the set of all vectors in the codomain that are orthogonal to every vector in the null space of T.

## How do you prove that range T' = ##(\text{null}\ T)^0##?

To prove that range T' = ##(\text{null}\ T)^0##, you need to show two inclusions: (1) that every vector in the range of T' is orthogonal to every vector in the null space of T, and (2) that every vector orthogonal to the null space of T is in the range of T'. This involves understanding the relationships between T, T', and their respective null spaces and ranges.

## Why is the relationship range T' = ##(\text{null}\ T)^0## important?

This relationship is important because it provides a deep connection between the properties of a linear transformation and its transpose. It helps in understanding the structure of vector spaces and dual spaces, and it is fundamental in various areas of mathematics, including functional analysis and the theory of differential equations.

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