Lecture 5 - Continuing Duality

Recall the idea of a linear functional:

Definition

The vector space $L (V, F)$ is called the dual space of $V$ . It is denoted:
$V^{'}$
then $V$ and $V^{'}$ are isomorphic, assuming $V$ is finite-dimensional.

Note that when $V$ is finite dimensional:

dim (V^{'}) = dim (V)

so then $V$ and $V^{'}$ are isomorphic. However, if it's infinite-dimensional, then it's possible where no isomorphism exists.

The idea of finding these isomorpisms was to find the basis vectors and map from one basis to the other. As such, we'll define the dual basis:

dual basis

Given a basis $β = {v_{1}, . . ., v_{n}}$ of $V$ , the dual basis of $β$ is $ϕ_{1}, . . ., ϕ_{n} \in V^{'}$ where:

ϕ_{j} (v_{k}) = {\begin{cases} 1 & k = j \\ 0 & k \neq j \end{cases} = χ (k = j)

for all $j$ . Here $χ$ returns 1 if the statement is true, and 0 when false.

Here the $ϕ_{j}$ is a valid linear functional, defining it by its action on a basis. We need to prove that it's actually a basis for $V^{'}$ below.

Proof
We already have $n$ $ϕ$ 's, which is the dimension of $V$ . As such, we'll just show LI.

Consider when:

a_{1} ϕ_{1} + \dots + a_{n} ϕ_{n} = \vec{0}

where $0$ is the linear functional in this case. Consider some arbitrary basis vector $v_{i} \in V$ . Then:

\begin{array}{r} (a_{1} ϕ_{1} + \dots + a_{n} ϕ_{n}) (v_{i}) = 0 = a_{1} ϕ_{1} v_{i} + \dots + a_{n} ϕ_{n} v_{i} \end{array}

But then this implies that all $a_{i} = 0$ since the only basis vector $v_{i}$ that allows something non-zero is the $i$ -th one. As such, then $a_{i} = 0$ for all $i$ , so then the set of vectors is LI.
☐

Basis gives

ϕ

Note that this whole process depended on the basis $v_{1}, . . ., v_{n}$ that we started with. If you change the basis, you're $ϕ$ 's will change as a result.

Dual Maps of $T$

As the picture shows, note that we can build a linear map from $V$ to $F$ by doing $T$ then $ϕ$ (which combined gives $ϕ \circ T$ . As such, we construct the dual map:

dual map

If $T \in L (V, W)$ then the dual map of $T$ is $T^{'} \in L (W^{'}, V^{'})$ defined by:

T^{'} (ϕ) = ϕ \circ T

for all $ϕ \in W^{'}$ .

As a fact, this map $T^{'}$ is linear.

As an example, define $T \in F^{n, n}$ by:

T ([\begin{matrix} 1 & 1 \\ a_{1} & \dots & a_{n} \\ 1 & 1 \end{matrix}]) = [\begin{matrix} 1 & 1 \\ a_{n} & \dots & a_{1} \\ 1 & 1 \end{matrix}]

Here it maps from $F^{n, n}$ to itself. Let $tr$ be the trace functional on $F^{n, n}$ . What is $T^{'} (tr)$ ? We can figure this out via:

T^{'} (tr) = tr \circ T

So what does this $tr \circ T$ do to a matrix like we have above. It's:

(tr \circ T) ([\begin{matrix} a_{11} & a_{1 n} \\ ⋮ & ⋮ \\ a_{n 1} & a_{n n} \end{matrix}]) = tr ([\begin{matrix} a_{1 n} & a_{11} \\ ⋮ & ⋮ \\ a_{n n} & a_{n 1} \end{matrix}]) = \sum_{i = 1}^{n} a_{i, n + 1 - i}

These are pretty funky, so let's do another example. Let $V, W$ be vector spaces. Let $e_{1}, e_{2}$ be a basis for $V$ and $f_{1}, f_{2}, f_{3}$ be a basis for $W$ . Suppose $T \in L (V, W)$ where $M (T, β_{V}, β_{W})$ is a 3x2 matrix:

[\begin{matrix} a & b \\ c & d \\ e & f \end{matrix}]

What about the dual map $T^{'}$ ? We know that we have $ϕ_{1}, ϕ_{2}$ as the dual basis for $β_{V}$ and $ψ_{1}, ψ_{2}, ψ_{3}$ is the dual basis for $W$ . Here $T^{'} \in L (W^{'}, V^{'})$ . Hence, the matrix must be instead 2x3 (hint: look at the dimensions for $W^{'}, V^{'}$ ).

The columns of the matrix represent to the image $T (e_{i})$ in this case for what parts of the basis compose the basis. So:

T (e_{1}) = a f_{1} + c f_{2} + e f_{3}, T (e_{2}) = b f_{1} + d f_{2} + f f_{3}

What is $M (T^{'}, β_{W}^{'}, β_{V}^{'})$ (the $ψ$ 's go first here)? Just look at the image of each basis vector:

T^{'} (ψ_{1}) = ψ_{1} \circ T

So for example:

T^{'} (ψ_{1}) (e_{1}) = (ψ_{1}) (T e_{1}) = (ψ_{1}) (a f_{1} + c f_{2} + e f_{3}) = a

T^{'} (ψ_{1}) (e_{2}) = (ψ_{1}) (T e_{2}) = (ψ_{1}) (b f_{1} + d f_{2} + f f_{3}) = b

Because the $ψ_{1}$ just gets the constant amount of $f_{1}$ as defined. So the first column of the matrix is:

[\begin{matrix} a & c & e \\ b & d & f \end{matrix}]

WHICH IS THE TRANSPOSE OMG!!!.

Dual Maps of T

Dual Maps of $T$