Lecture 6 - Matrices of Dual Maps

Recall from last time that when we have $v_{1}, . . ., v_{n}$ is a basis for $V$ , then the dual basis of $ϕ_{1}, . . ., ϕ_{m}$ is a dual basis for $V^{'}$ :

Suppose $ψ \in V^{'}$ . We know that there exist $α_{1}, . . ., α_{n}$ such that:

ψ = α_{1} ϕ_{1} + \dots + α_{n} ϕ_{n}

Can we somehow determine the $α$ 's? Both $ψ, ϕ_{i}$ are all functionals, so we can plug things into both sides, but the $ϕ_{i}$ spits out a 1 or 0 based on the basis vectors plug in. Doing so for each:

ψ (v_{i}) = α_{i} ϕ_{i} (v_{i}) = α_{i}

So yes we can determine the $α$ 's! And in fact then:

ψ = ψ (v_{1}) ϕ_{1} + \dots + ψ (v_{n}) ϕ_{n}

We used this when talking about the map yesterday via Lecture 5 - Continuing Duality#Dual Maps of $T$ . We used that:

T^{'} (ψ_{1}) = ψ_{1} \circ T

Where we used:

ψ_{1} \circ T (e_{1}) = a, ψ_{1} \circ T (e_{2}) = b, . . . ψ \circ T = a ϕ_{1} + b ϕ_{2}

Thus giving one of the columns of the matrix.

Other Properties of Dual Maps

We won't prove everything. A lot are in the book and are just symbol pushing.

Properties of Dual Maps

$(S + T)^{'} = S^{'} + T^{'}$
$(λ T)^{'} = λ T^{'}$
$(S T)^{'} = T^{'} S^{'}$

Proof
It's easy to show (1,2) since the dual map operation is linear. (3) is via Chapter 3 (cont.) - Products and Quotients of Vector Spaces#^c8c22b.
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annihilator

Given a subset $U$ of a vector space $V$ , the annihilator of $U$ is:

U^{0} = {φ \in V^{'} : φ (u) = 0 \forall u \in U}

You can think of $U^{0}$ are all items in $U$ that get "annihilated" down to $0$ , hence the name.

About annihilators

$U^{0}$ is a subspace of $V^{'}$ .

This shouldn't be too crazy. Clearly the zero vector is in the annihilator, and any two vectors that annihilate should still annihilate when added. Similar with scalar multiplication.

Proposition

When $U$ is a subspace:

dim (U) + dim (U^{0}) = dim (V)

Proof
Consider this outline:

We want to figure out what are: 1. $\text{null}(i')$ 2. $\text{range}(i')$

Here:

i^{'} : V^{'} \to U^{'}

we want to make sure that $ϕ$ maps everything in $U$ to 0. This is the annihilator! Hence:

null (i^{'}) = U^{0}

In a similar way:

range (i^{'}) = U^{'}

because if you want a functional from $U \to F$ then just do $ϕ$ and ignore all the other $v$ 's that aren't in $U$ to anything you want (to $π, 0, e, . . .$ ).

Using the FTOLM:

\begin{aligned} dim (V^{'}) & = dim (null (i^{'})) + dim (range) (i^{'})) \\ dim (V) & = dim (U^{0}) + dim (U^{'}) \\ = dim (U^{0}) + dim (U) \end{aligned}

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Some nice facts:

All about

null (T^{'})

Given $V, W$ are finite dimensional. Then:

$null (T^{'}) = range (T)^{0}$
$dim (null (T^{'})) = dim (null (T)) + dim (W) - dim (V)$

Proof
For (a), we'll do an iff. Say $ϕ \in null (T^{'})$ so then $T^{'} (ϕ) = \vec{0}$ is the zero functional. So then:

ϕ \circ T = \vec{0} \Leftrightarrow ϕ (T (v)) = 0 \forall v \in V

so then $ϕ \in range (T)^{0}$ as expected. This works both ways.

For (b), what is the $dim (null (T^{'}))$ ? Using (a), then it's the dimension of the annihilator of the range. Thus using the dimension of the annihilator:

\begin{aligned} dim (range (T)^{0}) & = dim (W) - dim (range (T)) \\ = dim (W) - (dim (V) - dim (null (T)) \\ = dim (W) + dim (null (T)) - dim (V) \end{aligned}

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Notice that if we take the fact that:

\begin{aligned} dim (null (T^{'})) & = dim (W) - dim (range (T)) \\ dim (range (T)) & = dim (W) - dim (null (T^{'})) \end{aligned}

Notice that $T^{'}$ is injective iff $T$ is surjective, because we can use the dimensions here to show that one is 0 or not.