Lecture 4 - Duality (intro)

Recall that:

\tilde{T}

Suppose $T \in L (V, W)$ . Define $\tilde{T} : V / (null (T)) \to W$ by:
$\tilde{T} (v + null (T)) = T v$

What we're doing here is if you collect all things that were in $null (T)$ , then that makes $\tilde{T}$ is injective. Thus, it's an isomorphism between $V / null (T)$ and $range (T)$ .

Really the injectivity comes from the fact that the only thing that get's mapped to ${\vec{0}}_{W}$ is anything in $\vec{0} + null (T) = null (T)$ . Like how a function cannot have multiple possible outputs, here we can only have multiple possible different inputs to the same output in $W$ .

Consider the Following

Suppose now that I have $U$ which is a subspace of $V$ . Let's suppose I have $T \in L (V)$ . The question is:

Question

When is the map $S : V / U \to V / U$ defined by:

S (\vec{v} + U) = T (\vec{v}) + U

well defined?

Consider when it's a problem. It's problem if we get two different outputs for the same input. Given $v_{1} + U = v_{2} + U$ , we need there images under $S$ to be equal. Checking this, notice that we must have $v_{1} - v_{2} \in U$ :

S (v_{1} + U) = S (v_{2} + U) \Rightarrow T (v_{1}) + U = T (v_{2}) + U \Rightarrow T (v_{1}) - T (v_{2}) \in U

So then $T (v_{1} - v_{2}) \in U$ . This means we need $U$ to be a $T$ -invariant subspace. This is equivalent to $null (T) \subseteq U$ (this is actually a stronger condition for the $T$ -invariance).

It's also an if and only if, interestingly enough.

Recall from last quarter that there was the following theorem:

Over

C

, every operator has an upper-triangular matrix

Suppose $V$ is a finite-dimensional complex vector space and $T \in L (V)$ . Then $T$ has an upper-triangular matrix with respect to some basis of $V$ .

We had a proof in the book that was a bit weird, but let's do a new one.

We use the fact that every operator $T$ on a finite-dimensional complex vector space has at least one eigenvalue. The idea was to take $T : V \to V$ , which has $λ_{1}$ as an eigenvalue, has an associated eigenvector $v_{1}$ . We start making the list ${v_{1}}$ . We consider $T v_{1} = λ_{1} v_{1}$ so then $T$ is invariant over $span (v_{1})$ .

As such, $S$ is well-defined, so then create $T_{1} : V / span (v_{1}) \to V / span (v_{1})$ , where $T_{1} (v + span (v_{1})) = T v + span (v_{1})$ . Notice the dimension of $V / span (v_{1})$ is $n - 1$ dimensional space.

But now this map $T_{1}$ itself has an eigenvalue since it's a finite-dimensional complex vector space and $T_{1} \in L (V / span (v_{1}))$ . It has an eigenvalue $λ_{2}$ with eigenvalue $v_{2} + span (v_{1})$ . Note that:

T_{1} (v_{1} + span (v_{1})) = λ_{2} v_{2} + span (v_{1})

by definition. As such, $v_{2}$ is not in the span of $v_{2}$ because if it were then the $λ_{2} v_{2} = \vec{0}$ but $v_{2}$ cannot zero since it's an eigenvector. As such, we have the list ${v_{1}, v_{2}}$ which is a LI list.

Now look at $T_{2} : V / span (v_{1}, v_{2}) \to V / span (v_{1}, v_{2})$ defined in the same way. We can repeat this process $n$ times.

We eventually get the list ${v_{1}, . . ., v_{n}}$ . It's a basis as the list is LI and is the right length. Note that also $T (v_{i}) \in span (v_{1}, . . ., v_{i})$ to show the UT part.

Notice that $v_{2}, . . ., v_{n}$ are not eigenvectors of $T$ . Rather, they are eigenvectors of $T_{2}, . . ., T_{n}$ respectively.

Duality

We defined

Definition

A linear functional on $V$ is an element of $L (V, F)$ .

for instance, the trace of a matrix is a functional. Another one is $ϕ : P (R) \to R$ defined by:

ϕ (p) = p (0) + p (1) + p (2)

Notice that $dim (L (V, F)) = dim (V)$ . Since these are the same, then these spaces are isomorphic

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.