Lecture 7 - Ending Duality

Last time we talked about:

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)

Using the second statement, we can use the dim(null(T)) via:

dim(null(T))=dim(V)−dim(range(T))

thus:

dim(null(T′))=dim(W)−dim(range(T))$$thus:

\text{dim}(\text{range}(T)) = \text{dim}(W) - \text{dim}(\text{null}(T'))

So then we can say that $T'$ is injective iff $T$ is surjective. As such: > [!theorem] > - $\text{dim}(\text{range}(T')) = \text{dim}(\text{range}(T))$ > - $\text{range}(T') = (\text{null}(T))^0$ We won't prove this one (it's very similar), but as a result we get that $T'$ is surjective, iff $T$ is injective. # Back to Chapter 6 (6.A-6.B) Review I feel confident in these parts, so I just listened in lecture. Refer to Chapter 6 - Inner Product Spaces for a deep dive, or start on Lecture 27 - Starting Inner Product Spaces up to Lecture 30 - Practice with Gram-Schmidt. Note that if $v \in V$ is fixed, then:

\braket{\cdot, v}

spansalllinearfunctionals$ϕ∈L(V,F)$.