Excerpt from text:
3.109 The range of T'
Suppose V and W are finite-dimensional and T $\in$ L(V,W). Then
range T' = $(null\;T)^0$
First suppose $\phi$ $\in$ range T. Thus there exists $\psi$ $\in$ W' such that $\phi$ = T'($\psi$). If v $\in$ null T, then
$\phi$(v) = (T'($\psi$))v = ($\psi$ $\circ$ T)(v) = $\psi$(Tv) = $\psi$(0) = 0
Hence $\phi$ $\in$ $(null\;T)^0$. This implies that range T' $\subset$ $(null\;T)^0$.
We will complete the proof by showing that $\mathbf range\;T'$ and $(null\;T)^0$ have the same dimension. To do this, note that
dim range T' = dim range T
dim range T' = dim V - dim null T
dim range T' = dim $(null\;T)^0$
Questions: I am not able to understand the intuition behind this theorem. $\mathbf range\; T'$ is the set of functionals from $\mathbf W'$ which take any $\mathbf w$ from $\mathbf W$ to $\mathbb F$ ($\mathbf w$ is produced by $\mathbf T$ here; T'($\phi$) = $\phi \circ T$ for $\phi$ $\in$ W'. $\mathbf Tv$ gives a $\mathbf w$ which is then given to $\phi$() ).
On the other hand, annihilator of $\mathbf (null\;T)$ is the set of all functionals from $\mathbf V'$ which take any vector from the set $\mathbf (null\;T)$ to 0. How can these two different sets be equal? Would'nt it mean that all functionals from T' are functionals that produce zero?
I am confused about the subset notation in the first part of the proof; " range T' $\subset$ $(null\;T)^0$ " . My current understanding says all functionals in V' which would take vectors from (null T) to zero will be part of range T' . That is, $(null\;T)^0$ $\subset$ range T' . What am I missing here?
(Note: I am assuming $\mathbf (null\;T)$ $\subset$ $\mathbf V$, since annihilator is defined with a subset/superset relation. Have I taken the right superset? )
Also, in the second part of the proof, the author just proves the dimensions of range T' and $(null\;T)^0$ are same. But, how can proving the dimensions being equal prove both spaces are equal and have same elements?
Could anyone kindly explain what I have misunderstood here?