(Short deadline) Linear Algebra
Let $V$ and $W$ be vector spaces over a field $F$, $T : V ? W$ a linear map, $X$ a subspace of $V$ and $Y$ a subspace of $W$.
Show that:
a) $(T^{1}(Y))^0=T^*(Y^0)$
b)$T(X) ? Y$ if and only if $T^?(Y^0) ? X^0$

What do you mean by Y^0 in this context?

Is it the dual of Y?

Hey. Annihilators.

T^* meaning adjoint.
Answer
a) Let $f\in T^{*}(Y^0)$. Then $f=T^*(g)$ where $g\in Y^0$. Now let $v\in T^{1}(Y)$. Then $Tv \in Y$. So $$g(Tv)=0 \implies f(v)=(T^*(g))(v)=g(Tv)=0$$
Therefore $f\in (T^{1}(Y))^0$. So $T^{*}(Y^0) \subset (T^{1}(Y))^0$.
Now let $f\in (T^{1}(Y))^0$. Note that $\ker T =T^{1}(0)\subset T^{1}(Y)$. So $f(\ker T)=\{0\}$. Now for any $w\in T(V)$ there is $v\in V$ so that $T^{1}(w)=v+\ker T$. So $$ f(T^{1}(w))=\{f(v)\}$$ Let us define $g\in W^*$ by $g(w):=f(u)$ where $u$ is some arbitrary element of $ T^{1}(Pw)$ and $P$ is some fixed projection onto $T(V)$. Note that by the above relation $g$ is welldefined. It is easy to see that $g$ is linear because $f,P$ are linear and the set $T^{1}(w)$ depends linearly on $w$. And for any $v\in V$ we have $$ g(Tv)=f(v) $$ since $v\in T^{1}(Tv)=T^{1}(PTv)$. So $f=T^*(g)$ and thus $T^{*}(Y^0) \supset (T^{1}(Y))^0$.

The following proof of the above fact also works when both $V,W$ are finite dimensional:
Now we know that $ T_{T^{1}(Y)} :T^{1}(Y)\to Y$ is onetoone and onto, so by ranknullity theorem we have $$ \dim Y + \dim \ker T = \mathrm{rank} T_{T^{1}(Y)} + \dim \ker T = \dim T^{1}(Y)$$ Note that $\ker T \subset T^{1}(Y)$ so $\ker T_{T^{1}(Y)} = \ker T$. So
$$ \dim (T^{1}(Y))^0 =\dim V  \dim T^{1}(Y)=\dim V  (\dim Y + \dim \ker T) \\ \;\\ = \dim V  (\dim Y + \dim V \mathrm{rank} T)=\mathrm{rank} T\dim Y =\mathrm{rank} T^* \dim Y\\\;\\ =\mathrm{rank} T^* (\dim W \dim Y^0)=\mathrm{rank} T^* \dim W^* +\dim Y^0\\\;\\ = \dim \ker T^* +\dim Y^0= \dim T^{*}(Y^0)$$
Note that in the last two equalities we applied the ranknullity theorem to the map $T^* :W^* \to V^*$ and the onto map $T^*_{Y^0} :Y^0 \to T^*(Y^0)$. We also used the fact that the dimension of a subspace plus the dimension of its annihilator is equal to the dimension of the space, and the rank of a map and its adjoint are the same.
So we must have $T^{*}(Y^0) = (T^{1}(Y))^0$ since the smaller subspace has the same dimension as the larger one.
b) Suppose $T(X)\subset Y$. Then $X\subset T^{1}(Y)$. So $X^0 \supset (T^{1}(Y))^0=T^{*}(Y^0)$. Conversely suppose $(T^{1}(Y))^0=T^{*}(Y^0)\subset X^0$. Then $$T^{1}(Y)\supset X\implies Y\supset T(X)$$ as desired.
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