Linear algebra

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ติดครับ ใครช่วยแนะนำหน่อย

Definition 1 Let $V$ be a vector space over the field $F$. The set $$V^* = \mathcal{L}(V,F)= \{T : V \rightarrow E| T \ \mbox{is a linear trasnformation}\}$$ is called the dual space of $V$.
Definition 2 Let $V$ be a vector space over a field $F$. For any subset $S$ of $V$, define $$S^o = \{f \in V^*| f(x) = 0 \ \forall x \in S\}.$$

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Let $V$ be a finite-dimensional vector space and $U, W$ is subspace of $V$.
1) If $V = U \oplus W$, then $V^* = U^o \oplus W^o.$

Definition 3 Let $T : V \rightarrow W$ be a linear map. Define $T^t : W^* \rightarrow V^*$ by $$T^t(f) = f \circ T$$ for any $f \in W^*.$
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Show that $T$ is $1-1 $ iff $T^t$ is onto and $T$ is onto iff $T^t$ is $1-1$. Furthermore, if $T : V \rightarrow W$ is a linear map where $V$ is finite-dimensional, rank $T$ = rank $T^t$.

[MINGA]

1) Let $V=U\oplus W.$ Since both $U^\circ$ and $W^\circ$ are subspaces of $V$ and $U^\circ \cap W^\circ = \{0\},$ we then get $U^\circ\oplus W^\circ \subseteq V^*.$ We only need to show $V^* \subseteq U^\circ\oplus W^\circ.$

Let $f\in V^*.$ Write $f = f_W + f_U$ where
\[ f_W(v) :=f(v_W),\quad f_U(v) := f(v_U) \]
for $v=v_U+v_W \in U\oplus W.$ So $f = f_W+f_U \in U^\circ\oplus W^\circ.$

2) $T$ is 1-1 iff $T^t$ is onto.
($\Rightarrow$) Let $T:V\rightarrow W$ be 1-1. So the inverse $T^{-1}:T(V)\rightarrow V$ exists. We want $T^t$ to be onto, so we need to show $\forall g\in V^*~ \exists f\in W^*,~f\circ T = g.$

Since $T(V)$ is a subspace of a finite dimensional vector space $W,$ we can write $W=T(V)\oplus W'.$
Let $g\in V^*.$ Define $f\in W^*$ by
\[ f(w) = g(T^{-1}(w_{T(V)})) \]
for any $w= w_{T(V)}+w' \in T(V)\oplus W'.$ Then $f\circ T = g.$

$(\Leftarrow)$ Assume $T^t$ is onto.
Let $\{v^*_i\}$ be a basis of $V^*$. Since $T^t$ is onto, there is $w^*_i\in W^*$ s.t. $w^*_i \circ T = v^*_i$ for all $i.$

Let $v\in V$ be such that $T(v)=0.$ Then
\[ v^*_i(v) = w^*_i(T(v)) =0 \]
for all $i.$ This means $v=0,$ and $T$ is 1-1.

3) $T$ is onto iff $T^t$ is 1-1.
$(\Rightarrow)$ Assume $T$ to be onto. Want to show that for any $f\in W^*,$
\[\text{if } f\circ T = 0, \text{ then } f=0 . \]

Let $f\in W^*$ be such that $ f\circ T = 0.$ Let $w\in W.$ Since $T$ is onto, there is $v\in V$ s.t. $T(v) =w.$

So, $f(w) = f(T(v)) =0.$ Hence $f=0.$

$(\Leftarrow)$ Suppose that $T$ is not onto. Write $W=T(V)\oplus W'.$ Let $\{w^*_i\}$ be a basis of $W'.$ Then $w_i^*\circ T = 0.$ So $T^t$ is not 1-1.

4) $\text{rank } T = \text{rank } T^t$
\begin{align}
\text{rank } T^t &= \dim W^*-\ker T^t \\
&= \dim W - \dim \{ f\in W^* \mid f\circ T=0 \} \\
&= \dim W - \dim \{ f=f_{T(V)}+f_{W'} \in (W')^\circ \oplus T(V)^\circ \mid f_{T(V)}=0 \} \\
&= \dim W - \dim T(V)^\circ \\
&= \dim W - ( \dim W - \dim T(V) ) \\
&= \text{rank }T
\end{align}