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Let $V$ and $W$ be $F$-vector space, and $V^*$ and $W^*$ be the dual space of $V$ and $W$, respectly. Let $T:V \to W$ be a linear transformation. Define $T^*:W^* \to V^*$ by $T^*(f)=f\circ T$ for all $f\in W^*$. Show that $T$ is onto if and only if $T^*$ is one to one.

My attempt:

$T$ is onto $\implies$ $T^*$ is one to one: Since $T$ is onto, given $w\in W$, $\exists v\in V$, such that $T(v)=w$. Then, if $f,g\in W^*$ such that $T^*(f)=T^*(g)$, then $f\circ T\equiv g\circ T$,so $f\circ T(v)=g\circ T(v)$ $\forall v\in V$. For that given $w$ we earlier mentioned $f(w)=f(T(v))=g(T(v))=g(w)$, and since $w$ is arbitrary, we have $f\equiv g$.

But the other direction of the proof I got stuck:

$T^*$ is one to one $\implies$ $T$ is onto : Since $T$ is one to one, $\ker(T)=\{0\}$..., and I can't go further, thanks for help!

Thanks martini, I have a typo, it should be proving $T^*$ is one to one $\implies$ $T$ is onto, does it use the same way?

Bernard
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扁頭科學麵Kevin Lee
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1 Answers1

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Let $(v_i)$ be a basis of $V$, as $T$ is one-to-one $(Tv_i)$ is a basis of $T[V]$, the image of $T$. Extend it by $(w_j)$ to a basis of $W$.

Now let $f \in V^*$ be given, define $g \in W^*$ by linear extension of $$ g(Tv_i) = f(v_i), \qquad g(w_j) = 0 $$ Then $g \in W^*$ and for every $i$, we have $$ (T^*g)(v_i) = g(Tv_i) = f(v_i) $$ as $T^*g$ and $f$ agree on a basis of $V$, we have $T^*g = f$. So $T^*$ is onto.

Now we prove that $T$ is onto, if $T^*$ is one-to-one:

As $T^*$ is one-to-one, we have that $f \circ T = 0$ implies $f = 0$ for $f \in W^*$, that is any linear functional, which vanishes on $T[V]$, vanishes on the whole of $W$, but this is only possible for $T[V] = W$. For, suppose $T[V] \subsetneq W$, choose a basis $(w_i)_{i\in I}$ of $T[V]$ and extend it to a basis $(w_i)_{i\in I \cup J}$. Define $f \in W^*$ by linear extension of $$ f(w_i) = 0, i\in I, \qquad f(w_i) = 1, i \in J $$ Then $f|_{T[V]} = 0$, that is $f \circ T = 0$, but $f \ne 0$. Hence $T[V] = W$.

martini
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