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Let V and W be nonzero vector spaces over the same field, and let $T:V \rightarrow W$ be linear map. Prove that $T^t$ is onto if and only if T is one-to-one.

Here is the theorem: Let V and W be finite-dimensional vector spaces over F with ordered bases $\beta$ and $\gamma$, respectively. For any linear map $T:V \rightarrow W$, the mapping $T^t:W^* \rightarrow V^*$ defined by $T^t(g)=gT$ for all $g \in W^*$ is a linear map with the property that $[T^t]^{{\beta}^*}_{{\gamma}^*}=([T]^\gamma_\beta)^t$.

I want to prove the reverse direction.

Assume T is one-to-one, and I want to prove that $T^t$ is onto. How am I supposed to construct a linear functional such that T is one-to-one? Any hint is appreciated.

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  • $\begingroup$ What is $T^t$ here? $\endgroup$ Commented Apr 9, 2020 at 23:18
  • $\begingroup$ @AdamMartens transpose of T $\endgroup$ Commented Apr 9, 2020 at 23:18

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Let $f\in V^*$, $f\neq 0$, consider a basis $(v_2,...)$ of $Ker(f)$, $v_1:f(v_1)=1$, $(T(v_1),T(v_2),...)$ is independent. Complete to a basis $(T(v_1),T(v_2),...,w_1,...)$ of $W$, write $h(T(v_1))=1, h(T(v_i))=0,i>1, h(w_i)=0$.

$h\circ T=f$, to see this, write $v=\sum_ia_iv_i$, $f(v)=a_i, h(T(v))=h(a_1T(v_1)+...)=a_1$.

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