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I'm having trouble showing this:

Let $T : V → W$ be a linear map of finite dimensional vector spaces. Prove that $T$ is surjective (respectively, injective) if and only if $T^*$ is injective (respectively, surjective).

I've been using the dimension theorem to show relatedness through cardinality, but this has unearthed nothing. I was thinking that a method using canonical isomorphisms might be in order. I suspect the answer is just an $iff$-chain of theorems.

Just as an aside, is there an "induces" arrow?

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  • $\begingroup$ Do you mean $\Leftrightarrow$? That's $\Leftrightarrow$. There's also $\Leftarrow$ and $\Rightarrow$, and if you like them longer, there's $\Longleftrightarrow$, which is $\Longleftrightarrow$, for example. $\endgroup$ Commented May 9, 2013 at 3:34
  • $\begingroup$ No, totally not related to this question. Just something I write a lot in other areas of mathematics. $\endgroup$ Commented May 9, 2013 at 3:35
  • $\begingroup$ Can you give me an example of a context in which it comes up? $\endgroup$ Commented May 9, 2013 at 3:36
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    $\begingroup$ You may find that someone quickly writes you a complete answer to this particular question. However, in general, with questions like this it is best to tell us exactly where you are getting stuck, what you have tried, try to quantify what your difficulty is, etc... this can help us give you better answers. $\endgroup$ Commented May 9, 2013 at 3:37
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    $\begingroup$ And once you've proved $T$ is surjective iff $T^*$ is injective, the other one follows immediately since the canonical isomorphisms from $V \to (V^*)^*$ and $W \to (W^*)^*$ take $T$ to $(T^*)^*$. $\endgroup$ Commented May 9, 2013 at 3:51

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So, I see now that if $f \in W^*$ and $T^*(f) = 0=f\circ T$, then because $T$ is surjective, we have for $w\in W$ there exists $v\in V$ such that $Tv=w$. Thus, $(f \circ T)(v) = f(T(v)) = f(w)=0$, so for all $w\in W$ we have that $f(w)=0$, so that $f=0$. Hence, it follows that $T^*$ is injective, right?

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  • $\begingroup$ Dear Barisa, This is correct. Cheers, $\endgroup$ Commented May 9, 2013 at 5:09
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For the second part I see now that for $g \in V^*$ and $f \in W^*$ we must have $T^*(f) = g$. Let $v_1,\dots,v_n$ be a basis of $V$; then the list $Tv_1,\dots,Tv_n$ is linearly independent in $W$, so we can extend it to a basis $Tv_1,\dots,Tv_n,w_1,\dots,w_k$ of $W$. Now we can define $f : W \rightarrow F$ by setting $f(Tv_i) = g(v_i)$ and $f(w_i) = 0$. Then $T^*(f) : V \rightarrow F$ and $T^*(f)(v_i) = (f \circ T)(v_i) = f(Tv_i) = g(v_i)$, so $T^*(f) = g$, as desired.

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  • $\begingroup$ I recommend this: Bach, J.S.: Fantasia and Fugue, for organ in G minor ("Great"), BWV 542: Fugue $\endgroup$ Commented May 9, 2013 at 5:05
  • $\begingroup$ Dear Barisa, The body of the argument looks correct, but the first sentence is a bit strange: you are choosing $g$ and trying to construct $f$, but this is not what your first sentence says. Regards, $\endgroup$ Commented May 9, 2013 at 5:11

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