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This exercise is from Linear Algebra Done Right by Sheldon Axler (Chapter 3.F numbers 6)

Suppose $V$ is a finite-dimensional$V$ector space and $v_1$,...$v_m \in$V$$. Define a linear map $\Gamma:$V$' -> F^m$ by

$\Gamma(\phi) = (\phi(v_1),\dots,\phi(v_m))$.

(a) Prove that $v_1$,...$v_m$ spans $V$ if and only if $\Gamma$ is injective.

(b) Prove that $v_1$,...$v_m$ is linearly independent if and only if $\Gamma$ is surjective.

My attempts

(a) We first try to prove that, if $\Gamma$ is injective, $v_1, \dots, v_m$ spans $V$. If $\Gamma$ is injective, the only element of $V$' that $\Gamma$ annihilates is the zero linear functional. So we have that, for each $\phi_j$ in the basis of $V'$, $\Gamma(\phi_j)$ is not equal the zero vector. This implies that all the elements of a basis for $V$ are in $(v_1,...,v_m)$. We can say that $v_1, \dots, v_m$ spans $V$.

Now we have to prove that, if $v_1$,…$v_m$ spans $V$, $\Gamma$ is injective. The list $v_1$,…$v_m$ can be reduce to a basis of $V$: $v_1$,…$v_n$, with $m \geq n$. Let’s consider $\Gamma(\phi)$ and $\Gamma(\xi)$. Suppose $\Gamma(\phi) = \Gamma(\xi)$. If we observe the behaviour of $\Gamma$ on the basis of $V$, we can say that $\phi_1(v_1) = \xi_1(v_1), \dots,\phi_n(v_n)=\xi_n(v_n)$. We can conclude that $\phi = \xi$ by the theorem about the unicity of linear maps. So $\Gamma$ is injective.

(b) We demonstrate that, if $\Gamma$ is surjective, $v_1, \dots, v_m$ is linearly independent. The surjectivity of $\Gamma$ ensures us the fact that $span\{(\phi(v_1),...\phi(v_m))\}$ = ($F^m$). Hence, standard basis of ($F^m$) is in $(\phi(v_1),...\phi(v_m))$. It exists a linear functional $\psi$ such that $\psi(v_k) = (v_k)$ in the standard basis of ($F^m$) $(k = 1,…m)$.

Regarding the prove in the opposite direction, we can make reference again to the standard basis of $V$’. In this case, $\phi_1$,…$\phi_m$ results linearly independent. As consequence [$\phi_j$($v_1$),…$\phi_j$($v_m$)] is linearly independent for each $j =1,...m$. A list of m linearly independent vectors in a vector space of dim m generates all the space. So, $\Gamma$ is surjective.

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    $\begingroup$ Welcome to Math Stack Exchange. Please use MathJax to format posts. In particular, use it to type mathematical symbols and Greek letters rather than using Unicode characters, e.g. type \$\in\$ for $\in$ instead of using ∈, or type \$\Gamma\$ for $\Gamma$ instead of using Γ. $\endgroup$
    – Robin
    Oct 16, 2023 at 19:40

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Some fixes of your attempts.

(a) We first try to prove that, if $\Gamma$ is injective, $v_1, \dots, v_m$ spans $V$. If $\Gamma$ is injective, the only element of $V$' that $\Gamma$ annihilates is the zero linear functional.

Your statement after this point is unclear: "so we have that, for each $\phi_j$ in the basis of $V'$, $\Gamma(\phi_j)$ is not equal the zero vector" is correct but not obviously useful. The statement "this implies that all the elements of a basis for $V$ are in $(v_1,...,v_m)$" doesn't make sense.

Picking things up from the quoted section: we know that the only element $\phi \in V'$ that satisfies $\phi(v_1) = \cdots = \phi(v_m) = 0$ is the zero functional. Now, suppose for the purpose of contradiction that $v_1,\dots,v_m$ does not span $V$. It follows that for every subspace $U \subset V$ with $\dim(U) = \dim(V) - 1$, it does not hold that $\operatorname{span}(v_1,\dots,v_n) \subset U$; otherwise, the quotient map $\pi:V \to V/U \cong \Bbb F$ would define a non-zero linear functional that maps each $v_i$ to zero. Thus, $\dim(\operatorname{span}(v_1,\dots,v_n)) > \dim(V) - 1$, which implies that $\dim(\operatorname{span}(v_1,\dots,v_n)) = \dim(V)$ and hence $\operatorname{span}(v_1,\dots,v_n) = V$.

Your proof of the converse is mostly correct but it can be made a bit cleaner and has nonsensical statements like "if we observe the behaviour of $\Gamma$ on the basis of $V$" (note that $\Gamma$ acts on $V'$, not on $V$, so it doesn't have any "behaviour" on elements of $V$).

Now we have to prove that, if $v_1$,…$v_m$ spans $V$, $\Gamma$ is injective. The list $v_1$,…$v_m$ can be reduce to a basis of $V$: $v_1$,…$v_n$, with $m \geq n$....

Consider any $\phi \in \ker(\Gamma)$. Because $\Gamma(\phi) = 0$, we can say that $\phi(v_1) = \cdots = \phi(v_n) = 0$. By the theorem about the unicity of linear maps, it follows that $\phi$ is the zero-functional, so that $\ker(\Gamma) = \{0\}$. So, $\Gamma$ is injective.

For (b):

(b) We demonstrate that, if $\Gamma$ is surjective, $v_1, \dots, v_m$ is linearly independent. The surjectivity of $\Gamma$ ensures us the fact that $\{(\phi(v_1),...\phi(v_m))\color{red}{: \phi \in V'}\} = F^m$.

This part makes sense, after a minor correction. I'm not sure what "Hence, standard basis of $F^m$ is in $(\phi(v_1),...\phi(v_m))$" is supposed to mean. As a hint towards one approach to completing this argument, consider the map $f \circ \Gamma$ where $f \in (\Bbb F^m)'$ is the map $$ f(x_1,\dots,x_n) = c_1x_1 + \cdots + c_n x_n. $$ Because $\Gamma$ is surjective, $f \circ \Gamma$ is only zero if $f = 0$. Perhaps this framing will help you with the opposite direction as well. What does $(f \circ \Gamma)(\varphi)$ look like? How does this related to the definition of linear independence?

For your current proof of the opposite direction, it is not necessarily true that the elements $\Gamma(\phi_1),\dots,\Gamma(\phi_m)$ are linearly independent for some "standard basis" of $V'$.

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  • $\begingroup$ Thanks for the help. I try to clarify why I thought "this implies that all the elements of a basis for V are in ($v_1$,...,$v_m$)". Let’s consider ($v_1$,...,$v_n$) a basis for V. If we consider a dual basis of V’ as $ϕ_k$ (k=1,…n) such that $ϕ_k$($v_j$) = 1 (j =1,…n) if k = j and 0 if k ≠ j, it results, from injectivity of Γ, that ($v_1$,...,$v_n$) ∈ ($v_1$,...,$v_m$). Regarding "Hence, standard basis of $F^m$ is in (ϕ($v_1$),...ϕ($v_m))", I wanted to adfirm that, for Γ is surjective, every element of a basis of $F^m$ results from the application of Γ on some ϕ. $\endgroup$
    – datfq
    Oct 17, 2023 at 19:21
  • $\begingroup$ @datfq In order to consider a dual basis, you need to first establish that $(v_1,\dots,v_n)$ is a basis, which you had not established at that point in the proof. Also, you mean $(v_1,\dots,v_n) \subset (v_1,\dots,v_m)$ (rather that $\in$). $\endgroup$ Oct 17, 2023 at 19:40
  • $\begingroup$ @datfq The second part of your comment is difficult to understand. It sounds like you're saying the following: if we consider a particular basis $w_1,\dots,w_m$ of $\Bbb F^m$ (for instance, the "standard basis" consisting of the columns of the identity matrix), then for every $j$ there exists a functional $\phi$ such that $\Gamma(\phi) = v_j$. If that's the case, then yes I agree. However, it's not clear how this leads you to the conclusion that the vectors $v_1,\dots,v_n$ are linearly independent. Perhaps you have an idea involving dual bases in mind for this. $\endgroup$ Oct 17, 2023 at 19:45
  • $\begingroup$ I think that, in both cases, I made the mistake to consider that, if a linear functional has got that behaviour on the vectors, can be considered part of a dual basis associated with the vectors. $\endgroup$
    – datfq
    Oct 17, 2023 at 19:55
  • $\begingroup$ It can be, in a certain sense. If $v_1,\dots,v_k \in V$ and $\phi_1,\dots,\phi_k$ are such that $\phi_i(v_j) = 1$ if $i = k$ and $0$ otherwise, then we can deduce that $v_1,\dots,v_k$ are linearly independent. If $U = \text{span}(v_1,\dots,v_k)$, then the restricted maps $\phi_1|_U,\dots,\phi_k|_U$ form a basis for $U'$ that is dual to the basis $(v_1,\dots,v_k)$ of $U$. However, all of that would require proof. $\endgroup$ Oct 17, 2023 at 20:00

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