# {Hint} Injectivity and surjectivity of a linear map Γ: V' -> $F^m$ with V finite dimensional and $v_1$,...$v_m$ ∈ V.

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.

• 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. Oct 16, 2023 at 19:40

(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 $$V'$$.

• 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$results from the application of Γ on some ϕ. Oct 17, 2023 at 19:21 • @datfq In order to consider a dual basis, you need to first establish that$(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$). Oct 17, 2023 at 19:40 • @datfq The second part of your comment is difficult to understand. It sounds like you're saying the following: if we consider a particular basis$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$are linearly independent. Perhaps you have an idea involving dual bases in mind for this. Oct 17, 2023 at 19:45 • I think that, in both cases, I made the mistake to consider that, if a linear functional has got that behaviour on the vectors, can be considered part of a dual basis associated with the vectors. Oct 17, 2023 at 19:55 • It can be, in a certain sense. If$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. Oct 17, 2023 at 20:00
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