Suppose $V$ and $W$ are finite dimensional vector spaces, and that $f:V\mapsto W$ is a linear map. Suppose $\{e_1,...,e_n\}\subset V$ and that $\{f(e_1),...,f(e_n)\}$ is a basis of $W$. Then which of the following are true?

I. $\{e_1,...,e_n\}$ is a basis of $V$

II. There exists a linear map $g:W\mapsto V$ such that $g\circ f=Id_V$

III. There exists a linear map $g:W\mapsto V$ such that $f\circ g=Id_W$

(A) I only

(B) II only

(C) III only

(D) I and III

(E) II and III

The solutions manual says it is (B), but I think it should be (C) since we can regard $W$ to be isomorphic to some subspace of $V$?

Thanks in advance!

  • $\begingroup$ If (C) should be the answer in your opinion, statement II should be wrong. Why would that be the case? $\endgroup$ Commented Aug 28, 2019 at 0:36
  • $\begingroup$ Let $V$ be $\mathbb{R}^3$, $W$ be $\mathbb{R}^2$ and $f$ projection onto the first two factors. Then the assumption of the theorem holds, as does III, but I and II do not hold. $\endgroup$ Commented Aug 28, 2019 at 1:14

3 Answers 3


Consider III. Suppose it is true. $$w \in W \implies w=fg(w)=f(g(w))\\\implies w \in \;\;\text{range}\; f$$ Hence $ \text{range}\; f=W$, concluding $f$ is onto.

Is $f$ onto ?

  • $\begingroup$ We derived no contradiction but does this constitute as a proof? $\endgroup$ Commented Aug 28, 2019 at 0:54
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    $\begingroup$ I mean, If $f$ is onto, then III is true and conversely if III is true then $f$ must be onto $\endgroup$ Commented Aug 28, 2019 at 0:54
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    $\begingroup$ Actually I learn this result in Axler's linear algebra. In that book, Axler state this as a exercise, namely, " If $T:V \to W$ is a linear map, then $T$ is onto $\iff$ there exists a linear map $S:W \to V$ such that $TS=I_W$" $\endgroup$ Commented Aug 28, 2019 at 0:59
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    $\begingroup$ something something ...every epimorphism in the category of $k$-vector spaces splits... $\endgroup$ Commented Aug 28, 2019 at 1:02
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    $\begingroup$ @RyleeLyman: Yes!...I agree..... $\endgroup$ Commented Aug 28, 2019 at 1:05

For I, we can exhibit a counterexample. Let $f : \mathbb{R}^n \to \mathbb{R}$ be given by $f(a_1, \ldots, a_n) = a_1$ with $n > 1$. Clearly this is a linear map. Let $e_1, \ldots, e_n$ be the standard basis on $\mathbb{R}^n$. Then $f(e_1) = 1$ is a basis for $\mathbb{R}$, yet $e_1$ is not a basis for $\mathbb{R}^n$.

For II, we can also exhibit a counterexample. Consider the same linear map given in the first counterexample. Suppose there did exist a linear map $g : \mathbb{R} \to \mathbb{R}^n$ such that $g \circ f = Id_{\mathbb{R}^n}$. Then $$ (0,1,0 ,\ldots, 0) = e_2 = Id_{\mathbb{R}^n} (e_2) = (g \circ f)(e_2) = g(f(e_2)) = g(0) = (0, \ldots, 0) $$ which is a contradiction.

For III, we can prove that it must be the case. Suppose $f: V \to W$ is a linear map as in the problem. Take $w \in W$. Then, since $\{f(e_1), \ldots, f(e_n)\}$ is a basis for $W$, there exist $w_1, \ldots, w_n$ such that $$ w = \sum_i w_i f(e_i) = \sum_i f(w_i e_i) = f\left(\sum_i w_i e_i\right) $$ Clearly $v \equiv \sum_i w_i e_i \in V$, so we may conclude that there exists $v \in V$ such that $f(v) = w$. In other words, $f$ is a surjective linear map. Since $f$ is surjective, there must exist a linear map $g : W \to V$ such that $f \circ g = Id_W$. (I previously asserted that this fact was trivial to prove, but it really isn't. In fact, I don't know of a way to prove it that doesn't require the axiom of choice! Nevertheless, it is true). This proves III.

To conclude, I'd have to agree that (C) is the right answer.

  • $\begingroup$ Just want to point out that I was able to prove the existence of $g$ without the axiom of choice, but it still isn't what I'd call a trivial proof. $\endgroup$ Commented Aug 28, 2019 at 19:12
  • $\begingroup$ I don't know if the existence of $g$ is that tough. Define $g$ on the set $\{f(e_1),\dots,f(e_n)\}$ by mapping $f(e_i)$ to $e_i$. Extend linearly to yield a linear map $g\colon W \to V$, and then check that the double composition $f\circ g$ is the identity map of $W$. $\endgroup$ Commented Aug 29, 2019 at 18:50
  • $\begingroup$ That's exactly right. I suppose it's a matter of taste what one calls trivial. I try to avoid using the word trivial if the proof involves any original ideas. The choice to map $f(e_i)$ to $e_i$ and extend linearly is right on the edge for me. In this case I leaned toward calling it non-trivial because you might have the bad idea (like I did) to establish the existence of $g$ (say with the axiom of choice) and then prove that $g$ is linear or that at least one such $g$ is. To put it a different way, I called it non-trivial because I had the wrong idea for the proof at first glance. $\endgroup$ Commented Aug 30, 2019 at 13:26
  • $\begingroup$ The correct proof (your proof), also caught me a little by surprise. To show the existence of a right inverse for a surjective function, you need the axiom of choice. To show the existence of a linear right inverse for a linear surjective function, which a priori seems like it should be harder than merely finding a right inverse, you only need to make finitely many choices after picking a basis. What's more, if the vector spaces are infinite dimensional, we're back to needing the axiom of choice. $\endgroup$ Commented Aug 30, 2019 at 13:31
  • $\begingroup$ You're right, it is a surprising—and useful—property of vector spaces! (and free constructions more generally) $\endgroup$ Commented Aug 30, 2019 at 13:34

For sure the I. it's false infact V could be $\mathbb{R^{n+1}}$ and $f(e_{n+1})=0$ and the hypothesis still true. The II. could be false infact if as in the first example $f(e_{n+1})=0$ we have that $g(f(e_{n+1}))=g(0)=0$ so the rank of $g(f(v))$ is $n$ and not $n+1$. So the right one is the III. infact if you choose $g$ such that $g(f(e_n))=e_n$ we have that $f(g(v))$ on $W$ in the base ${f(e_1),f(e_2),...,f(e_n)}$ is the identity.

If the answer it's wrong i will fix it tomorrow.

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    $\begingroup$ the assumption is that the $f(e_k)$ form a basis for $W$, so none of them are zero. $\endgroup$ Commented Aug 28, 2019 at 0:51
  • $\begingroup$ that's false because no one says that the dimension of V is n (look for example at the point I), so the first n $f(e_k)$ are a basis the rest could be everything. $\endgroup$ Commented Aug 28, 2019 at 0:54
  • $\begingroup$ Oh I see. You aren't wrong, but your explanation is confusing: the statement of the theorem doesn't mention any $e_k$ past $n$. In fact, a priori one might think the $e_k$ could be a linearly dependent set in $V$, so it's not clear how one could complete them to a basis. Of course our assumption implies that this is possible, but you need to tell us what you are assuming. $\endgroup$ Commented Aug 28, 2019 at 1:22

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