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Suppose $V$ and $W$ are finite dimensional vector spaces, and that $f~:~V \to W$ is a linear map.

Suppose $\{e_1, \dots, e_n\} \subset V$ and that $\{f(e_1), \dots , f(e_n)\}$ is a basis of $W$.

Then which of the following are true?

  • I. $\{e_1, \dots, e_n\}$ is a basis of $V$.
  • II. There exists a linear map $g~:~W \to V$ such that $g \circ f = \text{Id}_V$
  • III. There exists a linear map $g~:~W \to V$ such that $f \circ g = \text{Id}_W$

I know I is not right and thought that III would be the correct one because $f$ is surjective and it should have a right inverse. But it turns out that II is the only correct option and I have no clue how that could be possible. Any hints/explanations would be greatly appreciated.

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  • $\begingroup$ Visit this page for information on how to type with MathJax. $\endgroup$ – JMoravitz Feb 20 at 1:36
  • $\begingroup$ Who told you that II is the only correct statement? They're wrong, for exactly the reason that you say. $\endgroup$ – user3482749 Feb 20 at 1:37
  • $\begingroup$ Thank you for your reply. I got this question from U Chicago for their GRE practice.math.uchicago.edu/~min/GRE/files/week2.pdf its question 19 and the answer key is at the bottom. I do not think that they would make such a mistake. $\endgroup$ – Bor Kari Feb 20 at 1:41
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    $\begingroup$ We all make mistakes. $\endgroup$ – Lubin Feb 20 at 2:22
  • $\begingroup$ Since the map is surjective (and you've indicated the spaces are finite-dimensional), it must be that $dim(W)\leq dim(V)$. Whenever the inequality is strict, we see that there can exist no surjective $g$ from $W$ to $V$ and II must necessarily be violated. III is true because of $f$s surjectivity $\endgroup$ – Cardioid_Ass_22 Feb 20 at 6:04
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The condition that $e_1,\dots, e_n$ is mapped to a basis $f(e_1,),\dots,f(e_n)$ means that the map is surjective as you figured out yourself. It is easy to write down a counterexample for II: Consider $f:\mathbb R^2\to \mathbb R$ given by the matrix $$A=\begin{pmatrix}1\\ 0\end{pmatrix}.$$ Let $e_1,e_2$ be the standard basis vectors. Then, $Ae_1=A(1,0)=1$ and $Ae_2=A(0,1)=0.$ Thus the kernel of $A$ is spanned by $e_2$ and $A$ cannot be injective.

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