# Does there always exist left inverses for linear transformations for finite dimensional vector spaces?

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.

• Visit this page for information on how to type with MathJax. – JMoravitz Feb 20 at 1:36
• Who told you that II is the only correct statement? They're wrong, for exactly the reason that you say. – user3482749 Feb 20 at 1:37
• 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. – Bor Kari Feb 20 at 1:41
• We all make mistakes. – Lubin Feb 20 at 2:22
• 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 – Cardioid_Ass_22 Feb 20 at 6:04

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.