Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

On a finite dimensional vector space, the answer is yes (because surjective linear map must be an isomorphism). Does this extend to infinite dimensional vector space? In other words, for any linear surjection $T:V\rightarrow V$, AC guarantees the existence of right inverse $R:V\rightarrow V$. Must $R$ be linear?

How about $T:V\rightarrow W$ linear surjection in general?

share|cite|improve this question
A "surjective linear map must be an isomorphism". For example, the map from $\mathbb R^{207}$ to $\{0\}$, right? – Andrés Caicedo Aug 24 '12 at 3:09
@AndresCaicedo: thanks for your comment, i think i should put W=V. – Timothy Aug 24 '12 at 3:32

2 Answers 2

up vote 1 down vote accepted

No. Let $V = \text{span}(e_1, e_2, ...)$ and let $T : V \to V$ be given by $T e_1 = 0, T e_i = e_{i-1}$. A right inverse $S$ for $T$ necessarily sends $v = \sum c_i e_i$ to $\sum c_i e_{i+1} + c_v e_1$ but $c_v$ may be an arbitrary function of $v$.

share|cite|improve this answer

Let $T$ be any map of $V$ to $W$ that is onto but not one-to-one. A right inverse for $T$ is any $R: W \to V$ such that $T(R(x)) = x$ for every $x \in W$. In particular, for any $x$ such that there are $y_1 \ne y_2$ with $T(y_1) = T(y_2) = x$, you are free to make $R(x) = y_1$ or $R(x) = y_2$. If $x = u + v$ (and neither $u$ nor $v$ is $0$, so neither is $x$), at least one of these choices will not be $R(u) + R(v)$. So there will always be a nonlinear right inverse.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.