Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $V$ be a vector space. Let $f : V \rightarrow V$ be a bijection. Define two new operations $+_f$ and $\cdot_f$ as follows. If $v$ and $w$ are two vectors in $V$, $v +_f w$ is defined to be the vector $f^{-1}(f(v) + f(w))$ where $f^{-1}$ is the inverse function of $f$. If $a$ is a scalar and $v$ is a vector in $V$, $a \cdot_f v$ is defined to be the vector $f^{-1}(af(v))$. Prove that $V$ together with the new addition of vectors, $+_f$, and the new multiplication of vectors by scalars, $\cdot_f$, is also vector space.

share|cite|improve this question
What have you tried? Your problem is to prove that $V$ is a vector space under the new operation. What would you need to show in order to have proved this? – MJD Mar 26 '12 at 17:42
Google "transport of structure". – Bill Dubuque Mar 26 '12 at 17:48
Noone seems to figure this out. Bummer..! – user28873 Apr 11 '12 at 22:07
Actually Bill's comment is a nearly complete answer. Furthermore this is really nothing more than a simple exercise in verification that the definition of "vector space" holds with the new operations. – Asaf Karagila Apr 11 '12 at 22:48

Replace your problem by the following: Let $X$ be a set and $V$ be a vector space, and let $f:X\to V$ be a bijection. Define operations $\oplus$ and $\otimes$ on $X$ as follows: $\ldots$

It will turn out that you have just transported the vector space structure of $V$ back to $X$, or what is the same thing: The vectors $v\in V$ have got new names $x:=f^{-1}(v)\in X$, but otherwise the operations remain unchanged.

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.