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

There is vector space $F$ of all real functions. $f(x) \in F$.

Now suppose that scalar multiplication of vector space $F$ is modified so that it is now defined as $cf(x) = f(cx)$.

In this case, which axiom of vector space is broken?

share|cite|improve this question
Hi! Have you tried just writing down the axioms and seeing whether you can contradict them one at a time? In general, it's best to tell us what you've tried and exactly where you're stuck. – Sharkos Feb 10 '14 at 12:35
up vote 1 down vote accepted

Hint: What would happen with something like $(c+d)f(x)$?

share|cite|improve this answer

Given two field elements $c$ and $d$, and given a vector $f$, we must have: $$(c+d)\,f=c\,f+d\,f.$$ But, instead, we get the vector $$ x\mapsto f((c+d)\,x)\qquad\text{as our $(c+d)\,f$} $$ and the vector $$ x\mapsto f(c\,x)+f(d\,x)\qquad\text{as our $c\,f+d\,f$} $$ The two are not necessarily equal.

For instance consider the square function $x\mapsto x^2$. $$((c+d)\,x)^2\quad\text{is not the same as}\quad (cx)^2+(dx)^2\quad\text{for all $x$.}$$

share|cite|improve this answer

When one multiplies a vector by$~0$, the result should be the zero vector; however, this definition gives $0f:x\mapsto f(0)$, a constant function, but not necessarily the zero function (neutral element for addition).

Now $\def\vv{\vec v}0\vv=\vec0$ is not a vector space axiom, at least not one in the usual list. However it can be proved using the fact that $\vec0$ is the unique solution to the vector equation $\vec x+\vec x=\vec x$, together with the identity $0\vv+0\vv=(0+0)\vv=0\vv$. The former property only involves vector addition, so it certainly (still) holds in the example. The latter part uses only the distributive law $a\vv+b\vv=(a+b)\vv$ (the simplification $0+0=0$ does not involce the vector space structure), so it is this law that must fail. Indeed if $\vv$ is for instance the constant function $1$, then $0\vv=\vv$, and $0\vv+0\vv\neq0\vv$ (LHS is constant$~2$, RHS is constant$~1$).

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.