My linear algebra book (Linear Algebra Done Right by Sheldon Axler) has the following problem as exercise 1.6:

Give an example of a nonempty subset $U$ of $\mathbb{R}^2$ such that $U$ is closed under addition and under taking additive inverses (meaning $-u \in U$ whenever $u \in U$), but $U$ is not a subspace of $\mathbb{R}^2$.

It seems to me that such a set cannot exist, since the only subspace condition it's not mandated to fulfill is containing $0$, and for any $u \in U$, I can negate it to get $-u$ and then get $u + (-u) = 0$.

What is going on?

  • $\begingroup$ $0$ is not the issue here. Try $\mathbb{Q}^2$. $\endgroup$ May 3, 2015 at 16:14

1 Answer 1


Hint: Consider all ordered pairs $(a,b)$ where $a$ and $b$ are integers. Or, a little simpler, all ordered pairs $(a,0)$ where $a$ ranges over the integers.

  • $\begingroup$ Ah, I forgot that $U$ needs to be a vector space in its own right. Thanks. $\endgroup$
    – Eli Rose
    May 3, 2015 at 16:23
  • $\begingroup$ You are welcome. Closure under scalar multiplication fails. $\endgroup$ May 3, 2015 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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