I need to prove that the vector space of $\mathbb{R}^2$ with the following operations:

  • $x + y = (x_1 + 2y_1, 3x_2 - y_2)$
  • The usual scalar multiplication of $cx = (cx_1, cx_2)$

The answers in my book say that axiom $\textbf{4}$ fails to hold that is that:

For each vector $x \in V$ there exists the additive inverse $-x$ such that $x + -x = 0$.

I found the zero vector as such:
$x + y = (x_1 + 2y_1, 3x_2 - y_2)$
$x_1 + 2y_1 = 0, y_1 = 0$
$3x_2 - y_2 = 0, y_2 = -3x_2$

So the zero vector is $(0, -3x_2)$. For getting the additive inverse of any vector we have:

$x_1 + 2y_1 = 0$ and $3x_2 - y_2 = -3x_2$

And both these questions have solutions to them, though the inverse for the vector $0, 0$ is just itself. (but this is ok, right?). Can someone please enlighten me as to what I'm doing wrong?


1 Answer 1


There is no zero vector, that is, no vector $y$ such that $x+y=x$ for all $x$.

Added: Suppose that $y=(y_1,y_2)$ is such that $(x_1,x_2)+(y_1,y_2)$ for all $(x_1,x_2)$. Then since $$(x_1,x_2)+(y_1,y_2)=(x_1+2y_1, 3x_2-y_2),$$ it follows that $x_2=3x_2-y_2$ for all $x_2$. This is impossible.

  • $\begingroup$ Is the zero vector not $(0, -3x_2)$? $\endgroup$
    – q.Then
    Feb 11, 2016 at 7:28
  • $\begingroup$ If a vector $y$ had the specific property mentioned in my comment above, it would have to be an explicit ordered pair, like $(2,4)$. Your candidate is not of that shape. If you look at the definition of addition, you will find that there is no $y$ that simultaneously works for all $x$. By the way, there are several other reasons the set of ordered pairs with the given operations is not a vector space. $\endgroup$ Feb 11, 2016 at 7:34

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