Why does vector sum $(x_1,x_2)+'(y_1,y_2)=(x_1+2y_1, 3x_2-y_2)$ and $c(x_1, x_2)=(cx_1,cx_2)$ fail to hold the axiom of vector space?

Is it because $(x+'y)+z=(x_1+2y_1, 3x_2-y_2)+(z_1, z_2)=(x_1,x_2)+(2y_1+z_1, -y_2+z_2)\ne x+(y+z)$

But I don't see how this is not right. Could anyone point out what is wrong here?

  • 2
    $\begingroup$ You should not mix ordinary vector addition with the vector operation +'. Check if $(\mathbf x +' \mathbf y) +' \mathbf z = \mathbf x +' (\mathbf y +' \mathbf z)$. That is not what you're doing. $\endgroup$
    – KCd
    Jan 24, 2016 at 16:08
  • $\begingroup$ I don't see how $+'$ generally works $\endgroup$
    – 1412
    Jan 24, 2016 at 16:09
  • 1
    $\begingroup$ You gave us a formula for it! So it "works" by using it: substitute in actual numbers for the coordinates on the left and the formula on the right tells you what it is. Do you not see how the function $f(a,b) = a+3b$ "works"? $\endgroup$
    – KCd
    Jan 24, 2016 at 16:10

2 Answers 2


One of the axioms of vector spaces is associativity of vector addition. Thus, $(x+y)+z=x+(y+z)$ must be true in order for this to be a vector space. Let $x=(x_1, x_2)$, $y=(y_1, y_2)$, and $z=(z_1, z_2)$.

$$(x+y)+z=(x_1+2y_1, 3x_2-y_2)+(z_1, z_2)=(x_1+2y_1+2z_1, 3(3x_2-y_2)-z_2)=(x_1+2y_1+2z_1, 9x_2-3y_2-z_2)$$ $$x+(y+z)=(x_1, x_2)+(y_1+2z_1, 3y_2-z_2)=(x_1+2(y_1+2z_1), 3x_2-(3y_2-z_2))=(x_1+2y_1+4z_1, 3x_2-3y_2+z_2)$$

As you can see, $(x+y)+z \neq x+(y+z)$, so this is not a vector space.

  • $\begingroup$ But how about $cx_1, cx_2$, does that not hold either? $\endgroup$
    – 1412
    Jan 24, 2016 at 16:30
  • $\begingroup$ I don't think anything's wrong with $c(x_1, x_2)=(cx_1, cx_2)$. However, for it to be a vector space, all of the axioms must hold, not just some of them. $\endgroup$ Jan 24, 2016 at 16:37
  • $\begingroup$ Well, suppose $c\cdot f=c+f$, how do you check associativity and commutativity? Do you find a new function or a new scalar? $\endgroup$
    – 1412
    Jan 24, 2016 at 16:40
  • $\begingroup$ I'm assuming $c$ and $f$ are scalars since you're multiplying them, so that doesn't give us a formula for vector addition for which we can check associativity and commutativity. To check associatvity and commutativity, you have to use the formula of vector addition given to you from the vector space. $\endgroup$ Jan 24, 2016 at 16:43
  • $\begingroup$ what if it is $cf(x)$, I can only think of distributivity, but having no idea about associativity $\endgroup$
    – 1412
    Jan 24, 2016 at 16:53

Indeed associativity fails. But it is easier to show that commutativity fails, that in general we do not have $u+'v=v+'u$.

Let $u=(1,0)$ and $v=(0,1)$. Then $u+'v=(1,-1)$ and $v+'u=(2,3)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .