# Is V a vector space over the field of real numbers with the following operations?

$V=R^2$, with operations:

addition ($\oplus$): $(x_1, y_1) \oplus (x_2, y_2) = (x_1 + x_2, y_1 + y_2)$

multiplication($\circ$): $c \circ (x, y) = (cx, y)$

This is exercise II.1.4 from Hoffman-Kunze's Linear Algebra book. I was able to show that all the axioms are satisfied. Can someone confirm if that is correct?

$\textbf{EDIT}$: I made a mistake in checking one of the distributive properties. It is not a vector space.

• No, it is wrong. – Phira Jun 20 '18 at 15:09
• No it's not, but you should show your work so that we can tell you where you're wrong. – Arnaud Mortier Jun 20 '18 at 15:09
• Thanks guys. False alarm. Made a careless mistake in checking one of the distributive property. – student Jun 20 '18 at 18:12

The operation $\oplus$ has $(0,0)$ as neutral element, because $$(x,y)\oplus(0,0)=(0,0)\oplus(x,y)=(x,y)$$ If we had a vector space, it would be true that $$0\circ(x,y)=(0,0)$$ for every $(x,y)$. What about $0\circ(1,1)$?

• I was able to spot my mistake, thanks! – student Jun 22 '18 at 15:36

Distributivity of multiplication over addition requires that

$2c\circ(x,y) = (c+c)\circ(x,y) = c\circ(x,y) + c\circ(x,y)$

but

$c\circ(x,y) = (cx,y)$

and

$c\circ(x,y) + c\circ(x,y) = (cx,y) + (cx,y) = (2cx,2y) \ne 2c\circ(x,y)$

As per the comments, check if $(a+b)\mathbf v=a\mathbf v+b\mathbf v$, i.e., "distributivity of scalar multiplication with respect to field addition".

• Why was this voted down? – wjm Jun 20 '18 at 15:20
• I guess because it's more of a comment, or because the OP is currently a poorly documented do-my-homework-plz question, to which it is better not to give answers if we want to avoid getting more of such questions...? – Arnaud Mortier Jun 20 '18 at 15:23
• That's sort of subjective, but thanks for explaining as now I know I didn't make a mathematical mistake. – wjm Jun 20 '18 at 15:25
• Thanks @Raptor. – student Jun 20 '18 at 18:13
• @ArnaudMortier: This is summer time. I am trying to go through HK on my own. I am not even a math major to begin with. – student Jun 20 '18 at 18:14