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I'm trying to prove an axiom of a vector space - namely, distributivity of scalar multiplication over addition. In the answer section of the book, it says that the property holds. When I do it, it doesn't seem that it does. This is what I have:

The Property

In order for a set to be a vector space it has to satisfy,

$$\forall \vec x, \vec y \in V: \forall c \in \Bbb R,\ c(\vec x + \vec y) = c \vec x + c \vec y $$

For my problem, vector addition has been defined as, $$(x_1,x_2)\ +' \ (y_1, y_2) = \ (x_1 + 2y_1, 3x_2 - y_2)$$

What I Have

Let $c$ be a scalar s.t. $c \in \Bbb R$. Then, $$c(\vec x \ +' \vec y) = c((x_1, x_2) \ +' \ (y_1, y_2)) = c(x_1 +2y_1, 3x_2 - y_2) = (c(x_1 + 2y_1) \ + \ c(3x_2 - y_2)) \\ = (cx_1 + c2y_1, c3x_2 - cy_2) = \ (cx_1, c3x_2) \ +' \ (c2y_1, cy_2) = c(x_1, 3x_2) \ +' \ c(2y_1, y_2)$$

Which is not equal to $c \vec x + c \vec y$. So it seems the property doesn't hold for this definition of vector addition. What am I doing wrong? Thanks in advance.

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2 Answers 2

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Your second from the last equality is wrong. You should have: \begin{align*} c(\vec x \ +' \vec y) &= c((x_1, x_2) \ +' \ (y_1, y_2)) \\ &= c(x_1 +2y_1, 3x_2 - y_2) \\ &= (c(x_1 + 2y_1) , \ c(3x_2 - y_2)) \\ &= (cx_1 + c2y_1, c3x_2 - cy_2) \\ &= ( (cx_1) + 2(cy_1), 3(cx_2) - (cy_2)) \\ &= (cx_1, cx_2) \ +' \ (cy_1, cy_2) \\ &= c(x_1, x_2) \ +' \ c(y_1, y_2) \\ &= c\vec x \ +' c\vec y. \\ \end{align*}

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  • $\begingroup$ I think I get your answer. Quantities 2 and 3 disappear b/c 2 x c and 3 x c become another scalar c? $\endgroup$
    – Romaion
    Commented Sep 17, 2017 at 16:35
  • $\begingroup$ @Romaion yes, the scalars $2$ and $3$ should disappear. Try simplifying $c(\vec{x} \ +' \ \vec{y})$ and $c\vec{x} \ +' \ c\vec{y}$ from both ends and see what you get. $\endgroup$ Commented Sep 17, 2017 at 16:37
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$\newcommand{\vect}[1]{{\bf #1}}$

\begin{eqnarray} c(\vect{x} ~+'~ \vect{y}) &=& c((x_1, x_2) ~+'~ (y_1, y_2)) \\ &=&c(x_1 + 2y_1, 3x_2 - y_2) \\ &=& (cx_1 + 2cy_1, 3cx_2- cy_2) \\ &=&(cx_1, cx_2) ~+'~(cy_1, cy_2) \\ &=& c\vect{x} ~+'~ c\vect{y} \end{eqnarray}

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