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Problem 43:

Show that $\V$ is not a vector space over $\R$ under the operations defined below:

$(x,y)+(z,t) = (x+z,y+t)$ and $\alpha(x,y)=(\alpha^2x, \alpha^2y)$

Problem 43: Let $\V = \{(x,y):x,y \in \R\}$

Show that $\V$ is not a vector space over $\R$ under the operations defined below:

$(x,y)+(z,t) = (x+z,y+t$) and $\alpha(x,y)=(\alpha^2x, \alpha^2y)$

Solution: Given $\V = \{(x,y):x,y \in \R\}$

I. Properties under addition
i. Closure Property
Let \begin{align*} (x,y),(z,t) \in \V\\ \Rightarrow x,y,z,t \in \R \Rightarrow x+z,y+t \in \R\\ \therefore (x,y)+(z,t) = (x+z,y+t) \in \V \end{align*} $\Rightarrow \V$ is closed under addition.

ii. Associativity:
Let \begin{align*} (x,y),(z,t),(x_3,y_3)\in \V\\ \end{align*} Now, \begin{align*} \left[(x,y)+(z,t)\right]+(x_3,y_3) = \left[(x+z,y+t)\right]++(x_3,y_3)\\ = \left[(x+z)+x_3,(y+t)+y_3\right]\\ = \left[x+(z+x_3),y+(t+y_3)\right]\\ = (x,y)+\left[(z+x_3),(t+y_3)\right]\\ = (x,y)+\left[(z,t)+(x_3,y_3)\right]\\ \end{align*} $\Rightarrow$ Addition is associative in $\V$.

iii. Existence of additive identity:
For all $(x,y) \in \V, \exists (0,0) \in \V$
Now \begin{align*} (x,y)+(0,0)=(x+0,y+0) = (x,y)\\ (0,0)+(x,y)=(0+x,0+y) = (x,y)\\ \end{align*} $\Rightarrow (0,0)$ is the additive identity in $\V$.

iv. Existence of additive inverse:
Let $(x,y) \in \V$, be any element $(-x,-y)\in \V$
\begin{align*} \ \because x,y \in \R \Rightarrow -x,-y \in \R\\ \end{align*} Now \begin{align*} (x,y)+(-x,-y)= (x+(-x),y+(-y))=(0,0)\\ (-x,-y)+(x,y)= ((-x)+x,(-y)+y)=(0,0)\\ \end{align*} $\Rightarrow (-x,-y)$ is the additive inverse of $(x,y)$ for each $(x,y) \in \V$.

v. Commutative Law:
Let $ (x,y),(z,t) \in \V$, Now \begin{align*} (x,y)+(z,t)= (x+z,y+t) = (z+x,t+y) = (z,t)+(x,y) \end{align*} $\Rightarrow$ addition is commutative in $\V$.

II. Properties under Scalar Multiplication
vi. Closure under Scalar Multiplication
Let $\alpha\in \R$ and $(x,y) \in \V$ s.t. $x,y \in \R$ \begin{align*} \alpha(x,y)=(\alpha^2x, \alpha^2y) \in \V \therefore \alpha (x,y) \in \V \end{align*} $\Rightarrow \V$ is closed under Scalar Multiplication.

vii.Distributivity of Addition of Real Numbers
Let $\alpha, \beta\in \R$ and $(x,y) \in \V$ s.t. $x,y \in \R$ \begin{align*} (\alpha+\beta)(x,y)= ((\alpha+\beta)^2 x,(\alpha+\beta)^2 y) \end{align*} And \begin{align*} \alpha(x,y)+\beta(x,y)= (\alpha^2 x,\alpha^2 y)+(\beta^2 x,\beta^2y) = ((\alpha^2+\beta^2) x, (\alpha^2+\beta^2)y) \end{align*} $\Rightarrow (\alpha+\beta)(x,y) \ne \alpha(x,y)+\beta(x,y)$
$\Rightarrow $ Distributivity of Addition of Real Numbers does not exists.

Hence, $\V$ is not a vector space over reals.

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