Skip to main content

Problem 42: Show that V is not a vector space under the above defined operations.

Problem 42: Let R be the field of reals and V be the set of all ordered pairs$(x,y)$, where x, y are reals. Define.

$(x_1,y_1)+(x_2,y_2) = (x_1+x_2,y_1+y_2) \forall (x_1,y_1),(x_2,y_2) \in \V$ and

$\alpha(x,y)=(\alpha x,y) \forall \alpha \in \R$ and $(x,y) \in \V$.

Show that V is not a vector space under the above defined operations.

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

I. Properties under addition
i. Closure Property
Let \begin{align*} (x_1,y_1),(x_2,y_2) \in \V\\ \Rightarrow x_1,y_1,x_2,y_2 \in \R \Rightarrow x_1+x_2,y_1+y_2 \in \R\\ \therefore (x_1,y_1)+(x_2,y_2) = (x_1+x_2,y_1+y_2) \in \V \end{align*} $\Rightarrow \V$ is closed under addition.

ii. Associativity:
Let \begin{align*} (x_1,y_1),(x_2,y_2),(x_3,y_3)\in \V\\ \end{align*} Now, \begin{align*} \left[(x_1,y_1)+(x_2,y_2)\right]+(x_3,y_3) = \left[(x_1+x_2,y_1+y_2)\right]++(x_3,y_3)\\ = \left[(x_1+x_2)+x_3,(y_1+y_2)+y_3\right]\\ = \left[x_1+(x_2+x_3),y_1+(y_2+y_3)\right]\\ = (x_1,y_1)+\left[(x_2+x_3),(y_2+y_3)\right]\\ = (x_1,y_1)+\left[(x_2,y_2)+(x_3,y_3)\right]\\ \end{align*} $\Rightarrow$ Addition is associative in $\V$.

iii. Existence of additive identity:
For all $(x_1,y_1) \in \V, \exists (0,0) \in \V$
Now \begin{align*} (x_1,y_1)+(0,0)=(x_1+0,y_1+0) = (x_1,y_1)\\ (0,0)+(x_1,y_1)=(0+x_1,0+y_1) = (x_1,y_1)\\ \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_1,y_1),(x_2,y_2) \in \V$, Now \begin{align*} (x_1,y_1)+(x_2,y_2)= (x_1+x_2,y_1+y_2) = (x_2+x_1,y_2+y_1) = (x_2,y_2)+(x_1,y_1) \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 x,y) \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) x,y) = (\alpha x + \beta x, y) \end{align*} And \begin{align*} \alpha(x,y)+\beta(x,y)= (\alpha x,y)+(\beta x,y) = (\alpha x+ \beta x, 2y) \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$.

Labels:

Comments

Post a Comment

Popular posts from this blog

Problem 98: If 6 of 18 new buildings in a city violate the building code, what is the probability that a building inspector, who randomly selects 4 of the new buildings for inspection, will catch
a. none of the buildings that violate the building code?
b. 1 of the new buildings that violate the building code?
c. 2 of the new buildings that violate the building code?
d. at least 3 of the new buildings that violate the building code?

Problem 98: If 6 of 18 new buildings in a city violate the building code, what is the probability that a building inspector, who randomly selects 4 of the new buildings for inspection, will catch a. none of the buildings that violate the building code? b. 1 of the new buildings that violate the building code? c. 2 of the new buildings that violate the building code? d. at least 3 of the new buildings that violate the building code? Solution: a. none of the buildings that violate the building code? Given that $x=0,n=4,a=6, N=18$ Hypergeometric distribution, $h(x;n,a,N)=\frac{\left(\begin{array}{c}a\\ x\end{array}\right)\left(\begin{array}{c}N-a\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$ $P(x=0) = h(0;4,6,18)=\frac{\left(\begin{array}{c}6\\ 0\end{array}\right)\left(\begin{array}{c}18-6\\ 4-0\end{array}\right)}{\left(\begin{array}{c}18\\ 4\end{array}\right)} = 0.16176.$ b. 1 of the new buildings that violate the building code? Given ...
Post a Comment
Read more

Problem 105: If a bank receives on the average α = 6 bad checks per day, what are the probabilities that it will receive (a) 4 bad checks on any given day?
(b) 10 bad checks over any 2 consecutive days?

Problem 105: If a bank receives on the average α = 6 bad checks per day, what are the probabilities that it will receive (a) 4 bad checks on any given day? (b) 10 bad checks over any 2 consecutive days? Solution: Poisson distribution, $f(x;\lambda )=\frac{{\lambda }^{x}{e}^{-\lambda }}{x!}$ a. 4 bad checks on any given day Given that $x=4, \lambda = \alphaT = 6\dot 1 = 6$ $f(4;6 )=\frac{{6 }^{4}{e}^{-6 }}{4!} = 0.134$ b. 10 bad checks over any 2 consecutive days Given that $x=10, \lambda = \alphaT = 6\dot 2 = 12$ $f(10;12 )=\frac{{12 }^{10}{e}^{-12 }}{10!} = 0.104837$
Post a Comment
Read more

Problem 56: Let $S = \{x \in \R^3 : x = (r-2s,3r+s,s), r,s \in \R \}$. Establish that $S$ is a subspace of $\R^3$.

Problem 56: Let $S = \{x \in \R^3 : x = (r-2s,3r+s,s), r,s \in \R \}$. Establish that $S$ is a subspace of $\R^3$. Solution: $S = \{x \in \R^3 : x = (r-2s,3r+s,s), r,s \in \R \}$. S is certainly nonempty. Let $x, y \in S$. Then for some $r, s, u,v \in \R$, $x = (r-2s,3r+s,s)$ and $y = (u-2v,3u+v,v)$. Hence, $x + y = (r-2s,3r+s,s) + (u-2v,3u+v,v)$ $= (r-2s+u-2v,3r+s+3u+v,s+v) = ((r+u)-2(s+v),3(r+u)+(s+v),(s+v))$ $= (k-2l,3k+l,l)$, where $k = r + u, l=s+v$. Consequently, $S$ is closed under addition. Further, if $c \in \R$, then $cx = c(r-2s,3r+s,s) = (c(r-2s),c(3r+s),cs) $ $= (cr-2cs,3cr+cs,cs) = (a-2b,3a+b,b)$, where $a = cr,b=cs$. Therefore $S$ is also closed under scalar multiplication. It follows that $S$ is a subspace of $\R^3$.
Post a Comment
Read more