Question: May a vector space have more than one zero vector?justify. Let $$v = {(a_1, a_2):a_1, a_2 \in R}$$. for $$(a_1, a_2)$$ , $$(b_1, b_2)$$ = $$(a_1 + 2b_1, a_2 + 3b_2)$$ and $$c(a_1, a_2)$$ = $$(ca_1, a_2)$$ .Is v a vector space over f with these operations? justify your answer.

Solution:
Let a vector space have two zeros say 0 and 0^’. From $v_{s3}$ 0 is the identity under addition
Hence $$\forall x \in V$$
x + 0 = 0 + x = x…………(1)
also from $v_{s3}$ 0’ is identity under addition.
x + 0’ = 0’ + x = x……….(2)
From (1) and (2)
x + 0 = x + $$0'$$ = x .
Hence there is unique zeros in vector space.

Second part

Here given
v = $${(a_1, a_2):a_1, a_2 \in R}$$ for $$(a_1, a_2),\ (b_1, b_2) \in V$$ and $$c \in R$$
Define $$(a_1, a_2) + (b_1 , b_2) = (a_1 + 2b_1, a_2 + 3b_2)$$ and $$c(a_1, a_2) = (ca_1, a_2)$$ $$V_{s1}:$$ for $$(a_1, a_2), (b_1, b_2) \in V$$ $$(a_1, a_2) + (b_1, b_2) = (a_1 + 2a_2, b_1 + 3b_2) \neq (a_2 + 2a_1, b_2 + 3b_1) \neq (b_1, b_2) + (a_1, a_2)$$
$$V_{s1}$$ does not hold.
$$V_{s2}$$: for $$(a_1, a_2), (b_1, b_2) \in V\ and\ (c_1, c_2) \in V$$
$$[(a_1, a_2) + (b_1, b_2)] + (c_1, c_2) = [ a_1 + 2b_1, a_2 + 3b_2] + (c_1, c_2) \neq$$ $$(a_1 + 2b_1 + c_1, a_2 + 3b_2 + 3b_3) \neq (a_1, a_2) + [ b_1 + 2c_1, b_2 + 3c_2] \neq (a_1, b_2) + [(b_1, b_2) + (c_1, c_2)]$$
$$V_{s2}$$ does not hold.

$$V_{s3}$$:if there exist a element z = $$(d_1, d_2) \in V$$ such that $$(a_1, a_2) + (d_1, d_2) =(a_1, a_2)$$
$$(a_1 + 2d_1, a_2 + 3d_2) = (a_1, a_2)$$
$$a_1 + 2d_1 = a_1 \Rightarrow d_1 = 0$$
$$a_2 + 3d_2 = a_2 \Rightarrow d_2 = 0$$
This show that $$(a_1, a_2) + (0, 0) = (a_1, a_2)$$

$$v_{s4}$$:Let $$\forall (a_1, b_1) \exists (-a_1, -a_2) \in V$$ such that $$(a_1, a_2) + (-a_1, -a_2) = [a_1 +(-2a_1), a_2 + (-3a_2)] = (-a_1, -2a_2) \neq (0, 0)$$.
Hence $$v_{s4}$$ does not satisfy.

$$V_{s5}: 1 \in F\ 1.(a_1, b_1) = (1.a_1, b_1) = (a_1, b_1)$$
$$V_{s6}: \forall x, y \in F\ xy(a_1, b_1) = x(ya_1, b) = x[y(a_1, b1)]$$
$$V_{s7}: \forall x, y \in F\ (x+y) (a_1, b_1) = [(x+y)a_1, b_1] = (xa_1 + ya_1, b_1)$$
also $$x(a_1, b_1) + y(a_1, b_2) = (xa_1, b_1) + (ya_1, b_1) = (xa_1 + ya_1, b_1 + b_2) = (xa_1 + ya_1, 2b_1)$$
$$(x+y)(a_1, b_1) ≠ x(a_1, b_1) + y(a_1, b_1)$$
Hence $$v_{s7}$$ does not satisfy.

$$V_{s8}$$: $$\forall x \in F$$ $$x[(a_1, b_1) + (a_2, b_2)] = x(a_1, b_1) + x(a_2, b_2)$$ = $$(xa_1, b_1) + (xa_2, b_2)$$
All the condition of vector space does not satisfy hence it is not vector space.

Tags:

Categories:

Updated: