Question: Does ax = bx imply a = b in any vector space? Justify. Let v = $${(a_1, a_2):a_1, a_2 \in f}$$ where f is field. Define addition of elements of v coordinate wise, and for $$c \in f$$ and $$(a_1, a_2) \in V$$ . Define $$c(a_1, a_2) = (a_1, 0)$$. Is vector space over F with these operations? Justify your answer by taking all condition of vector space.

Solution: In any vector space ax = bx implies that a = b false
It is only true if we take x = 0
i.e ; ax = bx \
x(a – b) = 0 either x = 0 or ( a-b) = 0
take x = 0 and any a≠b for x.

Let us suppose $$(a_1, a_2), (b_1, b_2) \in V$$ then
$$V_{s1}: (a_1, a_2) + (b_1, b_2) = (a_1 + b_1, a_2 + b_2) = (b_1 + a_1, b_2 + a_2) = (b_1, b_2) + (a_1, a_2)$$
$$V_{s2}: [(a_1, a_2) + (b_1, b_2)] + (c_1, c_2) = (a_1, a_2) + [(b_1 , b_2) + (c_1, c_2)] \forall (c_1, c_2) \in f$$
$$V_{s3}: \exists (d_1, d_2) \in V\ s.t\ (a_1, a_2) + (d_1, d_2) = (a_1, a_2)$$
$$(a_1, a_2) + (0, 0) = (a_1, a_2)$$
$$(d_1, d_2) = (0, 0)$$
$$V_{s4}: \forall (a_1, a_2) \in V \exists (-a_1, -a_2) \in V$$
s.t $$(a_1, a_2) + (-a_1, -a_2) = (0, 0)$$
$$V_{s5}: \forall \in F 1.(a_1, a_2) = (a_1 ,0) = (a_1, 0) ≠ (a_1, a_2)$$
Hence $$v_{s5}$$ does not satisfy.
$$V_{s6}: \forall x, y \in F\ xy(a_1, a_2) = x(ya_1, 0)$$
$$V_{s7}: \forall x, y \in F$$
$$(x + y)(a_1, a_2)$$
= $$[(x + y)a_1, 0]$$
= $$[xa_1 + ya_1, 0]$$
= $$(xa_1, 0) + (ya_1, 0)$$
$$V_{s8}: \forall x \in f$$
$$x[(a_1, a_2) + (b_1, b_2)]$$ =$$x[a_1 + b_1, a_2 + b_2]$$
=$$[x(a_1 + b_1), 0]$$
=$$(xa_1, 0) + (xb_1, 0)$$
=$$x(a_1, a_2) + x(b_1, b_2)$$\

$$V_{s5}$$ does not satisfy it is not vector space.

Question: Does ax = bx imply x = y in any vector space? Justify. Let v denote the set of ordered pairs of real numbers.if $$(a_1, a_2)$$ and $$(b_1, b_2)$$ are element of v and $$c \in R$$ , define $$(a_1 , a_2) + (b_1 , b_2) = (a_1.b_1, a_2 + b_2)$$ and $$c(a_1, a_2) = (a_1, ca_2)$$. Is v a vector space over R with these operations ? Justify your answer by showing all conditions of vector space.

Solution: In any vector space ax = ay implies that x = y is false.
It is only true if we take a = 0
i.e; ax = ay a(x – y) = $$0 \in V$$
either a = 0 , or ( x – y) = 0 if we take a = 0 and any x ≠ y for a.\

Second part
Here v denote the set of ordered pairs of real numbers. If $$(a_1, a_2) and\ (b_1, b_2)$$ are element of v and $$c \in R$$,define $$(a_1 , a_2) + (b_1 , b_2) = (a_1.b_1, a_2 + b_2)$$ and $$c(a_1, a_2) = (a_1, c.a_2)$$.
$$V_{s1}: (a_1, a_2) + (b_1 , b_2) = (a_1.b_1, a_2 + b_2) = (b_1.a_1 , b_2 + a_2) = (b_1, b_2) + (a_1, a_2)$$
$$V_{s2}: \in [(a_1, a_2) + (b_1, b_2)] + (c_1, c_2) = (a_1.b_1, a_2 + b_2) + (c_1, c_2)$$\
= $$(a_1.b_1.c_1 , a_2 + b_2 + c_2) = (a_1, a_2) + (b_1.c_1 , b_2 + c_2) = (a_1, a_2) + [(b_1, b_2) + (c_1 , c_2)]$$
$$V_{s3}: \exists (d_1, d_2) \in V$$
s.t $$(a_1, a_2) + (d_1, d_2) = (a_1, a_2)$$
Or,$$(a_1.b_1 , a_2 + b_2 ) = (a_1 , a_2) \Rightarrow a_1.b_1 = a_1 \Rightarrow b_1 = 1$$
Also,$$a_2 + b_2 = a_2 \Rightarrow b_2 = 0$$
Hence $$(d_1, d_2) = (1, 0)$$
so, $$(a_1, a_2) + (d_1, d_2) ≠ (a_1 , a_2)$$
$$v_{s3}$$ does not satisfy.
$$V_{s4}: \forall (a_1 , a_2) \exists (-a_1, -a_2) \in V$$ s.t $$(a_1, a_2) + (-a_1 , -a_2) = [a_1.(-a_1) , b_1 + (-b_1) ] = (-a_1^2, 0)≠0$$
Hence $$v_{s4}$$ also not satisfy.
$$V_{s5}: for 1\in R 1.(a_1, a_2) = (a_1, 1.a_2) = (a_1, a_2)$$
$$V_{s6}: \forall x, y \in R$$
$$xy(a_1, a_2) = x(a_1, ya_2) = x[y(a_1, a_2)$$
$$V_{s7}: \forall x, y \in R$$
$$(x + y) [(a_1, a_2) + (b_1 , b_2)] = (x + y)(a_1, a_2) + (x + y) (b_1, b_2)$$
L.H.S = $$(x + y)(a_1.b_1, a_2 + b_2)$$
= $$[a_1.b_1 , (x + y)(a_2 + b_2)]$$
=$$[a_1.b_1 , (x + y) a_2 + (x + y) b_2]$$
R.H.S = $$[ a_1 , (x + y)a_2] + [b_1 , (x + y)b_2]$$
= $$[a_1.b_1 , (x + y)a_2 + (x + y) b_2]$$
L.H.S = R.H.S
$$V_{s8}: \forall x \in R$$
$$x[(a_1, a_2) + (b_1, b_2)] = x(a_1, a_2) + y(b_1, b_2)$$ L.H.S
= $$x[ a_1.b_1 , a_2 + b_2]$$
= $$[a_1.b_1 , x(a_2 + b_2)]$$
= $$[a_1.b_1 , xa_2 + xb_2]$$
R.H.S
= $$(a_1 , xa_2) + (b_1 , xb_2) = (a_1.b_1 , xa_2 + xb_2)$$
L.H.S = R.H.S
Here $$v_{s3}$$ and $$v_{s4}$$ are not satisfy. Hence it is not vector space.

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