let V be set of all functions with down R which are such that f(t)=Acost +Bsin t,t c R, where A and B are real numbers. define addition in V by (f+g)(t)=f(t)+g(t) and scalar multiplication by (¥f)(t)=¥(f(t)),¥€R.prove that V is a vector space over R
To prove that V is a vector space over R, we need to show that it satisfies the vector space axioms.
1. Closure under addition:
Let f, g ∈ V. We need to show that f + g ∈ V.
Given that f(t) = Acos(t) + Bsin(t) and g(t) = Ccos(t) + Dsin(t), where A, B, C, and D are real numbers.
Then (f + g)(t) = f(t) + g(t) = Acos(t) + Bsin(t) + Ccos(t) + Dsin(t).
Rearranging terms, we have (f + g)(t) = (A + C)cos(t) + (B + D)sin(t).
Since (A + C) and (B + D) are real numbers, we can see that (f + g)(t) is a function in V. Hence, V is closed under addition.
2. Closure under scalar multiplication:
Let f ∈ V and α ∈ R. We need to show that αf ∈ V.
Given that f(t) = Acos(t) + Bsin(t), where A and B are real numbers.
Then (αf)(t) = α(f(t)) = α(Acos(t) + Bsin(t)).
Applying scalar multiplication, we have (αf)(t) = αAcos(t) + αBsin(t).
Since αA and αB are real numbers, we can see that (αf)(t) is a function in V. Hence, V is closed under scalar multiplication.
3. Additive identity:
Let 0 be the function defined as 0(t) = 0 for all t ∈ R.
For any f ∈ V, we have (f + 0)(t) = f(t) + 0(t) = f(t) + 0 = f(t).
Since f(t) = Acos(t) + Bsin(t) for some real numbers A and B, we can see that f + 0 = f. Hence, 0 is the additive identity element in V.
4. Additive inverse:
For any f ∈ V, let -f be the function defined as -f(t) = -f(t) for all t ∈ R.
Then (f + (-f))(t) = f(t) + (-f)(t) = f(t) - f(t) = 0.
Hence, every function f in V has an additive inverse element -f in V.
5. Associativity of addition:
For any f, g, and h ∈ V, we have ((f + g) + h)(t) = (f + g)(t) + h(t) = (f(t) + g(t)) + h(t).
By the associativity of addition in R, we can rearrange terms as (f(t) + g(t)) + h(t) = f(t) + (g(t) + h(t)).
Hence, ((f + g) + h)(t) = f(t) + (g(t) + h(t)).
6. Commutativity of addition:
For any f and g ∈ V, we have (f + g)(t) = f(t) + g(t) = g(t) + f(t).
Hence, (f + g)(t) = (g + f)(t).
7. Distributivity:
For any f and g ∈ V, and α ∈ R, we have (α(f + g))(t) = α(f + g)(t) = α(f(t) + g(t)).
Expanding, we have (α(f + g))(t) = αf(t) + αg(t).
Similarly, we have ((α + β)f)(t) = (α + β)f(t) = αf(t) + βf(t).
And for α, β ∈ R and f ∈ V, we have ((αβ)f)(t) = (αβ)f(t) = α(βf(t)).
Hence, V satisfies the distributivity properties.
By satisfying all the vector space axioms, we can conclude that V is a vector space over R.