Question:a. Is the empty set a subspace of every vector space ? Justify and let s = $${(a, b, (a + b):a, b \in R)}$$ . Is a subspace of $$R^3$$ under the usual operation.
b. Let I = (-a , a) a > 0 be an open interval in R let v = Rt the space of all real valued function define on I. Let $$v_e = {f\in v : f(-x) = f(x)} \forall x \in I$$ the set of all even function on I. And let $$v_o = {f \in v :f(-x) = -f(x)\forall x \in I}$$, the set of all odd function on I. Then show that v = $$v_e \bigoplus v_0.$$ \

Solution:
The answer is no. The empty set is empty in the sence that it does not contain any elements. Thus a zero vector is not member of the empty set. Without zero we can not say that it is subspace of vector space.
Here s = $$[a, b, (a + b): a, b \in R]$$
$$\forall a \exists –a$$
$$V_{ss1}: a + (-a) = 0 \in s$$
$$V_{ss2}: a + b \in s$$
$$V_{ss3}: for c \in R$$
s.t $$ca \in s$$.
Hence three condition of vector sub space are satisfy so, s is vector subspace.\

For solution of b.
Here given two function are define by $$v_e = [f \in v : f(-x) = f(x) \forall x \in I]$$ and $$v_0 = [f \in v :f(-x) = -f(x) \forall x\in I]$$
First we show that $$v_e$$ and $$v_o$$ are subspace of vector space.
For even:
Here $$v_e = [ f \in v : f(-x) = f(x) \forall x \in I]$$
$$V_{ss1}$$: Since constant function is even . 0 is even function.
$$0 \in v_e$$
$$V_{ss2}: \forall x \in I$$
Let $$f_1(x), f_2(x) \in v_e$$
Define, $$f(x) = f_1(x) + f_2(x)$$
Now,$$f(-x) = f_1(-x) + f_2(-x)$$
=$$f_1(x) + f_2(x)$$
=$$f(x)$$
$$V_{ss3}$$: $$\forall c \in R$$
$$E(x) = c f(x)$$ $$(E(x),f(x) \in v_e, a\in f)$$
Now, $$E(-x) = cf(-x) = cf(x) = E(x)$$
Therefore $$E(x) = E(-x)$$
Hence $$cf(x) \in v_e$$
Therefore the set of all even function on I i.e; $$v_e$$ is subspace of V.

For odd:
Here $$v_o = [f \in v :f(-x) = -f(x) \forall x\in I]$$
$$V_{ss1}$$: Since constant function is odd . 0 is odd function.
$$0 \in v_0$$
$$V_{ss2}: \forall x \in I$$
Let $$f_1(x), f_2(x) \in v_o$$
Define, $$f(x) = f_1(x) + f_2(x)$$
Now,$$f(-x) = f_1(-x) + f_2(-x)$$
=$$-[f_1(x) + f_2(x)]$$
=$$-f(x)$$
$$V_{ss3}: \forall c \in R$$
$$O(x) = c f(x) (O(x), f(x) \in v_o, a\in f)$$
Now, $$O(-x) = cf(-x) = -cf(x) = -O(x)$$
Therefore $$O(-x) =-O(x)$$
Hence $$cf(x) \in v_o$$
From above result we can say that $$v_0$$ is the subspace of V for direct we have to show that $$V_e \cap v_o = 0$$
$$V_e + V_o = V$$
Also, we need to ahow that these property are unique for this we proceeed as follows,
Let U be any elenment of $$V_e \cap v_o$$. It means that U is both even and odd function.
Now, for even function.
$$U(-x) = U(-x)$$……(1)
for odd function.
$$U(-x) = -U(x)$$…….(2)
From (1) and (2) we can write,
$$U(x) = -U(x)$$
$$2U(x) = 0$$
$$U(x) = 0$$.…..(3) It means that $$V_e \cap v_o = {0}$$. Intersection contains only zero element.
Also,Let $$f \in N$$
Clearly,
$$f(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2}$$
$$= g(x) + h(x)$$
Where,
$$g(x) = \frac{f(x) + f(-x)}{2}$$
$$h(x) = \frac{f(x) - f(-x)}{2}$$
Now,
$$g(-x) = \frac{f(-x) + f-(-x)}{2}$$
$$=\frac{f(-x) + f(x)}{2}$$
$$g(x)$$
so g(x) is even
and,$$h(-x) = \frac{f(-x) - f[-(-x)]}{2}$$
$$= \frac{f(-x) - f(x)}{2}$$
$$-h(x)$$
Therefore $$f = g + h$$. Where g is even and h is odd.
Uniqueness
Clearly,
$$p(x) = \frac{p(x) + p(-x)}{2} + \frac{p(x) - p(-x)}{2}$$
$$= m(x) + n(x)$$
Now, $$m(-x)= \frac{p(-x) + p(x)}{2}$$
$$= \frac{p(x) + p(-x)}{2}$$
$$= m(x)$$ .So m(x) is even.
Similarly,
$$(-x) =\frac{p(-x) p(x)}{2}$$
$$-[\frac{p(x) - p(-x)}{2}]$$
$$= - n(x)$$
Therefore n(-x) = n(x).So n is the odd function.
p = m + n.Where m is even and n is odd function.
From above result we see,
$$v_e \bigoplus v_0.$$…..(4)
From (3) and (4) we can say that $$v_e \bigoplus v_0.$$

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