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 = \(v_e \bigoplus v_0.\) \

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,\(f(-x) = f_1(-x) + f_2(-x)\)
=\(f_1(x) + f_2(x)\)
\(V_{ss3}\): \(\forall c \in R\)
\(E(x) = c f(x)\) \((E(x),f(x) \in v_e, a\in f)\)
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,\(f(-x) = f_1(-x) + f_2(-x)\)
=\(-[f_1(x) + f_2(x)]\)
\(V_{ss3}: \forall c \in R\)
\(O(x) = c f(x) (O(x), f(x) \in v_o, a\in f)\)
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 \(f \in N\)
\(f(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2}\)
\(= g(x) + h(x)\)
\(g(x) = \frac{f(x) + f(-x)}{2}\)
\(h(x) = \frac{f(x) - f(-x)}{2}\)
\(g(-x) = \frac{f(-x) + f-(-x)}{2}\)
\(=\frac{f(-x) + f(x)}{2}\)
so g(x) is even
and,\(h(-x) = \frac{f(-x) - f[-(-x)]}{2}\)
\(= \frac{f(-x) - f(x)}{2}\)
Therefore \(f = g + h\). Where g is even and h is odd.
\(p(x) = \frac{p(x) + p(-x)}{2} + \frac{p(x) - p(-x)}{2}\)
\(= m(x) + n(x)\)
Now, \(m(-x)= \frac{p(-x) + p(x)}{2}\)
\(= \frac{p(x) + p(-x)}{2}\)
\(= m(x)\) .So m(x) is even.
\((-x) =\frac{p(-x) p(x)}{2}\)
\(-[\frac{p(x) - p(-x)}{2}]\)
\(= - n(x)\)
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 \(v_e \bigoplus v_0.\)