**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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