**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.

**Question:** 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 of a :**

The answer is no. The empty set is empty in the sense 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 how that these property are unique for this we proceed as follows,

Let U be any element 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,

$n(-x) =\frac{p(-x) p(x)}{2}-[\frac{p(x) - p(-x)}{2}]$

$ = - 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$