# Homework Assigment 1 of linear Algebra

Solution set 1

Excercise. 1.Does every vector space contain a zero vector? Justify.Let s = {0,1} and f = R in f(S, R) . Show that f = g and f + g = h where f(t) = 2t + 1 , $g(t) = 1 + 4t - 2t^2$ and h(t) = 5t + 1.

Solution:Yes, every vector space contaion a Zero vector . Zero vector is a identity under addition. To be vector space it is necessary to satisfy identity property.

Second part

Here s = { 0, 1} and f = R in F (S, R)

also,
f(t) = 2t + 1, g(t) = 1 + 4t – 2$t^2$ and h(t) = 5t + 1 .

Take t = 0

f(0) = 2.0 + 1 = 1

	g(0) = 1 + 4.0 – 2.(0)^2 = 1

h(0) = 50 + 1 = 2

hence f(0) = g(0) so f(t) = g(t)


Take t = 1

f(1) = 2.1 + 1 = 3

	g(1) = 1 + 4.1 – 2.(1)^2 = 3

h(1) = 51 + 1 = 6

hence f(1) + g(1)  = h(1)

so f(t) + g(t) = h(t)  verified.


2.May a vector space have more than one zero vector?justify. Let $v = {(a_1, a_2):a_1, a_2 \in R}$. for $(a_1, a_2)$, $(b_1, b_2)$ = $(a_1 + 2b_1, a_2 + 3b_2)$ and $c(a_1, a_2)$ = $(ca_1, a_2)$.Is v a vector space over f with these operations?justify your answer.

Solution:
Let a vector space have two zeros say 0 and 0^’ . From $v_{s3}$ 0 is the identity under addition
Hence $\forall x \in V$
x + 0 = 0 + x = x…………(1)
also from $v_{s3}$ 0’ is identity under addition.
x + 0’ = 0’ + x = x……….(2)
From (1) and (2)
x + 0 = x +$0^’$ = x .
Hence there is unique zeros in vector space.

Second part

Here given
v = ${(a_1, a_2):a_1, a_2 \in R}$ for $(a_1, a_2)$, $(b_1, b_2) \in V$ and $c \in R$
Define $(a_1, a_2) + (b_1 , b_2)$ = $(a_1 + 2b_1, a_2 + 3b_2)$ and $c(a_1, a_2)$ = $(ca_1, a_2)$
$V_{s1}:$ for $(a_1, a_2)$, $(b_1, b_2) \in V$ $(a_1, a_2) + (b_1, b_2)$ = $(a_1 + 2a_2, b_1 + 3b_2)$ $\neq$ $(a_2 + 2a_1, b_2 + 3b_1)$ $\neq$ $(b_1, b_2) + (a_1, a_2)$
$V_{s1}$ does not hold.
$V_{s2}$: for $(a_1, a_2)$, $(b_1, b_2) \in V$ and $(c_1, c_2) \in V$
$[(a_1, a_2) + (b_1, b_2)] + (c_1, c_2)$ = $[ a_1 + 2b_1, a_2 + 3b_2] + (c_1, c_2)$ $\neq$ $(a_1 + 2b_1 + c_1, a_2 + 3b_2 + 3b_3)$ $\neq$ $(a_1, a_2)$ + $[ b_1 + 2c_1, b_2 + 3c_2]$ $\neq$ $(a_1, b_2)$ + $[(b_1, b_2) + (c_1, c_2)]$
$V_{s2}$ does not hold.
$V_{s3}$:if there exsist a element z = $(d_1, d_2) \in V$ such that $(a_1, a_2) + (d_1, d_2)$ = $(a_1, a_2)$
$(a_1 + 2d_1, a_2 + 3d_2)$ = $(a_1, a_2)$
$a_1 + 2d_1$ = $a_1$ $\Rightarrow d_1 = 0$
$a_2 + 3d_2$ = $a_2 \Rightarrow d_2 = 0$
This show that $(a_1, a_2) + (0, 0)$ = $(a_1, a_2)$
$v_{s4}$:Let $\forall$ $(a_1, b_1)$$\exists (-a_1, -a_2) \in V\ such that (a_1, a_2) + (-a_1, -a_2) = [a_1 +(-2a_1), a_2 + (-3a_2)] = (-a_1, -2a_2) \neq (0, 0) .\ Hence v_{s4} does not satisfy.\ V_{s5}: 1 \in F 1.(a_1, b_1) = (1.a_1, b_1) = (a_1, b_1)\ V_{s6}: \forall x, y \in F$$xy(a_1, b_1)$ = $x(ya_1, b)$ = $x[y(a_1, b1)]$
$V_{s7}$:$\forall x, y \in F$ (x+y) $(a_1, b_1)$ = $[(x+y)a_1, b_1]$ = $(xa_1 + ya_1, b_1)$
also $x(a_1, b_1)$ + $y(a_1, b_2)$ = $(xa_1, b_1)$ + $(ya_1, b_1)$ = $(xa_1 + ya_1, b_1 + b_2)$ = $(xa_1 + ya_1, 2b_1)$
(x+y)$(a_1, b_1)$ ≠ $x(a_1, b_1) + y(a_1, b_1)$
Hence $v_{s7}$ does not satisfy.
$V_{s8}$: $\forall x \in F$ $x[(a_1, b_1) + (a_2, b_2)]$ = $x(a_1, b_1) + x(a_2, b_2)$ = $(xa_1, b_1) + (xa_2, b_2)$
All the condition of vector space does not satisfy hence it is not vector space.

3.Does ax = bx imply a = b in any vector space? Justify.let v = ${(a_1, a_2):a_1, a_2 \in f}$ where f is field. Define addition of elements of v coordinate wise, and for $c \in f$ and $(a_1, a_2) \in V$ . Define $c(a_1, a_2)$ = $(a_1, 0)$ .Is vector space over F with these operations? Justify your answer by taking all condition of vector space.

Solution: In any vector space ax = bx implies that a = b false
It is only true if we take x = 0
i.e ; ax = bx \
x(a – b) = 0 either x = 0 or ( a-b) = 0
take x = 0 and any a≠b for x.

Let us suppose $(a_1, a_2)$, $(b_1, b_2) \in V$ then
$V_{s1}$: $(a_1, a_2)$ + $(b_1, b_2)$ =$(a_1 + b_1, a_2 + b_2)$ = $(b_1 + a_1, b_2 + a_2)$ = $(b_1, b_2) + (a_1, a_2)$
$V_{s2}$: $[(a_1, a_2) + (b_1, b_2)] + (c_1, c_2)$ = $(a_1, a_2) + [(b_1 , b_2) + (c_1, c_2)]$ $\forall (c_1, c_2) \in f$
$V_{s3}$: $\exists (d_1, d_2) \in V$ s.t $(a_1, a_2) + (d_1, d_2)$ = $(a_1, a_2)$
$(a_1, a_2) + (0, 0)$ =$(a_1, a_2)$
$(d_1, d_2)$ = $(0, 0)$
$V_{s4}$:$\forall (a_1, a_2) \in V \exists (-a_1, -a_2) \in V$
s.t $(a_1, a_2) + (-a_1, -a_2)$ = $(0, 0)$ $V_{s5}$: $\forall \in F 1.(a_1, a_2)$ = $(a_1 ,0)$ = $(a_1, 0)$ ≠ $(a_1, a_2)$
Hence $v_{s5}$ does not satisfy. $V_{s6}$:$\forall x, y \in F$ $xy(a_1, a_2)$ = $x(ya_1, 0)$ $V_{s7}$: $\forall x, y \in F$
$(x + y)(a_1, a_2)$
= $[(x + y)a_1, 0]$ =$[xa_1 + ya_1, 0]$ =$(xa_1, 0) + (ya_1, 0)$

$V_{s8}$: $\forall x \in f$
$x[(a_1, a_2) + (b_1, b_2)]$ = $x[a_1 + b_1, a_2 + b_2]$ =$[x(a_1 + b_1), 0]$ =$(xa_1, 0) + (xb_1, 0)$ =$x(a_1, a_2) + x(b_1, b_2)$
$V_{s5}$ does not satisfy it is not vector space.
4.Does ax = bx imply x = y in any vector space? justify. let v denote the set of ordered pairs of real numbers.if $(a_1, a_2)$ and $(b_1, b_2)$ are element of v and $c \in R$ , define $(a_1 , a_2)$ + $(b_1 , b_2)$ =$(a_1.b_1, a_2 + b_2)$ and $c(a_1, a_2)$ = $(a_1, ca_2)$ . Is v a vector space over R with these operations ? Justify your answer by showing all conditions of vector space
Solution
In any vector space ax = ay implies that x = y is false.
It is only true if we take a = 0
i.e; ax = ay a(x – y) = $0 \in V$
either a = 0 , or ( x – y) = 0 if we take a = 0 and any x ≠ y for a.
Second part
Here v denote the set of ordered pairs of real numbers. If $(a_1, a_2)$and $(b_1, b_2)$ are element of v and $c \in R$ ,define $(a_1 , a_2)$ + $(b_1 , b_2)$ = $(a_1.b_1, a_2 + b_2)$ and $c(a_1, a_2)$ = $(a_1, c.a_2)$.
$V_{s1}$:$(a_1, a_2) + (b_1 , b_2)$ = $(a_1.b_1, a_2 + b_2)$ = $(b_1.a_1 , b_2 + a_2)$ =$(b_1, b_2) + (a_1, a_2)$
$V_{s2}$: $\in [(a_1, a_2) + (b_1, b_2)] + (c_1, c_2)$ = $(a_1.b_1, a_2 + b_2) + (c_1, c_2)$ = $(a_1.b_1.c_1 , a_2 + b_2 + c_2)$ = $(a_1, a_2) + (b_1.c_1 , b_2 + c_2)$ = $(a_1, a_2) + [(b_1, b_2) + (c_1 , c_2)]$
$V_{s3}$: $\exists$$(d_1, d_2) \in V$
s.t$(a_1, a_2) + (d_1, d_2)$ = $(a_1, a_2)$
Or,$(a_1.b_1 , a_2 + b_2 )$ =$(a_1 , a_2) \Rightarrow a_1.b_1 = a_1 \Rightarrow b_1 = 1$
Also,$a_2 + b_2$ = $a_2 \Rightarrow b_2 = 0$
Hence$(d_1, d_2)$ = $(1, 0)$
so, $(a_1, a_2) + (d_1, d_2)$ ≠ $(a_1 , a_2)$
$v_{s3}$ does not satisfy.
$V_{s4}$: $\forall (a_1 , a_2) \exists (-a_1, -a_2) \in V$ s.t $(a_1, a_2) + (-a_1 , -a_2)$ = $[a_1.(-a_1) , b_1 + (-b_1) ]$ =$(-a_1^2, 0)≠0$
Hence $v_{s4}$ also not satisfy.
$V_{s5}$: for $1\in R$ $1.(a_1, a_2) = (a_1, 1.a_2) = (a_1, a_2)$
$V_{s6}$: $\forall x, y \in R$
$xy(a_1, a_2)$ =$x(a_1, ya_2)$ = $x[y(a_1, a_2)$
$V_{s7}$:$\forall x, y \in R$
$(x + y) [(a_1, a_2) + (b_1 , b_2)]$=$(x + y)(a_1, a_2) + (x + y) (b_1, b_2)$
L.H.S = $(x + y)(a_1.b_1, a_2 + b_2)$
= $[a_1.b_1 , (x + y)(a_2 + b_2)]$
=$[a_1.b_1 , (x + y) a_2 + (x + y) b_2]$
R.H.S = $[ a_1 , (x + y)a_2] + [b_1 , (x + y)b_2]$
= $[a_1.b_1 , (x + y)a_2 + (x + y) b_2]$
L.H.S = R.H.S
$V_{s8}$: $\forall x \in R$
$x[(a_1, a_2) + (b_1, b_2)]$ =$x(a_1, a_2) + y(b_1, b_2)$ L.H.S
= $x[ a_1.b_1 , a_2 + b_2]$
= $[a_1.b_1 , x(a_2 + b_2)]$
= $[a_1.b_1 , xa_2 + xb_2]$
R.H.S
= $(a_1 , xa_2) + (b_1 , xb_2)$ = $(a_1.b_1 , xa_2 + xb_2)$
L.H.S = R.H.S
Here$v_{s3}$ and$v_{s4}$ are not satisfy .Hence it is not vector space.

5.
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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