Linear Algebra Assignment: Vector Spaces, Subspaces, and Spans

This document presents a solution to a Linear Algebra assignment, addressing key concepts such as vector spaces, subspaces, and spans. The solution begins by examining whether certain operations on vector components hold true, demonstrating counterexamples where necessary. It then proves that the set of all functions from a set X to a field F constitutes a vector space and establishes that a given set W is a subspace of F(R, C). Furthermore, the solution shows that the sum of two subspaces, W1 and W2, is a subset of the vector space V containing them. It also proves that Span (S1 ∩ S2) is a subset of Span(S1) ∩ Span(S2) and provides examples to illustrate the relationship between them. Finally, the solution addresses the task of finding a subspace U such that P(F) = U + W, where P(F) is the vector space of all polynomials over F and W is a specified subspace.

1.
(a) c* (x1, x2) = (0, cx2)
let say, c=2, x1= i+j, x2= 2i-3j
L.H.S. = c*(x1, x2)
= 2*(i+j, 2i-3j)
= (2i+2j, 4i-6j)
R.H.S.= (0, cx2)
= (0, 2*(2i-3j))
= (0, 4i-3j)
Here, L.H.S.≠ R.H.S.
Hence we can say that c* (x1, x2) = (0, cx2) is not always true.
(b) c * (X1, x2) = (cx1, c2x2)
Let’s say, c = 3, x1 = i-2j, x2= 4i-j
L.H.S. = c * (X1, x2)
= 3*(i-2j, 4i-j)
= (3i-6j, 12i-3j)
R.H.S. = (cx1, c2x2)
= (3(i-2j), 9(4i-j))
= (3i-6j, 36i-9j)
Here, L.H.S.≠ R.H.S.
Hence we can say that c * (X1, x2) = (cx1, c2x2) is not always true.
(c) c * (x1, x2) = { ( x 1, x 2 ) if c =1
( cx 1 , cx 2 ) otherwise
This axiom is always true.
C*(x1, x2) = (cx1, cx2)
At c=1,
C*(x1, x2) = 1*(x1,x2) = (x1, x2)
2.
(a) Let X be a non-empty set and F a field.
Vector pointwise operations are:
Sum: f + g  f(x) + g(x)
Product: α(f)  αf(x)
Therefore, set of all functions from x to F follows the vector rules, hence it is a vector space.

(b) ʄ (t) ∈ F(R, C)
ʄ (-t) ∈ F(R, C)
( ʄ (t) - ʄ (-t)) ∈ F(R, C)
ʄ (t) - ʄ (-t) ∈ W
W ∈ F(R, C)
Hence W is subspace of F(R, C).
3.
W1 and W2 are subspaces of vector V.
W1 ∈ V
W2 ∈ V
W1 + W2 ∈ V
V is composed of n subspaces. Hence V consists all of them.
V = W1 ∪ W2 ∪ W3 ∪ …. Wn …………..{Equ. 1}
Hence W1 ∪ W2 is subsection of W1 ∪ W2 ∪ W3 ∪ …. Wn
W1 ∪ W2 ∈ (W1 ∪ W2 ∪ W3 ∪ …. Wn)
W1 ∪ W2 ∈ V
Now,
W1 ∪ W2 = W1 + W2
W1 + W2 ∈ (W1 ∪ W2 ∪ W3 ∪ …. Wn) ……………{Equ.3}
From Equ.1 and Equ.3,
W1 + W2 ∈ V
4.
(a) Let v ∈ Span(S1∩S2).
Then, by definition of span, v can be expressed as a linear combination
v = c1v1+ c2v2+ … +cmvm
{where (m >=0), v1, v2, …. vm are from set S1 ∩ S2, and c1, c2, …. cm are scalars.
The element vi is from S1, so v ∈ Span(S1), and they also lie in S2, so v ∈ Span(S2) as well.

Thus, v ∈ Span(S1) ∩ Span(S2).
Since v was an arbitrary element of Span (S1 ∩ S2), we have proved that Span (S1 ∩ S2) ⊆
Span(S1) ∩ Span(S2).
(b) Example1: If S1 = S2, then Span (S1 ∩ S2) = Span(S1) = Span(S2) = Span(S1) ∩ Span(S2).
(As a concrete example, we could take V = R, S1 = S2 = ∅.)
Example2: Let V = R, S1 = {1} and S2 = {2}, Then S1∩S2 = ∅, so Span (S1 ∩ S2) = Span (∅) = {0},
while Span(S1) = R = Span(S2) = Span(S1) ∩ Span(S2).
5.
Let P(F) denote the vector space of all polynomials over F and consider the subspace
W = {p(x) ∈ P(F) : p(0) = p(1) = 0}. Find a subspace U such that P(F)= U + W.
W= {0} =∅