Linear Algebra and Taylor Series

This assignment delves into concepts of linear algebra and calculus. It explores linear independence and spanning sets of vectors, examines matrix representations of linear transformations, and investigates the dual space. Additionally, it presents problems involving Taylor series expansions, demonstrating their applications in approximating functions.

Q1a) ||x – y||2 = ∫
0
1
¿ ¿
We say that vector v1….vk∈ τ
are linear independent if a0v0 + a1v1…..akvk = ⃗ 0 scalars ai ∈ S
but we have
(-1)v1 + a2v2 + …anvn = ⃗ 0
Since this is a non trival linear combination, S is linearly dependent
If n = 1 and v1 ∈ span (span/{v1} then v1 ∈ span ({}) = {⃗ 0}
b)
f(t) = 1 + t + t2/2
F(t) = (t – a)0/0! * f(a) + (t – a)1/1! + (t – a)2/2! + …
F1(t) = 1 + t
F1(a) = 1 + a
Calculate the second order differential
F11(t) = 1
F11(a) = 1
Substituting the value
F(t) = (t – a)0/0! * f(a) + (t – a)1/1! + (t – a)2/2! + …
Taylor series at a = 0
F(t) = t2 + t2/3 + 2t6/15 + ….
Expression for f(t0) using Taylor series
F(t0) = f(t0) + E1(0)
Writing equation for f(t0 + h) using Taylor series
F(t0+h) – f(t0) = f(t0) + hf1(t0) ) E1(h)
Substituting
F(t0+h) – f(t0) = f(t0) + hf1(t0) ) E1(h) – {f(t0) + E1(0)}
F1(t0) = 1/h((f(t0 + h) – f(t0)) + {E1(h) – E1(0)}

Obtaining truncation error from above equation
E = {E1(h) – E1(0)}
c)
a distance is a function
d:X×X⟶R+
while a norm is a function:
∥⋅∥X⟶R+
So a norm always induces a distance by:
d(x,y)=∥x−y∥
However, the other way around is not always true. For a distance to come from a norm, it needs
to verifiy:
d(αx,αy)=|α|d(x,y)
If we take the discrete distance on any space:
d(x,y)= {1if x= y
0 if x ≠ y
Then this distance does not verify the condition, for example for α=2
d) Now plot x(t) and ^x(t) and comment on the difference this norm makes.
function test
clc
clear all
x = -pi:0.01:pi;
subplot(1,2,1)
plot(x, func1(x), x, func2(x))
hold on
plot(x, func3(x), 'r-', 'LineWidth', 2)
title('Using standard function definition')
subplot(1,2,2)
plot(x, func1(x), x, func2(x))
hold on
func4 = @(x) func1(x) - func2(x);
plot(x, func4(x), 'r-', 'LineWidth', 2)
title('Using anonymous function')

function val = func1(x)
val = sin(x);
function val = func2(x)
val = sin(2*x);
function val = func3(x)
val = func1(x) - func2(x);
Q2a) show linear independence suppose
∑
ij
aijτ ij = OT
For each k = 1,…n applied both side to VR
∑
ij
aijτ ij(vk) = OT(vk) = ⃗ O⃗
O = ∑
j=1
m
∑
i=1
n
aijτ ij = ∑
j=1
m
akjwj
But wj form a basis so all ak,1…ak,m = 0

Showing spanning purpose τ : V →W is linear, then i= 1…n the vector τ ¿vi) is w
τ ¿vi) = ai,1w1 + …ai,mwm
Hence the transformation
ˇτ = ∑
ij
aijτ ij
We claim that this is equal τ
τ ¿vi) = ai,1w1 + …ai,mwm
ˇτ (vk) = ∑
ij
aijτ ij(vk) = ∑
j=1
m
∑
i=1
n
aijτ ij(vk)
= ∑
j=1
m
akjwj
= ak,1w1 +…ak,mwm
b) X = [ x 1,1 x 1,2 .. x 1, N
x 2,1 x 2,2 .. x 2, N
x N , 1 x N , 1 .. x N , N ] = [T]βγ[v]β
Suppose A and B are 2 matrices such as v ∈V
A[v]β = [T(v)]γ = B[v]β
Take v = vk, then [v]β = ek and A[v]β is the kth column of A
Therefore, A and B have all the column, this means A = B
c) since ɸ is an isommer ɸ ( β ) isa basis of V ¿∗¿¿
ɸ(vi)(fvk) = fvk(vi)
which is 1 if I = k and 0 otherwise. This mean ɸ(β) = β**
if W is a subspace, let w ∈W ,letting f ∈ W ⟘
ɸ(w)(f) = f(w) =0
so w ∈ ( w❑ )⟘ since ɸ is an isomorphism ɸ(w) is a subspace of (w⟘)⟘

d)
X =
[ a 0,0 a 1,2 ..a 0 , N
2−1
a 1,0 a 1,1 .. x 2 , N
a N /2 aN
2−1 ,1 .. x N
2−1 , N
2−1 ] =( [T]βγ)τ
Show that is linear if g1, g2 ∈W ¿ and C ∈ F theneach v ∈V
T**(ag1+Ψ(v) = (ag1 + Ψ)(T(v)) = ag1(T(v)) + Ψ(T(v))
= aT*(g1)(v) + T*(Ψ)(v) = (aT*(g1) + T*(Ψ)(v)
Let β = {v1,…vn}, γ = {w1….wm}, β* = {fv1…fvn}
And γ* = {gw1…gwn}
[T]βγ = (aij) , then recall from the lemma that for any f ∈V ¿ we have
F = f(v1)fv1 + ….+ f(vn)fvn
The coefficient of fv1 for T*(gwk) is
T*(gwk)(Vi) = gwk(T(Vi)) = gwk(a1,iw1 + ….am,iwm) = ak,i
This is the kth column of [T* ]γ*β*
.