Linear Algebra Assignment Solutions

This document contains solutions to Linear Algebra assignments and problems. It includes topics such as basis of P2, orthogonal basis of P2, inner product, and more.

Running head: ASSIGNMENT 1
Linear Algebra
Student Name
Institution

ASSIGNMENT 2
Question 1
Question 1(a)
Let B= { x2, x,1} et S = { x2+x,2x-1,x+1} be two basis of P2
Let iB and is be the coordinate map induced on P2
B= { x2, x,1} T: P2 ⟶ P2
S= S = { x2+x,2x-1,x+1}
B and S are two basis of P2
[T ]B , S = [ 1 2 0
−1 3 5
2 2 −2 ] = {xϵ R3| T x= 0}
[ 1 2 0
−1 3 5
2 2 −2 ] [ x1
x2
x3 ] = [ 0
0
0 ]
x1 - 2x2 = 0
x1 = 2 x2 ……………………. Equation (1)
- x1 +3 x2 +5 x3 = 0 ……………………… Equation (2)
2x1 + 2x2-2 x3 =0 ……………………….. Equation (3)
From Equation (1) and Equation (2)
-2 x2 +3 x2 +5 x3 = 0
x2 +5 x3 = 0
From Equation (1) and Equation (3)
4 x2 +2x2 -2 x3 = 0
6 x2 -2 x3 = 0
3 x2 - x3 = 0

ASSIGNMENT 3
Therefore, x3 = 3 x2 = 3( x2
2 ¿
x3 = 3 x2
2
x2 = x1
2 ⟶ x1 (1, 1
2 , 3
2 )
Null ( [T ]B ,θ) = {c(1, 1
2 , 3
2 ) ; cϵ R }
Null [ T BS ] = Span
{( 1
1
2
3
2 ) }
To find I B (T ) ,
[ 1 2 0
−1 3 5
2 2 −2 ]3∗3 [ x1
x2
x3 ]3∗1
= [ x1+2 x2
−x1+3 x2 +5 x3
2 x1 +2 x2 −2 x3 ]
= x1
[ 1
−1
2 ] + x2
[ 2
3
2 ] + x3
[ 0
5
−2 ]
Thus
[ 0
5
−2 ] = −2 [ 1
−1
2 ] + 1 [2
3
2 ]
= x1
[ 1
−1
2 ] + x2
[ 2
3
2 ] - 2 x3
[ 1
−1
2 ] + x3
[2
3
2 ]
= ( x1 -2 x3)
[ 1
−1
2 ] + ( x2 + x3)
[2
3
2 ]

ASSIGNMENT 4
I B (([T ])B , s) span {( 1
−1
2 ) , ( 1
3
2 ) }
Therefore,
(T BS)= Span {( 1
−1
2 ), (2
3
2 ), (0
5
2 ) }
Question 1(b)
T: M 3∗3 ⟶ P2
T
(a b c
d e f
g h i ) = (a-b) x2 +c(x-1)+d
T
(1 0 0
0 0 0
0 0 0 ) = x2
T
(0 1 0
0 0 0
0 0 0 ) = (-1) x2
T
(0 0 0
0 0 0
1 0 0 ) = 1(X-1)
T
(0 0 0
0 0 0
0 0 1 ) = 1
Now,
T
(a b c
d e f
g h 1) = 0, a=1, e=1,i=1 and b,c,d,f,g,h = 0

ASSIGNMENT 5
So the basis of Ker(T) = ( 1 0 0
0 1 0
0 0 1 )
And the basis of Image of T (Im(T) )
= {1,X, X2}
Question 2
We have shall check the following inner product for < , ∙ ,> ; let u,v,wϵ v and λ is a scalar then;
a) <u,v>1 =¿ T ( u ) , T ( v )> ¿
¿ T ( v ) ,T ( w ) >¿ ¿ ¿
= ¿ v , u>¿ ¿ [∵ <,> is an inner product on V]
b) <u+v,w>1 = ¿ T ( u+ v ) ,T ( w ) >¿
= ¿ T ¿
= ¿ T ¿
=<u,w>1 =+ <v,w>1
c) <λu+v>1 = ¿ T ( λu ) , T ( v ) >¿
= ¿ λT ¿ [∵ T is linear]
= ¿ T ¿
=<u,w>1 =+ <v,w>1
d)
<u,u>1 = <T(u),T(w) ≥ 0 and <T(u) , T(u)> = 0 iff T(u) = θ
where θis the null element of u=θ [∵ T is injective ]
so, <u,u> ≥ 0 for all v and <u,u>1 for all uϵ V and <u,u>1 = 0 iff u=0
hence <,>1 satisfies all axioms of inner product and it is a inner product on V
If T is positive definite matrix, then T will give n inner product

ASSIGNMENT 6
<f,g> = ∫
0
1
(f ( t ) −f ' ( t ) )( g ( t ) −g' ( t ) )dt
<f,f> = ∫
0
1
(f ( t )−f ' ( t ) )2 dt
If <f,f> = 0 ⟹ f ( t )−f ' ( t ) = 0
⟹ f ( t )=f ' ( t )
⟹ f ( t )=et ∀ t and f ϵ pn
But et ∉ pn thus not an inner product since <f,f> = 0 ≇ f = 0 hence false
Question 3
Question 3(a)
(f,g) = ∫
0
∞
e−t f(t)g(t) dt
P(x) = x3
The basis of P2 is {1,x,x2}
We need to determine the orthogonal basis of P2
Let
v1 = 1 ………………….. Equation 1
v2 = x ………………….. Equation 2
v3 = x3 ………………….. Equation 3
Therefore by Gram Schmidt orthogonalization process, we get
w1 =1 …………………………………… Equation 4
v1 = 1 ……………………………….. Equation 5
w2 = v2- (v2 , w1 )
(w1 , w1)…………………………..Equation 6

ASSIGNMENT 7
From equation 1 and 2,
w2 = x- ( x ,1)
(1,1) .1
w2 = x- 1
(x,1) = ∫
0
∞
e−t .t.1 dt = [ −t e−t−e−t ] ∞
0 = 1
(1,1) = ∫
0
∞
e−t .1 dt = [ −e−t ] ∞
0 = 1
w3 = v3- ( v3 , v1)
( w1 , w1)* w1 - (v3 , w2 )
( w2 , w2)
= x2 – ( x2 ,1)
(1,1) - ¿ ¿(x-1)
Therefore,
¿) = ∫
0
∞
e−t .t2 dt= 0+2 = 2
¿) = ∫
0
∞
e−t .t2 (t−1) dt
= ∫
0
∞
t3 e−tdt - ∫
0
∞
t2 e−tdt
= 0+(3*2)-2
=4
(x-1,x-1) = ∫
0
∞
e−t .(t−1)(t−1) dt
= ∫
0
∞
e−t .(t2−2 t+1) dt
= 2-2+ [ e−t ] ∞
0

ASSIGNMENT 8
=2-2+1
= 1
Therefore,
w3 = x2- 2
1- 4
1 (x-1)
= x2- 2−4 (x−1)
= x2-2-4x+4
= x24x+2
Thus, orthogonal basis of P2 IS {1,x-1, x24x+2}
Orthogonal projection of P(x) = x3 onto P2 = Projw1P(x) Projw2P(x)+Projw3P(x)
= ( P ( x ) , w1)
( w1 , w1 ) w1 + ( P ( x ) , w2)
(w2 , w2 ) w2+( P ( x ) , w3)
( w3 , w3 ) w3
(P ( x ) , w1) = ( x3, 1)
= ∫
0
∞
e−t .t3 dt
= 0+(3*2)
= 6
(P ( x ) , w2) = ( x3, x-1)
= ∫
0
∞
e−t .t3 (t−1)dt
= ∫
0
∞
e−t t4dt - ∫
0
∞
e−t t3dt
= 0+(24-6)
= 18
(P ( x ) , w3) = (x3, x2−4 x+2)

ASSIGNMENT 9
= ∫
0
∞
e−t .t3 (t2 −4 t +2)dt
= (0+5∗18)-(4*18+12)
= 0+(24-6)
= 18+12
= 30
(w3 , w3 ¿ = ∫
0
∞
e−t .(t2−4 t+2)(t2−4 t+2)dt
= ∫
0
∞
e−t .¿-8t +2t2 -8t +4t) dt
= ∫
0
∞
e−t .¿-8t +2t2 -8t +4t) dt
= ∫
0
∞
e−t .¿) dt
= 24-8*6+20*2-16*1+4
= 4
Therefore, orthogonal projection of P(x) = x3 onto P2 is
= 6*1+18 w2 + 30
4 w3
= 6+18(x-1)+ 15
2 ¿ ¿ -4x+2)
= 6+18x-18+ 15
2 x2 -30x+15
= 15
2 x2 -12x+3
Question 3(b)
∈ = {f*(P2|f(0) = 0}

ASSIGNMENT 10
Let a+bx+(x2+a x3 ∈ E)
⟹ a+b(0) +c(0)2 +d(0)3-E
a+b(0) +c(0)2 +d(0)3-E = 0
a=0
Therefore, the elements of the form +( x2+a x3 ∈ E) and it is spanned by x, x2, x3 . Therefore basis
of E is (x,x2, x3) .
Question 4
Question 4
In order to prove u0 is unique in T(v) = < u0 , v> ∀ r ϵ v
Let u0 is not unique ⟹ T(v) = < u0 , v>
T(v) = < u1 , v>
⟹ < u0 , v> = < u1 , v>
⟹ < u0-u1, 0> = 0
So by property of inner product ⟹ u0-u1 (u0=u1) hence u0 is unique
Now to find qϵ p2 ∀ P ϵ P2
P(0) = ∫
−1
1
q (t)p(t) dt
Let q(t) = A+Bt+C t2
P(0) = ∫
−1
1
( A +Bt +C t2)p(t) dt
Let P(t) = 1 ⟹ P2 ( x) ⟹ 1 = ∫
−1
1
( A +Bt +C t2) dt
1= 2A+ 2C
3 ……………………………………………. Equation 1

ASSIGNMENT 11
P(t) = t ⟹ 0 = ∫
−1
1
( At + B t2+C t3 )dt
P(0) = 0 ⟹ 0 = 0+ 2 B
3 +0 ⟹ B=0 ……………………. Equation 2
Let P(t) = t2 ⟹ P2
P(0) = 0
0 = ∫
−1
1
( A t 2 +B t3 +C t4 )dt
0 = 2A ( t3
3 )
'
+0 +2C( t5
5 )
'
0 = 2 A
3 + 2C
3 = 0 ⟹ 5A = -3C ………………. Equation 3
From equation 1,
1= 2 A+ 2C
3
⟹ 1= −8
15 C
C= −15
8
A= 9
8
Therefore,
q(t) = 9
8 + 0.t- 15
8 t2
q(t) = 9
8 - 15
8 t2
Question 5
Question 5(a)

ASSIGNMENT 12
Given, A= [ 3 2 2
2 3 2
2 2 3 ]
The characteristic equation of A is, |3−x 2 2
2 3−x 2
2 2 3−x|x=0
i.e (3-x)*[(3−x)2−4] – 2*[6-2x-4] +2*[4-+2x] = 0
(3-x)*[5−6 x+ x2] + 8x -8 =0
x3 - 9x2 +15x -7 =0
(x−1)2 (X-7) = 0
X= 1, 1, 7
Thus the Eigenvalues of A are 1, 1, and 7 then the Eigenvectors are given as:
c
[
−1
1
0 ] +d [−1
0
1 ] Where c, d ≠ 0
The corresponding Eigenvector to the above Eigenvalue 7 is :
c
[1
1
1 ]
Let P = [−1 −1 1
1 0 1
0 1 1 ] and a non-singular matrix
Then,
AP = [3 2 2
2 3 2
2 2 3 ][−1 −1 1
1 0 1
0 1 1 ] = [ −1 −1 7
1 0 7
0 1 7 ]
PD = [−1 −1 1
1 0 1
0 1 1 ][1 0 0
0 1 0
0 0 1 ] = [−1 −1 7
1 0 7
0 1 7 ]

ASSIGNMENT 13
Thus, AP = PD
i.e A= PDPT where P-1= PT
Question 5(b)
The maximum value of G(u)
G
[u1
u2
u3 ] = 3 ( u1
2+ u2
2+ u3
2 ) +3 (u1 u2+u2 u3 +u3 u1 )
u1
2 +u2
2 +u3
2
Then the maximum value of G(u) is at (-1,0,1), (0,-1,1), (-1,1,0),(1,0,-1),(1,-1,0),and (0,1,-1)