Linear Algebra Assignment: Vector Spaces, Norms, and Inequalities

This document presents a comprehensive solution to a linear algebra homework assignment. The solution addresses several key concepts, including evaluating inner products, calculating vector norms, and determining distances and angles between vectors within a vector space. It demonstrates the application of integration by parts to solve definite integrals, calculates the length of vectors using the norm formula, and determines the distance between vectors using the norm of their difference. Furthermore, the solution verifies the Cauchy-Schwarz inequality by computing both sides and demonstrating the inequality holds true. The assignment also proves the parallelogram law using vector algebra and provides a solution for finding an orthonormal basis of a given set of vectors. Finally, it solves a system of linear equations using matrix operations, demonstrating a complete understanding of linear algebra principles.

3. (a)
⟨ f , g ⟩ =∫
a
b
f ( x ) g ( x ) dx
¿∫
−1
1
x2 e−x dx
integrating by parts we have ,letf =x2 , g'=e− x ⟹ f ' =2 x ,
g=−e−x .
∫
−1
1
x2 e− x dx =−x2 e− x−∫−2 x e−x dx .
solving for ∫−2 x e−x dx
−2∫ x e− x dx=−2 x e−x−2∫−e− x dx .
solving for ∫−e−x dx=e− x .
plugging∈the parts we have
∫
−1
1
x2 e− x dx =−x2 e− x−2 x e−x−2 e− x ¿−1
1
¿ e−5 e−1=0.8788
(b)
we define the length of a vector f by
‖f ‖= √ ⟨ f , f ⟩
which is simply thenorm .
This can be evaluated ¿
¿ √ ∫
−1
1
x2 x2 dx= √ ∫
−1
1
x4 dx
¿ √ 1
5 x5 ¿−1
1
¿ √ 1
5 − (−1
5 )
¿ 0.632 5
(c) the distance between f and g can be obtained from
d ( f , g )=‖f −g‖= √ ⟨ f −g , f −g ⟩ .
we need ¿ plug∈the f ∧g functions∧integrate . i. e
d ( f , g )=‖x2−e− x
‖= √ ⟨ x2−e−x , x2−e−x ⟩
¿ √ ∫
−1
1
( x2−e− x ) ( x2−e−x ) dx

¿ √ ∫
−1
1
(x ¿¿ 4−2 x2 e−x +e−2 x )dx ¿
¿ √ [2 x2 e−x +4 x e− x+ 4 e− x− e−x
2 + x5
5 ]−1
1
¿ 1.5064
(d) The angle between f and g can be calculated from the formula
cos θ= ⟨ f , g ⟩
‖f ‖‖g‖
now we need ¿ the|value of ‖g‖since ⟨ f , g ⟩ ∧‖f ‖have been calculated
¿ a∧b above .
‖g‖= √ ⟨ g , g ⟩ = √ ∫
−1
1
e− x e− x dx= √∫
−1
1
e−2 x dx .
we now let u=−2 x ⟹ dx=−1
2 du
‖g‖= √ −1
2 ∫ eu du= √−1
2 eu= √ [ −1
2 e−2 x
]−1
1
=1.90442
Now cos θ=¿ 0.87888
0.6325 ×1.90442 ⟹ θ=cos−1
( 0.87888
0.6325 ×1.90442 )=43.14 ° ¿
4.
¿ verify the Cauchy−s chwarz inequality , we need ¿ show that
f ( x ) =cos x∧g ( x ) =sin x where ⟨ f , g ⟩ =∫
0
π
4
cos x sin x dx
satisfies ⟨ f , g ⟩ 2 ≤ ⟨ f , f ⟩ ⟨ g , g ⟩ . Lets start by evaluating the LHS i . e
⟨ f , g ⟩ 2
= (∫
0
π
4
cos x sin x dx )
2
let u=sin x ⟹ dx= 1
cos x du ⟹∫u du= u2
2 =( [ sin2 x
2 ] 0
π
4
)
2
=0.0625
⟨ f , f ⟩= √∫
0
π
4
cos x cos x dx= √∫
0
π
4
cos2 x dx which can be solved using thereduction formula ¿ get
√∫
0
π
4
cos2 xdx= [ cos x sin x
2 + x
2 ]0
π
4 =0.8017

⟨ g , g ⟩ = √ ∫
0
π
4
sin x sin x dx= √∫
0
π
4
sin2 xdx which can be solved using thereduction formula ¿ get
√∫
0
π
4
sin2 xdx =[ −cos x sin x
2 + x
2 ]0
π
4 =0.3777 .
Now we have
0.0625<0.8017 × 0.3777=0.3028
thus satsfying theinequality ∎
5. To prove the parallelogram law
‖u+v‖2 +‖u−v‖2=2‖u‖2 +2 ‖v‖2
we can expand‖u+v‖2= ⟨ u+v ,u+ v ⟩ =‖u ‖2 + ⟨ u , v ⟩ + ⟨ u , v ⟩ +‖v‖2 =‖u‖2+ 2 ⟨ u , v ⟩ +‖v‖2 …… … …… . ( i )
‖u−v ‖
2= ⟨ u−v , u−v ⟩ =‖u ‖
2− ⟨ u , v ⟩ − ⟨ u , v ⟩ +‖v‖
2=‖u‖
2−2 ⟨ u , v ⟩ +‖v‖
2 … … … … … . ( ii )
Now adding ( i )∧ ( ii ) we have
‖u+v‖2 +‖u−v‖2=‖u‖2 +2 ⟨ u , v ⟩ +‖v‖2 +‖u‖2−2 ⟨ u , v ⟩ +‖v‖2 =2‖u‖2 +2 ‖v‖2
8. B= { ( 7,24,0,0 ) , ( 0,0,1,1 ) , ( 0,0,1 ,−2 ) }
Take u1= ( 7,24,0,0 ) ,u 2= ( 0,0,1,1 ) ,u 3= ( 0,0,1,−2 )
let v 1=u 1 , e 1= v 1
‖v 1‖ = ( 7,24,0,0 )
√ 72 +242+ 0+0 = ( 7
25 , 24
25 , 0,0 )
v 2=u 2− ⟨ u 2 , e 1 ⟩ e 1= ( 0,0,1,1 )− (0 × ( 7
25 , 24
25 , 0,0 ) )= ( 0,0,1,1 )
e 2= v 2
‖v 2‖= ( 0,0,1,1 )
√0+0+1+1 = (0,0 , 1
√2 , 1
√2 )
v 3=u 3− ⟨ u3 ,e 1 ⟩ e 1− ⟨ u 3 , e 2 ⟩ e 2= ( 0,0,1 ,−2 ) −0− ⟨ ( 0,0,1 ,−2 ) , (0,0 , 1
√2 , 1
√2 ) ⟩ (0,0 , 1
√2 , 1
√2 )=(0,0 ,
e 3= v 3
‖v 3‖= (0,0 , 3
2 , −3
2 )
√0+ 0+( 3
2 )2
+ (−3
2 )2 =(0,0 , √2
2 ,− √ 2
2 )

B' = {( 7
25 , 24
25 , 0,0 ) , ( 0,0 , 1
√ 2 , 1
√ 2 ) , ( 0,0 , √ 2
2 ,− √ 2
2 ) }
9. X1
( 3
5
4
5
0 ) + X2
( −4
5
3
5
0 ) + X2 ( 0
0
1 )= ( 10
25
5 ) we can find the values of X1 , X 2 , X3
3
5 X1 − 4
5 X2+ 0 X3=10 … … … . ( i )
4
5 X1 + 3
5 X2 +0 X 3=25 … … ..(ii)
X3 =5 …… …(iii)
¿ ( i )∧ ( ii ) we have
3 X1−4 X2 =50 , 4 X1 +3 X2 =125 , solving X1=26 , X2=7 hence XB =
[ 26
7
5 ]
References:
Seymour (1976) .Schaum’s outline Linear Algebra 4th ed. Mac-GrawHill
Gilbert s.(1993) Introduction to linear algebra 2nd ed.
Sheldon A.(1995). Linear Algebra done right. Springer
Steven R.(1992).Advanced Linear algebra. Springer.