Math 115A (UCLA) Final Exam Solution: Integration and Infinite Series

This document presents the solutions to the Math 115A final exam administered at UCLA on March 19, 2020, focusing on integration, infinite series, and linear algebra. The exam comprised eight problems, with a total of 90 points. The solutions cover a range of topics including true/false questions, vector spaces, linear transformations, rank and nullity, orthonormal bases, orthogonal projections, and related concepts. Each problem is addressed with detailed explanations and justifications, demonstrating the application of key linear algebra principles. The document provides a comprehensive breakdown of the exam questions, making it a valuable resource for students studying linear algebra and preparing for similar assessments. It includes step-by-step solutions, covering topics like linear transformations, vector spaces, and inner products.

Integration And Infinite Series
Problem 1
a) True
b) True
c) False
d) True
e) False
Problem 2
a )
d ) 10
b)
a)
c)
b)
Problem 3
a )
R(T):{Tv:v∈V }
Let V be a finite a finite-dimensional vector of T(f) and T:V→W be a linear transformation.
Let v1,…….Vn be a linear transformation of f(T)
Let T(f)=T(a1,….an):a1v1 +……anVn is a linear transformation.
Given that T(f)= [ f (0) f (−3)
f (−10) f ( 0) ]
T:R2→W and Rank(T)=2
N(T):v∈V:Tv=0}
N(T)=T-1(0)
N(T0=T-1(0)=dim(N(T))=2
b)

T:R2→R2, T[ 0 −3
−10 0 ]
DimN(T)=2 AND dimR(T)=2
Hence N(T)=2 and R(T)=2
Problem 4)
Since =(1,x,x2) is the basis for P2(R) you can have;
R(T)=R(T2)= span[T2[β]]={T2(1),T(x2),T(x2}
=span{ [0 0
0 1 ], [0 0
0 1 ],[0 0
0 1 ],[−4 0
0 o ]
=span [ 0 0
0 1 ],[ 0 0
0 1 ],[ −4 0
0 1 ],
Hence V=N(T)*R(T)
But R(T)=N)(T)
The basis for R(T)
V=R(T)
Thus V=N(T) +R(T)
Problem 5
Problem a)
R(T):={Tv:v ∈V }
Consider V to be a finite dimensional vector and let v1,…………….Vn be a set of vectors in V.
Let T:V→W be a linear transformation and T: Rn →W
The standard basis for Rn= Consider the standard basis for Rn =
[1 0 0
0 1 0
0 0 0
0
0
1 ]
Y={v1,v2,v3,….vn} Which is the orthonormal basis of v
x→
∈V

X→=c1v1 +C2v2 +c3v3 +……ckvk
uix→=c1vix1 +c2vix2 +c3vix3+……………+civixi +…..+ckvlxk
v→iX=Ci
Problem b)
Let T:V→W be a linear transformation and T: Rn →W
The standard basis for Rn= Consider the standard basis for Rn =
[1 0 0
0 1 0
0 0 0
0
0
1 ]
Y={v1,v2,v3,….vn} Which is the orthonormal basis of v
x→
∈V
X→=c1v1 +C2v2 +c3v3 +……ckvk
uix→=c1vix1 +c2vix2 +c3vix3+……………+civixi +…..+ckvlxk
v→iX=Ci
U1=
[ 3
5
3
5 ] U2=
[ −4
5
3
5 ]
||u1||2= 9
25 + 16
25 =1
U1x2=−12
25 + 12
25=0
=
[ 3
5
−3
5
3
5
2
5 ]
Problem 6

Problem a)
Given that W:{(x,y,z) ∈R3;4x+2y-3z=0} and v=(3,2,4)
Let W be a subspace of a finite-dimensional inner product space V.
Let v be a vector in V
And let w be the orthogonal projection of v onto W .
Then w is closer to v than any other element of W ; more precisely, we have kv − w 0k > kv − wk for all
vectors w0 in W other than w.
Let T:V→W be a linear transformation and T: Rn →W
The standard basis for R3= Consider the standard basis for R3=
[1 0 0
0 1 0
0 0 0
0
0
1 ]
Y={v1,v2,v3,….vn} Which is the orthonormal basis of v
Distance between V an W;
=||v1-w|| +||v2=w|| +||v3-w||
=
[3
2
4 ]-
[ 4
−3
−3 ]
=
√12+ ( 5 )2+ ( 7 )2
¿ ¿
=
√1+25+49
¿ ¿
=√ 75
=8.660 units
Problem 6b
Β({1,x,x2} is the basis for P2(R)
Therefore;
R(T)=Span {T(1),T(X),T(x2)}\

Hence T is is invertible since it is a square matrix of order 2 and not a zero matrix.
ii)Span{[0 0
0 1 ], [0 0
0 1 ],[0 0
0 1 ],[ −3 0
0 o ]
=span {
[ 0 0
0 1 ],[0 0
0 1 ],[−3 0
0 0 ],}
You realize that the basis of N(T) =R(T) =2
Problem 7
Problem a)
Β ({1,x,x2+1} is the basis for P2(R)
R(T)=Span {T(1),T(X),T(x2) ,T(x2+1)}
= Span [ 0 0
0 1 ] , [ 0 0
0 1 ],[ 0 0
0 1 ],[ 1 1
1 2 ][ −3 0
0 o ]
=span [ 0 0
0 1 ],[ 0 0
0 1 ],[ −3 0
−3 0 ],
Hence N(T)=R(T)=2
Problem b
(f, g)= ∫
0
1
f (t)g(t)dt.
=∫
0
1
( x ¿¿ 2+ 1) dx ¿
=¿ +x +c]
=
4
3
β({1,x,x3/3+x} is the basis for P2(R)
R(T)=Span {T(1),T(X),T( x3
3 ) ,T( x3
3 +x)}

= Span [ 0 0
0 1 ] , [ 0 0
0 1 ],[0 0
0 1.3 ],[1 1
1 2 ][−1.3 0
0 o ]
=span [0 0
0 1 ],[0 0
0 1 ],[−1.3 0
−1.3 0 ],
Hence N(R)=R(T)=2
Problem 8
Problem a)
Given that Q :U→V ,T:V→X and S:V→W are linear transformation.
But;
The nullity(T) + rank(T)=dim (V)
Also;
Dim N(T)= 1 and dim R(T)= 1
Dim(V)=dim(W)
Hence
Given that dimU –dim W <3
dimU –dim W=2
Problem b)
Let S:U→V be an invertible transformation and let Q:W→Z be also an invertible transformation such
that;
Span{[0 0
0 1 ], [0 0
0 1 ],[0 0
0 2 ],[4 0
0 4 ]
=span [0 0
0 1 ],[0 0
0 1 ],[0 0
0 8 ],
You realize that the basis of N(Q) -R(W) =2
Rank (T)=Rank(TS)=rank (QTS)
But W=rank (TS)+Rank (T)

Hence Rank( W )=Rank (QTS)