程序代写代做代考 Math 215.01, Spring Term 1, 2021 Problem Set #7

Math 215.01, Spring Term 1, 2021 Problem Set #7
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1. Does
Explain.
2 1 0 Span0;1;0 = R ?
101
Proposition 4.2.14. u~1; u~1; : : : ; u~n 2 Rm. Let A be the m􏰅n matrix where the ith th column is u~i, and let B be an echelon form of A. We then have that Span(u~1; u~1; : : : ; u~n) = Rm if and only if every row of B has a leading entry.
Matrix:

210

0 1 0. 

101
By R2 \$ R3, we have

210 210

0 1 0=1 0 1: 

101 010
ByreplacingR2 withR2+(􏰃 )􏰄R1,wehave 2
1
3


210 210

1 0 1 = 0 􏰃1 1. 2 
010 010
By replacing R3 with R3 + (2) 􏰄 R2, we have

210 210

0 􏰃1 1 = 0 􏰃1 1. 22 
010 002
Since we have the matrix in echelon form and every row
has a leading entry and according to proposition 4.2.14, we
     
210
     
can conclude Span 0 ; 1 ; 0 = R3:      
     
101

2. Determine whether
1 2 4
􏰃3 ; 2 ; 􏰃4 5 4 14
is a linearly independent sequence in R3.
Proposition 4.3.3. Let u~1; u~1; : : : ; u~n 2 Rm. Let A be the m 􏰅 n matrix where the ith th column is u~i, and let B be an echelon form of A. We then have that (u~1; u~1; : : : ; u~n) is linearly independent if and only if every column of B has a leading entry.
Matrix :

124
  .
􏰃3 2 􏰃4 

5 4 14
By replacing R2 with R2 + 3 􏰄 R1, we have

124 124
  = .
􏰃3 2 􏰃4 0 8 8 

5 4 14 5 4 14
By replacing R3 with R3 + (􏰃5) 􏰄 R1, we have

124124

0 8 8=0 8 8. 

5 4 14 0 􏰃6 􏰃6
6
By replacing R3 with R3 +( )􏰄R2, we have 8


124124

0 8 8=0 8 8. 

0 􏰃6 􏰃6 0 0 0
Since we have the matrix in echelon form and not every
column has a leading entry and according to proposition
4.3.3, we can conclude
  
124
    ; ;  􏰃3 2 􏰃4      
is not
5 4 14 a linearly independent sequence in R3.

3. Let
1 􏰃1 2 􏰀 = 3;􏰃3;8:
209
(a) Show that 􏰀 is a basis of R3.
Proposition 4.3.3. Let u~1; u~1; : : : ; u~n 2 Rm. Let A be the m 􏰅 n matrix where the ith th column is u~i, and let B be an echelon form of A. We then have that (u~1; u~1; : : : ; u~n) is linearly independent if and only if every column of B has a leading entry.
Proposition 4.2.14. u~1; u~1; : : : ; u~n 2 Rm. Let A be the m 􏰅 n matrix where the ith th column is u~i, and let B be an echelon form of A. We then have that Span(u~1;u~1;:::;u~n) = Rm if and only if every row of B has a leading entry.
Matrix :

1 􏰃1 2

3 􏰃3 8. 

209
By replacing R2 with R2 + (􏰃3) 􏰄 R1, we have


1 􏰃1 2 1 􏰃1 2

3 􏰃3 8 = 0 0 2. 

209 209 By R2 \$ R3, we have

1 􏰃1 2 1 􏰃1 2

0 0 2=2 0 9: 

209 002
By replacing R2 with R2 + (􏰃2) 􏰄 R1, we have

1 􏰃1 2 1 3 2

2 0 9=0 2 5: 

002 002
Since we have the matrix in echelon form and every
row and column has a leading entry and according to proposition 4.3.3 and proposition 4.2.14, we can conclude 􏰀 is a basis of R3.
(b) Determine
Matrix :
 1   5  :
􏰃5
􏰀


1 􏰃1 2 1

3 􏰃3 8 5. 

2 0 9 􏰃5
By replacing R2 with R2 + (􏰃3) 􏰄 R1, we have

1 􏰃1 2 1 1 􏰃1 2 1

3 􏰃3 8 5=0 0 2 2. 

2 0 9 􏰃5 2 0 9 􏰃5 By R2 \$ R3, we have

1 􏰃1 2 1 1 􏰃1 2 1

0 0 2 2=2 0 9 􏰃5: 

2 0 9 􏰃5 0 0 2 2
By replacing R2 with R2 + (􏰃2) 􏰄 R1, we have

1 􏰃1 2 1 1 􏰃1 2 1

2 0 9 􏰃5=0 2 5 􏰃7: 

0022 0022
Then we have a system of following equations:
x 􏰃 y + 2z = 1 2y + 5z = 􏰃7 2z = 2: Propo- sition 4.2.12 tells use that if a matrix is in echelon form and every column, except the last column, has a leading entry,then the system is consistent and has a unique solution. Since we have the matrix in elch- elon form and every column, except the last column,

has a leading entry, we know the system is consis-
tent and has a unique solution. After caculation,
wehavex=􏰃7;y=􏰃6;z=1. Thus,wehave
   
1 􏰃7
    5    
 
= . 􏰃6  
 
􏰃5 1 􏰀

4. De􏰁neT:P1 !R2 byletting
T(a + bx) =
Show that T is a linear transformation.
De􏰁nition 5.1.1. Let V and W be vector spaces. A lin- ear transformation from V to W is a function T : V ! W with the following two properties: 1. T (~v1 + ~v2 ) = T (~v1) + T (~v2) for all ~v1; ~v2 2 V (i.e. T preserves addition). 2.T (c 􏰄 ~v ) = c 􏰄 T (~v ) for all~v 2 V and c 2 R (i.e. T pre- serves scalar multiplication). Let p1; p2 2 P1 be arbitrary and 􏰁x n1;n2;m1;m2 2 R such that p1 = n1 + n2x and p2 = m1 + m2x. Notice that
􏰈a 􏰃 b􏰉 b
:

T(p1 +p2)=T((n1 +n2x)+(m1 +m2x)) =T((n1 +m1)+(n2 +m2)x)

(n1 +m1)􏰃(n2 +m2)
 =
(n2 +m2)

n1 +m1 􏰃n2 􏰃m2
Since
Let c 2 R be arbitrary. Notice that
=  = 
n2 + m2
 
n1 􏰃 n2 n2
m1 􏰃 m2 m2
 + 
=T(n1 +n2x)+T(m1 +m2x)
= T (p1) + T (p2)
n1; n2; m1; m2 2 R are arbitrary, we have shown T


T(c􏰄p1)=T(c􏰄(n1 +n2x)) =T(c􏰄n1 +c􏰄n2)

c 􏰄 n1 􏰃 c 􏰄 n2
 =
c 􏰄 n2

= c 􏰄 T (p1)
Since p1; p2 2 P1 are arbitrary, we have shown T preserves scalar multiplication. Since T satis􏰁es the two properties in de􏰁nition 5.1.1, we can conclude T is a linear transfor- mation.
5. Comment on working with partner(s): Comment on the work you and your partner(s) accomplished together and what you accomplished apart.
n1 􏰃 n2 n2
= c 􏰄  =c􏰄T(n1 +n2x)
