# 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 Span0;1;0 = R ?

101

Proposition 4.2.14. u~1; u~1; : : : ; u~n 2 Rm. Let A be the mn 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. 22

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. DeneT:P1 !R2 byletting

T(a + bx) =

Show that T is a linear transformation.

Denition 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

perserves addition.

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(cp1)=T(c(n1 +n2x)) =T(cn1 +cn2)

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 satises the two properties in denition 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 =cT(n1 +n2x)