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)