Name
ID
MATH 311
Exam 2
Section 200
Solutions
1.
Spring 2001
1
/60
2
/20
3
/20
P. Yasskin
60 points Let P n be the vector space of polynomials of degree t n. Consider the linear map
I : P 1 ¯ P 2 given by
x
IŸp Ÿx 2 ; pŸt dt
1
Hint: For some parts it may be useful to write pŸt a bt P 1 and/or qŸx ) *x +x 2 P 2 .
a.
3 Identify the domain of I, a basis for the domain, and the dimension of the domain.
Ÿ DomŸI P 1
b.
basis £e 1 1,
dim DomŸI 2
3 Identify the codomain of I, a basis for the codomain, and the dimension of the codomain.
Ÿ basis £E 1 1,
CodomŸI P 2
c.
e2 t¤
E 2 x,
E3 x2 ¤
dim CodomŸI 3
5 Identify the kernel of I, a basis for the kernel, and the dimension of the kernel.
Ÿ x
2 ; pŸt dt 0
1
differentiate:
2pŸx 0
dim KerŸI 0
KerŸI £0 ¤
basis £empty ¤
x
x
OR: IŸa bt 2 ; a bt dt ¡2at bt 2 ¢ 1 2ax bx 2 " 2a " b 0
1
d.
®
ab0
5 Identify the range of I, a basis for the range, and the dimension of the range.
Ÿ x
dq
dq
differentiate:
2pŸx ®
pŸx 1
qŸx 2 ; pŸt dt
1
dx
2 dx
x
dq
Check: I 1
2 ; 1 1 dq dt qŸx " qŸ1 This qŸx only if qŸ1 0
2 dx
2 dt
If qŸx ) *x +x 2 then qŸ1 ) * + 0.
RanŸI q P 2 such that qŸ1 0
RanŸI £q ) *x " Ÿ) * x 2 ¤ £)Ÿ1 " x 2 *Ÿx " x 2 ¤ Span£1 " x 2 , x " x 2 ¤
dim RanŸI 2
basis £1 " x 2 , x " x 2 ¤
2
OR: IŸa bt 2ax bx " 2a " b 2aŸx " 1 bŸx 2 " 1 basis £x " 1, x 2 " 1 ¤
dim RanŸI 2
RanŸI Span£x " 1, x 2 " 1 ¤
e.
2 Is the function I one-to-one? Why?
Ÿ KerŸI £0 ¤
f.
®
I is 1 " 1
2 Is the function I onto? Why?
Ÿ I is not onto because dim CodomŸI 3 but dim RanŸI 2
g.
2 Verify the dimensions in a, b, c and d agree with the Nullity-Rank Theorem.
Ÿ dim KerŸI dim RanŸI 0 2 2 dim DomŸI 1
h.
5 Find the matrix of I relative to the standard bases: (Call it A .)
Ÿ e 1 1,
e 2 t for P 1
E se
E 1 1,
and
E 2 x,
"2 "1
x
IŸe 1 IŸ1 2 ; 1 dt 2Ÿx " 1 "2E 1 2E 2
1
x
A
®
IŸe 2 IŸt 2 ; t dt x 2 " 1 "E 1 E 3
E se
1
i.
0
0
1
6 Another basis for P 1 is f 1 1 2t, f 2 1 3t. Find the change of basis matrices between
the e and f bases. (Call them C and C .) Be sure to identify which is which!
f 1 1 2t e 1 2e 2
®
f 2 1 3t e 1 3e 2
esf
1 1
C
esf
C
®
2 3
f se
C "1 esf
3 "1
"2
1
6 Consider the polynomial q 3 4t. Find ¡q ¢ e and ¡q ¢ f , the components of q relative to the
e and f bases, respectively.
Ÿ q
¡ ¢e
k.
2
Ÿ f se
j.
E 3 x 2 for P 2
3
3 "1
C ¡q ¢ e q
¡ ¢f
3
"2
1
3 A polynomial r has components ¡r ¢ f 2
1
4
f se
Ÿ 4
5
"2
relative to the f basis. What is r?
r 2f 1 1f 2 2Ÿ1 2t 1Ÿ1 3t 3 7t
l.
5 Find the matrix of I relative to the f basis for P 1 and the E basis for P 2 . (Call it B .)
Ÿ E sf
"2 "1
B
E sf
m.
A
C
E se
esf
2
0
0
1
"4 "5
1 1
2 3
2
2
2
3
5 Find B by a second method.
Ÿ E sf
x
IŸf 1 IŸ1 2t 2 ; Ÿ1 2t dt 2Ÿx " 1 2Ÿx 2 " 1 "4E 1 2E 2 2E 3
1
x
IŸf 2 IŸ1 3t 2 ; Ÿ1 3t dt 2Ÿx " 1 3Ÿx 2 " 1 "5E 1 2E 2 3E 3
1
"4 "5
B
®
n.
E sf
2
2
2
3
2 relative to the f basis. Find ¡IŸr ¢ , the
E
1
components of IŸr relative to the E basis. What is IŸr ?
6 A polynomial r has components ¡r ¢ f Ÿ "4 "5
Ir
¡ Ÿ ¢E
B ¡r ¢ f E sf
2
2
2
3
2
1
"13
6
IŸr "13 6x 7x 2
7
2
o.
2 Find IŸr by a second method.
Ÿ x
IŸr IŸ3 7t 2 ; Ÿ3 7t dt 6Ÿx " 1 7Ÿx 2 " 1 "13 6x 7x 2
1
2.
20 points Consider a linear map L : R
n
¯
R whose matrix is A p
1 "2 0
3
2 "4 1
2
0
a.
2 What are n and p?
p3
6 Identify the kernel of L, a basis for the kernel, and the dimension of the kernel.
Ÿ 1 "2 0
3
0
2 "4 1
2
0
1 "4
0
0
0
x
y
1 "2 0
®
3
0
0
0
1 "4
0
0
0
1 "4
0
2s " 3t
w
®
®
4t
3
0
0
0
1 "4
0
0
0
0
0
0
"3
1
KerŸL Span
0
,
0
t
4
0
1
"3
2
1
basis 1 "2 0
2
s
z
0
,
0
dim KerŸL 2
4
0
c.
1 "4
0
Ÿ n4
b.
.
1
6 Identify the range of L, a basis for the range, and the dimension of the range.
Ÿ "2
1
RanŸL Span
"4
,
2
0
a
2
2
0
b
1
0
,
c
3
0
2 1
2
0
0 1 "4
0
,
2
"4
Are they independent?
2
"4
1
0
2
0
"4
®
,
3
1
1 0
3
1
3
1
1 0
1
0
0
1
,
0
1
Span
0
0
3
0
0 1 "4
0
0 1 "4
0
1 0
®
3
0
0 1 "4
0
0 0
0
0
"3t
a
®
b
c
4t
t
Not independent
3
1
RanŸL Span
2
0
,
2
1
0
,
0
1
dim RanŸL 2
1
2 Is the function L one-to-one? Why?
Ÿ KerŸL p £0 ¤
e.
basis 1
0
d.
1
®
L is not 1 " 1
2 Is the function L onto? Why?
Ÿ I is not onto because dim CodomŸI 3 but dim RanŸI 2
f.
2 Verify the dimensions in a, b and c agree with the Nullity-Rank Theorem.
Ÿ dim KerŸL dim RanŸL 2 2 4 dim DomŸL 3.
20 points Consider the parabolic coordinate system
y 2uv
x u2 " v2
a.
4 Describe the u-coordinate curve for which v 2.
(Give an xy-equation and describe the shape.)
Ÿ If v 2, then x u 2 " 4 y 4u. So x b.
4 Find e u , the vector tangent to the u-curve at the point Ÿu, v Ÿ3, 2 .
Ÿ e u c.
y2
" 4. This is a parabola which opens to the right.
16
x , y
u u
e u
Ÿ2u, 2v Ÿ3,2 Ÿ6, 4 4 Describe the v-coordinate curve for which u 3.
(Give an xy-equation and describe the shape.)
Ÿ If u 3, then x 9 " v 2 y 6v. So x 9 "
d.
4 Find e v , the vector tangent to the v-curve at the point Ÿu, v Ÿ3, 2 .
Ÿ e v e.
y2
. This is a parabola which opens to the left.
36
x , y
v v
Ÿ"2v, 2u e v
Ÿ3,2 Ÿ"4, 6 4 Compute e u , e v and e u e v . Find the angle between e u and e v .
Ÿ e u 36 16 52 ,
e v 16 36 52 and e u e v "24 24 0. The angle is 90°.
4