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MATH 520 MIDTERM IISPRING 2017
BROWN UNIVERSITYSAMUEL S. WATSON
This is a pencil-and-paper-only exam. You have two hours.
Problem 1(a) (8 points)
Find
2
41 0 0
1 2 0
3 2 1
3
5�1
Solution
Final answer:
Problem 1(b) (8 points)
Suppose A
�1v1
= e1
, A
�1v2
= e2
, and A
�1v3
= e3
, where v1
=
2
4�4
2
1
3
5, v
2
=
2
40
�1
1
3
5, and v
3
=
2
43
3
0
3
5. Find A.
(Note: e1
e2
, e3
denote the standard basis vectors in R3
.)
Solution
Final answer:
Solutions
lightly → we 's ::t→lbIioHIE3→kEttEED
EEK
Webuow Aei=F . for i= 1. 2,3,
so a=f¥9 !) Egd
Problem 2(a) (8 points)
Consider the linear transformation S
✓x
y
�◆=
3x
4y
�. Define T to be the 135-degree counterclockwise rotation
in R2
. Find the determinant of the composition S � T. Explain your reasoning.
Solution
Problem 2(b) (8 points)
Show that det(kA) = k
n
det A, if k is a real number and A is an n ⇥ n matrix.
Solution
Final answer:
det ST = ( detskdet )
= ¥91 . T← weaverotations preserve
areaand
orientation
= 12
dettkA) = I III.IIhe'd= k"
a,
- .. am - tiara "
. am
+ . . . ( n ! -2 other terms)= k
"
( A, , ,
- -. An
,n
- 9,294 ,
''
' Ann + ' - )
= bidet A ;
the point Is that each term in the rook expansionis multiplied by kn
,so the whole sum is multiplied
by W.
If
Problem 3(a) (10 points)
The matrices A =
2
41 4 5 8 2
0 1 �2 3 5
�2 �7 �12 �13 1
3
5and
2
41 0 13 �4 �18
0 1 �2 3 5
0 2 �4 6 10
3
5are row equivalent.
Find a basis for the row space of A and a basis for the column space of A.
Solution
Final answer:
Problem 3(b) (5 points)
Find the rank and the nullity of A.
Solution
Final answer:
@original midterm ]
Problem 4(a) (8 points)
Consider the set S = { f (x) = 0 for some x 2 [0, 1]} of continuous functions from [0, 1] to R which are zero some-
where. For example, the function
⇣x � 1
2
⌘3
is in S since it vanishes at x = 1
2
, but the function 1 + x
2
is not in S.
Show that S is closed under scalar multiplication in C([0, 1]) but is not closed under vector addition. So is S a
subspace of C([0, 1])?
Solution
Problem 4(b) (8 points)
Consider the subset of P4
consisting of all polynomials whose cubic and quadratic terms have the same coefficient.
For example, �1 + 3t
2 + 3t
3 + t
4
is in this set, while �1 + 2t
2 + 3t
3 + t
4
is not. Is this set a subspace of P4
? Explain
your reasoning.
Solution
If f ES,
then fCx)=O for some × . then for the same x,
we have ¢f) ( x7= cfftxD = c. 0=0 .so of ES
.
Note XES and txes but xtl - ×= I ¢ S. so
sisuot closed under vector addition & thus is not a subspace !
s
Yes , because 0 has the same
t3audt2coeffsCooth01hatbttct2ect3tdt4tai-BttEE2-Et3-ctt4-Caedt@eb7ttCcatt2tCcEH7fded7t9wox90p.q
€S ⇒ ptqes .
equal
Also H a isttuttcttdt 4) = katkbtt kcFtkct3+k¢t4ES.
W W
f)equal
Problem 5 (10 points)
Consider the linear transformation T : P7
! P7
, where T(p) = p
00. In other words, T acts by taking the second
derivative. So, for example, T(�3t
4 + t
2 � t) = �36t
2 + 2.
Find the range and the kernel of T. Feel free to describe these sets using either a verbal description or math notation,
as you prefer (as long as they are clearly specified). Explain your reasoning.
Solution
the range of T is Ps since p"
EPs whenever PEIR ,and
foray q€Ps , ffq ep ,and satisfies $q)" =p .
The bowel of T is IR ,since p
"= 0 iff pltkatbt
for some a. be R .
Problem 6(a) (6 points)
Show that {1 + t
2
, 1 + t + 2t
4} is a basis of the span of {1 + t
2
, 1 + t + 2t
4} (here 1 + t
2
and 1 + t + 2t
4
are polyno-
mials in the vector space of all polynomials, with the usual notions of addition and scalar multipliation). Find the
coordinates of 1 + 4t � 3t
2 + 8t
4
with respect to the basis.
Solution
Problem 6(b) (9 points)
Use Cramer’s rule to solve a b
c d
� x
y
�=
e
f
�
for x and y in terms of a, b, c, d, e, f , assuming that ad � bc 6= 0.
Solution
{ tt2,
let tut } is a basis of its span because its linearlyIndependent ( since al HE ) + bliettze ' ) =0 ⇒ a=b=O )
.
the coordinates are [ 34 ] since CZED (4) =8t4 and (3) IE) = -3T?
let !dX =
By= edtsfad
- be
laced of - ec
y =- =
-
law ad . be
Problem 7 (12 points)
Suppose that U and V are four-dimensional subspaces of a 10-dimensional vector space W. (a) Show that U [ V is
not necessarily a subspace of W. (b) Show that the span of U [ V, which is a subspace of W, is not equal to W.
Solution
U = Span ({ ei ,ez
, es , ey } )
V i Spun ({ es , ea , ei ,eo } )
Then UUV = { IER"= either Xs=Xa= . . . =x,o=O
or x , = . - = xy =o=Xq=Xo}
which is clearly not a subspace: (1/1,140,000,0)
+ ( 0000 1 1 1 1 00 ) EUW
The Span of Uu V is spawned by { Un . -
, Uy , 4. . ,vy} =L
where {u,. . . My } is a basis for U & { U
, ..
, 4 }is a basis for U
.To see this
,note that we
spanluw ) implies W = Ui . + um + u,
+ . . . turn
for some vectors u, .
. - in , Y , ...
,Vm with each uuin U
& each veinV.
Since each uu is in the span of L
& so is each vu ,we Spanl too . But a 10 D space
cannot have a spanning list of length less than10 , so Span UUV ¥ W
.