Name:

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

.

Addition space page 1

Addition space page 2

Addition space page 3

Addition space page 4

Addition space page 5