Math 54 Final Exam, Prof. SrivastavaDecember 14, 2018, 3:10pm–6:00pm, 150 Wheeler (170 minutes).

Name:

SID:

Instructions: Write all answers in the provided space. This exam includes three scratchspaces which will not be graded. Do not under any circumstances unstaple the exam. Writeyour name and SID on every page. Show your work — numerical answers without justifica-tion will be considered suspicious.

Calculators, phones, cheat sheets, textbooks, and your own scratch paper are not allowed.

UC Berkeley Honor Code: As a member of the UC Berkeley community, I act withhonesty, integrity, and respect for others.

Sign here:

Name of the student to your left:Name of the student to your right:

Question: 1 2 Total

Points: 24 24 48

Do not turn over this page until your instructor tells you to do so.

Name and SID:

1. (24 points) Fill in T (always true) or F (sometimes false) for each of the following.There is no need to provide an explanation. Two points each.

(a) If A is an m× n matrix, then the number of pivots in the REF of A is equal to the

number of nonzero singular values of A. T F

Solution: True. Both are equal to the rank of A.

(b) If V,W, Y are finite dimensional vector spaces and T : V → W and S : W → Y are

isomorphisms, then V and Y must have the same dimension. T F

Solution: True. Isomorphisms preserve dimension, so dim(V ) = dim(W ) =dim(Y ).

(c) If B = {b1, b2} and C = {c1, c2} are orthonormal bases of R2, then the change ofcoordinates matrix PC←B has the property that ‖PC←Bx‖2 = ‖x‖2 for every x ∈ R2.

Solution: True. If E is the standard basis then PC←B = PC←EPE←B =P−1E←CPE←B. Since PE←B and PE←C contain the columns of B and C theyare orthogonal matrices. Since inverses and products of orthogonal matrices areorthogonal, the conclusion follows.

(d) For every nonzero vector v ∈ R3, there is a basis of R3 containing v. T F

Solution: True. Just add a vector w /∈ span{v} and then a vector z /∈span{v, w} to get a basis.

(e) Every nonzero vector v ∈ R2 can be written as v = a+ b for some nonzero vectors

a, b with a · b = 0. T F

Solution: True. Choose any unit vector u such that u is not parallel or orthog-onal to v, and let a = (v · u)u and b = v − a. By the decomposition theoremthese vectors are necessarily orthogonal, and nonzero by the way we chose u.

(f) If V is a finite dimensional vector space and S = {v1, . . . , vk} spans V , then S

contains a basis of V as a subset. T F

Solution: True. If S is already a basis we are done, otherwise it is not linearlyindependent and we can remove some vector which is in the span of the previousones. Repeat this process until you get a linearly independent set. Since wedidn’t change the span, this set is a basis.

Math 54 Final Page 2 of 6 12/14/2018

Name and SID:

(g) If x is a solution to Ax = b then x ∈ Row(A). T F

Solution: False. If x is a solution to Ax = b then A(x + z) = Ax + 0 = b forany z ∈ Nul(A), but x+ z /∈ Row(A).

(h) If W is a subspace of Rn then the standard matrix of the orthogonal projection

onto W must be diagonalizable. T F

Solution: True. If W has dimension k and U is an n × k matrix containingan orthonormal basis of W as its columns then the standard matrix of theprojection is UUT . Since (UUT )T = UUT , the spectral theorem implies thatthis is diagonalizable.

(i) If A = AT and P is the orthogonal projection onto the column space of A, then

AP = PA. T F

Solution: True. Assume A is n×n. By the spectral theorem A is orthogonallydiagonalizable, so let u1, . . . , un be an orthonormal basis of Rn consisting ofeigenvectors of A, arranged so that Aui = λiui with λi 6= 0 for i ≤ k and Aui = 0for i > k. Observe that u1 . . . , uk ∈ Col(A) by the eigenvector equation, so forthese vectors we have

APui = Aui

andPAui = P (λiui) = λiui = Aui.

For ui with i > k we have

APui = A(0) = 0 = Aui

since ui ∈ Col(A)⊥ = Row(A)⊥ = Nul(A) since Col(A) = Row(A) because A issymmetric. Similarly, for these vectors

PAui = P (0) = 0 = Aui.

Altogether, we conclude that APui = PAui = Aui for all of the basis vectorsu1, . . . , un, which means that AP = PA = A.

(j) If A = AT and all the eigenvalues of A are positive, then there is a matrix B such

that B2 = A. T F

Solution: True. By the spectral theorem we have A = UDUT for some orthog-onal U and diagonal D which has positive entries λ1, . . . , λn > 0 Let C be thediagonal matrix with entries

√λ1, . . . ,

√λn, which are real since the eigenvalues

Math 54 Final Page 3 of 6 12/14/2018

Name and SID:

were positive, and note that C2 = D. Letting B = UCUT we see that

(UCUT )2 = UC(UTU)CUT = UC2UT = UDUT = A,

as desired.

(k) If A is invertible and σ is a singular value of A then 1/σ is a singular value of A−1.

T F

Solution: True. The SVD implies that A = UΣV T for orthogonal U, V anddiagonal Σ, which must be square since an invertible matrix has to be square,and which has positive diagonal entries. Now

A−1 = (UΣV T )−1 = (V T )−1Σ−1U−1 = V Σ−1UT ,

since U, V are orthogonal. But this is just the SVD of A−1, so A must havesingular values equal to the entries of Σ−1 which are the reciprocals of thesingular values of A.

(l) If A and B are similar square matrices, then every singular value of A is also a

singular value of B. T F

Solution: False. You should be suspicious because the singular values of A tellus how much A can stretch or shrink a vector, and this can be changed a lot by asimilarity transformation B = PAP−1 when P is not orthogonal. In particular,a square matrix A is orthogonal if and only if all of its singular values are equal

to one. Taking A = 1√2

[−1 11 1

]to be counterclockwise rotation by 45 degrees

and P =

[1 00 2

]we find that B = PAP−1 is not orthogonal, so it has singular

values other than one, which means that similarity does not preserve singularvalues.

Math 54 Final Page 4 of 6 12/14/2018

Name and SID:

2. Give an example of each of the following, explaining why it has the required property,or explain why no such example exists.

(a) (4 points) A 2× 3 real matrix A such that ATA and AAT are both invertible.

Solution: Does not exist. The reason is that rank(ATA) ≤ rank(AT ) sinceCol(ATA) ⊆ Col(AT ), but dim(Col(AT )) = rank(AT ) ≤ 2 since AT is 3× 2, soATA is a 3× 3 matrix with rank at most 2 and cannot be invertible.

(b) (4 points) A subspace W ⊆ R3 such that det(P ) 6= 0 and det(I −P ) 6= 0, where Pis the standard matrix of the orthogonal projection onto W .

Solution: Does not exist. Conceptual proof: Observe that I − P is the pro-jection onto W⊥, so rank(P ) + rank(I − P ) = rank(W ) + rank(W⊥) = 3. Thusone of these matrices must have rank less than 3, which means its determinantmust be zero.

Algebraic proof: det(P ) det(I − P ) = det(P (I − P )) = det(P − P 2) = det(P −P ) = 0, where P 2 = P since P is a projection.

(c) (4 points) A real 2 × 2 orthogonal matrix with all eigenvalues real and strictlynegative.

Solution: A =

[−1 00 −1

].

Math 54 Final Page 5 of 6 12/14/2018

Name and SID:

(d) (4 points) A second order differential operator T : F → F of the form

T (y) = y′′ + by′

with infinitely many eigenvalues, where b ∈ R and F is the vector space of infinitelydifferentiable functions y : R→ R.

Solution: The eigenvalue equation in this context is:

T (y) = y′′ + by′ = λy,

so the question is asking whether there is a b such that this is satisfied forinfinitely many λ and nonzero functions y. But this is just a second order ODE,which always has at least one real solution. So actually every λ ∈ R is aneigenvalue, regardless of the value of b. For a concrete example we may takeb = 0.

(e) (4 points) A second order differential equation y′′(t) + by′(t) + cy(t) = 0 which hasamong its solutions y1(t) = et(cos(3t)−3 sin(3t)) and y2(t) = et(3 cos(3t)− sin(3t)).

Solution: This is a ‘reverse engineering question’. Observe that any homoge-neous ODE with solutions containing et cos(3t) and et sin(3t) will have y1 andy2 among its solutions by taking linear combinations. The auxiliary polynomialr2 + br + c of such an ODE must have roots 1± 3i, so we have

r2 + br2 + c = (r − (1 + 3i))(r − (1− 3i)) = r2 − 2r + 10.

Thus the answer is y′′ − 2y′ + 10y = 0.

(f) (4 points) An inner product 〈·, ·〉 on R3 such that 〈v, w〉 > 0 for all pairs of nonzerovectors v, w ∈ R3.

Solution: Does not exist. For any nonzero vector v and w = −v, we have〈v,−v〉 = −〈v, v〉 = −‖v‖2 < 0 by the properties of inner products.

Math 54 Final Page 6 of 6 12/14/2018

3. (7 points) Suppose L'- is an infinite dimensional vector space, al,a.),'q € V arc vectors.

and ?: l,'-+ IR2 is a linetr.r trausformation such that srtch that

T(xy *2t,2* ".)

: l]r] and T(3q * 2t2 + xr) -

t-rll,l

VztV

LWlrat is tlre value of T(5ur + 2t;z - q)? ,\k J-L.tr

-3 2-l

I^21 I

1'L3O 034o O6+

+,] -I3i:lrrlr^L-

t3o{o-$

?zI

Z

-l

T fG v 1t>v,- \:) -- -r [T (v 1tz v, t3 v))t z (1 L:v s+ :

= -El rrl-l] =I ,'ltE] Iril

4. (7 points) Let Ma be the v'ector space of real 4 x 4 matrices rvith entrywise addition and

scaJar multiplication. and consider the linear transformation T : lla -+.r\la defined byaf (X) - AX rvhere

For rvhiclr valucs of h is T onLo? Explain 1'our reasouing.

lri -tt^[ (dururvr( of X V(, [r rLr X;,&.-h19 iJ hKQ M rYz r Xl , X.+ lo b, ,b,1b: ,4 .Sltrh +^SNt har,rC all ,,\t *ur r b>1 b3, b1,$.s wa,lrAy,,b, ,lrx*,, frx, Vo, *X+--b+ ha,,.r c, rd.*.s:E {rr aQ

b,,V tq r tr"\rrA yrt wr,t^3 *ln* it,A A(r^y,ng 6t +r

l-r zsal,: lf i i il

lo o o al

4PT\\ XIcq^,^bFa*1,

br) 5lQan lF+ S1h{r & ir 4x{, k r,.nlsi h avL { PvvR,-i-t z- o o

* lF i"' BLooot

Ax.,bha!(o[t ( qttb,

-\Inth 5l Final Pagc ir o1 12 1?/ r{/2018

Narrre anrl sro, fS'tf[h d^4a--?CP-,?,bb1uq'l

..Mft v

\irrrrr: ind SID:

5. (7 points) Let

0

Find vectors c € Row(,A) and 9 € Nui(,4) suclr rhat 1, : .r: + y.

ril

lr[,,

1

1 rl.:[i]RoN Ln)

ll-l

',, - Ii]

\,e f T]

u-= I-f

D(r)+tCt)-ttIo)

l4 tlzt' o'

r( t) r o(r; * t( o)

r*l'\--

I [-t) + trL!* tC)

l'1 I '-f ,'

\-X llt!;

tl .iitJH+'fl

Y'DJ', ,\

Pt)'x

[Scratch Space 1]

I il =l,fllil = lll,)

5

Ttirl.ll,!,1 1vilrr/sl

\= li/i)

vfri)

\lrrtlr 5l Firal l']age (i , rl 12 t2iL.t/2018

/1

-:-

l-I

Nilnrc ;r.nd SJI):

6. (7 points) Find the singular value decomposition A : UDV'L' (i.e.. where L/ antl V' arcorthogorral zrnd D is nonncgative diagonal) of the rnatlix

A\=uu

^:li1

0

,{'A= \fv' = f? ;]tl,l] =E ?]drt rklA -xI) = c*f,)X,-p=o

L__-GjIo-3 :) aj<t^wtW U-Tl,:J

)=t [i r] a

[Scratch Space 2]

^=1Ooi O=)r I

l

A\,=T3U I!l =l?l'T1lI u, ,fi,,]i il =1 Il

i95l=ii rll; ilTr?zA

u=I? t-i

\Jatlr ll -tiirrirl Page 7 o[ 12 1211 t 12018

i,1i?l

I

Naiuc irrrd STI) \ta .,r?o (.

Z- (a) (3 points) Find tr basis fol thc spu r: of roirl solutir,lrrs to the honr()gorrcolls cliffer-cntialecluation:

f'(t) - sy'(t) + 2oy :11.

Y'-lr 1)-o --o

(r -+)lr-c)''o

$'ctt*t'1c, ttk

1*r' ,

(b) (2 points) Find a solution to thc initial valuc problern for the equation in part (a)rvith hitial data y(0) : 0 irnd ,'(0) : 0

ULo) = Qtcu,o j' = 'lcn {{t+\<--<r

_q'( o) = 4q t S<-r- --O ,

4<^+2----o( c ,

-r-t1a,--1

Cr_4c-\--P

u t<) =b

\st

4

\iath 5.1 l'iral Page li of 12 t2 / r.t /20t8

I

Nirurc itnd SI l):

(c) (3 points) Find the general solution to the inhorrogeurxrus diflerential etluation

y"(L) - 9y'(t.) + 2Os 't..

Qws -- attb,--4ll

--b,

-14 *>otatt\r) =t-aA+ )Datr*)Ob--t

206,--, q- )o-{tt )ob'-O ,AV= ?.

$vUtt

5t

t_a\Y +agts) - *.t*fu'

ULt)' ctt'*c,<tv * L*t* +*

(d) (2 points) Find a solution to the initial value problern for thc cquation in pa^rt (c)*'ith initial data y(0) : 0 a.nd t/(0) : 0

jLoJ= cqtc,+fu --o \' tt) = \sn+Yct-+q

4cD--o

+c\1'?:,\ a.)-4€D -4

lQt*c,+s-rt*'-o?6

C>--'4=o _-L

(-t - L>- 3b -_J-lb

L+<){q)

9_+oD +oD

-ltmt** *qa q5t + 3_

\latlr i I Ijirrirl

uk)-

Pngt: !) ol 12

7o t-r 4eD

1217.t/2{t18

E<k\lq U,t," 4.affi21

\a, b-zZloG z7

8- Consider the hornogcrrcous sJ'stcrl of ODE:

x'(t): [il]",,,(a) (4 points) Fiud a b;r.sis for tlrc ver:tor sprrce y of real solutions (i.e., differ.cntiable

hrnctions x: IR -+ IR2 satisft.ilg thc crlrat,ion) of this systcrn.

A--I', i] drt (A-\1) -- [1-})[r ))- t -o

X-zaft-n7'->V'-'D

^5tr_r)=a *,=o,r

1-" l', ',1 ,t*'* Ilrl2=;. fl' l,l ,J,,'*rfl]

lLt)-- qn* Ill * .x*f \1*.,<.*Ill,,, I1l

) tll ,1";:)

(b) (1 point) \\rhat is thc dirncnsion of lu'?

w ff has two rrr*ors ri gt W 9(J

.\lath 5.1 Firral Page ll) oi 12 t? /t.t i2o1E

Narlc trncl SfO, t<\f\

-

Na.nrr: irnd SID: Fltat!^ 1ru^ ?o: )hG1z37

(c) (2 points) Consider the linear transformation T : V -+ IR2 defined bv

"(x) : 11e;'

Is 7 one to one? Explain why or rvhy not.

Ut t(tt)' qlil +c,4o[rJ > x€!-r\n ", \(o) = .,T_\]- "tll =p, \1 lfllT ti 0Y\t-fo-q^4 lrtcavp \kha- t\,r.t {4ns(ornaa,rGil

M^tri{ E ,,],,

blhrdn is rinveriu,b- $ DsTtrxa,r.

No *yn !r^,10/s I3,,] *g h-{U (^}^r, vr^drr.

[Scratch Space 3]

\.lzrtli irl l"irral l'}age 1l ol 12 t2 /14 /20t8

Narnc arrrl SID: \tq- Lls.rn b 2

9' Let F be thc vector spacc of pietcrvise contiluous fuuctiots rvith piecewisc contiluousderivative ot l-n,rl.(a) (5 points) Fild thc Fo

lL -- -[t a,rrlicr series of the flrnctiol f (:r) - ,. rlefilcr.l on l-n, zl.

=O Wca\rec { U oaa

Lecar^g- -( rs cil.

lrCDJ(h'II) + t.

o cDs(-YrTI +D

L

4n =oJl rat

v Jl xnntir")dx vt=T ;1r --oM [nr)41d..r -d'\ \t --t t\st"o)

-- f-*-tt^D + Jtcos(^r)4r)/1

-* (*,,, tnr)+ #,"inrt \ /:li)/-n

r-[

-7i^

(b) (2 points) Consider thc four di

-7r

of / onto theof F. Fintl thinner product

?'J. t =

c orthogoual plojectiU.4): f"t@oal-l ^xs\vr}cl\

h cps[nrD =]1-r)nZ?LD-stntq)h--t

rlensional subspi*e : span{sin(z), cos(;r:). sin(2Lr), c<-rs(2:r)}subspace S rvith respect to the

-tW

+

Aaz

o&\{ts

on

)d,

+

on F.

J1^o cD5 - S to'lgr-ii-

J-ll--n.

tfq" Jt\^{ ral6t.leSrs \h Par+G)-

Pml't 'f er) stin X * -| t-r>1'rnt 9zs\,^\-sln t>\)

I9+ oaa

.\larlr i 1 I"irr;rI Page 12 r,l. i 2 t2/ )1i2O18

-.---------._-.-