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Math W54 vector

Creation date：2024-05-11 17:24:03

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Math W54

• This is Midterm 2. The exam is out of 50 points and has 5 questions, where each
question is worth 10 points. You are required to solve all questions. Some of these
questions have multiple parts; next to each part, you will find the amount of points
the question is worth.
• You are required to show your work on each problem on this exam. Mysterious or
unsupported answers will not receive full credit. A correct answer,
unsupported by explanation or algebraic work will receive no credit; an incorrect
answer supported by substantially correct calculations and explanations will receive
partial credit.
• You must take this exam completely alone. Showing it to or discussing it with
anyone else is forbidden. Reproducing or sharing this material on any other websites
or platforms is also forbidden. You are permitted to consult your notes, the textbook,
any materials handed out in class, and any materials on the bCourses site. You may
not consult any other resources (for instance, public websites, Wolfram Alpha, and
calculators).
Math W54
Name:
Practice Midterm 2 July 26, 2022
1. (10 points) Do you think Mind your eigenbusiness would make a cool t-shirt?
Solution. Yes. There is no other right answer.
Page 2
Math W54
Name:
Practice Midterm 2 July 26, 2022
2. In this question, we develop some bounds for the rank of a product! Let A and B be
n× n matrices for simplicity.
(a) (2 points) Prove that rank(BA) ≤ min{rank(A), rank(B)}.
Solution. Let TA and TB be the unique linear transformations corresponding to A
and B, respectively.
This question has been solved in lecture; hence, we provide a different approach to
the solution.
Any vector that y that can be written as a linear combination of the columns of
BA as y = BAx can automatically be written as a linear combination of the
columns of B by y = B(Ax) = Bx′, for x′ = Ax. Thus col(BA) ⊆ col(B), so
rank(BA) ≤ rank(B).
Next, let X = AT and Y = BT . Then because rank(XY ) ≤ rank(X) as shown
already, taking transposes, we get rank(Y TXT ) ≤ rank(XT ). But because XT = A
and Y T = B, we have rank(BA) ≤ rank(A) as desired. It then follows that
rank(BA) ≤ min{rank(A), rank(B)}
(b) (4 points) What we realized by the previous part is that the rank of BA has some
kind of defect from the ranks of A and B. Along those lines, prove the following:
rank(BA) = rank(A)− dim(ker(B) ∩ im(A))
Solution 1 (Nice Rank-Nullity). First, rank(BA) = dim(im(TBTA)). By the
previous part rank(BA) ≤ rank(A), simply because im(TBTA) ⊆ im(TA). We need
to realize the defect now. We rearrange the theorem to say
rank(BA) + dim(ker(TB) ∩ im(TA)) = rank(A),
and make a clever use of the rank-nullity theorem. Specifically, consider a linear
transformation S : im(TA)→ im(TBTA), given by S(x) = Bx for x ∈ im(TA).
Then im(S) = im(TBTA) because y = S(x) for some x ∈ im(TA) if and only if
y = Bx. However, because x ∈ im(TA), x = Au for some u ∈ Rn. Thus y = BAu,
so y ∈ im(TBTA). We conclude then that y ∈ im(S) iff y ∈ im(TBTA), which
proves that im(S) = im(TBTA).
Next, ker(S) = ker(TB) ∩ im(TA), because ker(S) contains precisely the vectors in
im(TA) which get mapped to 0 under B. Now by rank-nullity,
dim im(S) + dimker(S) = dim(im(TA))
Then we know that
rank(BA) + dim(ker(TB) ∩ im(TA)) = rank(A).
Page 3
Math W54
Name:
Practice Midterm 2 July 26, 2022
Solution 2 (Extending Basis). This solution is from a more intuitive standpoint,
but as we will see, some rigor will be required. We wish to prove that
dim im(TBTA) + dim(ker(TB) ∩ im(TA)) = dim im(TA).
Notice that any vector in im(A) either goes to zero or goes to something non-zero
under TB. In the first case, it becomes a member ker(TB) ∩ im(TA). In the second
case, it becomes a member of im(TBTA), excluding zero.
Now to say something about dimension, we need to talk about bases. Any basis
for im(TA) would come from a basis β = {v1, . . .vp} for ker(TB) ∩ im(TA), and
another basis γ corresponding to a basis γ′ for im(TBTA).1
Thus, consider extending the basis β to make a basis for im(TA) as
B = {v1, . . .vp,u1, . . .uℓ}. Then dim im(TA) = p+ ℓ, while
dim(ker(TB) ∩ im(TA)) = p. If we prove that C = {T (u1), . . . T (uℓ)} is a basis for
im(TBTA), then we are done because
dim im(TBTA) + dim(ker(TB) ∩ im(TA)) = dim im(TA) = p+ ℓ.
This becomes our goal.
Linear independence. Suppose a1TB(u1) + · · · aℓTB(uℓ) = 0. By linearity,
TB(a1u1 + · · · aℓuℓ) = 0, so a1u1 + · · · aℓuℓ ∈ ker(TB) ∩ im(TA). But because β is a
basis for ker(TB) ∩ im(TA), what we have is
a1u1 + · · · aℓuℓ = b1v1 + · · ·+ bpvp.
Rearranging,
a1u1 + · · · aℓuℓ − b1v1 − · · · − bpvp.
Since B is a basis for im(TA) however, the set {v1, . . .vp,u1, . . .uℓ} is linearly
independent, so a1 = · · · = aℓ = b1 = · · · bp = 0. Thus C is a linearly independent
set.
Spanning. Take u ∈ im(TA) so that TB(u) ∈ im(TBTA). Then because B is a
basis for im(TA), we can write u as
u = b1v1 + · · ·+ bpvp + a1u1 + · · · aℓuℓ.
Applying TB, we obtain by linearity that
TB(u) = b1T (v1) + · · · bpT (vp) + a1T (u1) + · · · aℓT (uℓ)
= a1T (u1) + · · · aℓT (uℓ)
since v1, . . .vp ∈ ker(TB), hence T (v1), . . . , T (vp) = 0. Thus C indeed spans
im(TBTA).
(c) (4 points) Finally, something in the other direction: show that
rank(BA) + n ≥ rank(A) + rank(B).
1Notice that a very important detail is that we cannot just use the basis for im(TBTA), since this does
not live in the same space as im(TA) or ker(TB) ∩ im(TA). This is where the need for rigor comes in.
Page 4
Math W54
Name:
Practice Midterm 2 July 26, 2022
Solution. We use the rank-nullity theorem. First, subtracting 2n from both sides,
we get
rank(BA)− n ≥ (rank(A)− n) + (rank(B)− n).
Next, multiplying by (−1),
n− rank(BA) ≤ (n− rank(A)) + (n− rank(B)).
By the rank-nullity theorem, it suffices to prove that2
null(BA) ≤ null(A) + null(B).
This is doable. Any vector x ∈ nul(BA) has two possible trajectories:
(i) It either goes to 0 under TA, and then TBTA(x) = TB(0) = 0. In this case
x ∈ nul(A).
(ii) It goes to something non-zero first, but then TA(x) ∈ nul(B), so TBTA(x) = 0.
In this case, TA(x) = nul(B), or more specifically TA(x) ∈ nul(B) ∩ im(TA).
The idea here is that because nul(BA) is made up of vectors in nul(A) and
nul(B) ∩ im(TA), it is missing out on the vectors in nul(B) that are not in im(TA).
This is what causes the defect in null(BA), yielding the inequality.
We will once again talk about bases to conclude something about dimension. Let
β = {v1, . . .vp} be a basis for nul(A). Since nul(A) ⊆ nul(BA), we can extend β
to form a basis B = {v1, . . .vp,u1, . . .uℓ} of nul(BA). Thus null(BA) = p+ ℓ,
while null(A) = p.
We claim that {TA(u1), . . . TA(uℓ)}, vectors in nul(B) ∩ im(TA), form a linearly
independent set. If a1TA(u1) + · · · aℓTA(uℓ) = 0, then a1u1 + · · · aℓuℓ ∈ nul(A) by
linearity. But since {u1, . . .uℓ} are part of the independent set B, it follows that
a1 = · · · = aℓ = 0.
Now because u1, . . .uℓ ∈ nul(BA), we know that TBTA(u1), . . . TBTA(uℓ) = 0.
Thus TA(u1), . . . TA(uℓ) ∈ nul(B). Thus nul(B) contains a linearly independent set
of size ℓ, so any basis it has must be of size at-least ℓ. Thus null(B) ≥ ℓ, giving
nul(A) + nul(B) ≥ p+ ℓ = nul(BA)
as desired.
Looking back, parts (b) and (c) are hard enough to form their own separate questions
on an actual midterm.
2Notation thing: we are using nul(A) for the null-space of A, while null(A) = dimnul(A) (called the
nullity).
Page 5
Math W54
Name:
Practice Midterm 2 July 26, 2022
3. (10 points) Given similar matrices A =
1 2 34 5 6
7 8 9
 and B =
5 4 62 1 3
8 7 9
 . Find a basis
in R3 where the matrix for A is B.
Solution 1. Observe that we get B by swapping the first two rows and the first two
columns of A. In other words, B with its first two rows swapped = A with its first two
columns swapped. To swap the first two rows of B, we can multiply the elementary
invertible matrix:
P =
0 1 01 0 0
0 0 1

on the left of B. Now if we multiply this P on the right of A, it will swap the first two
columns of A. So we have:
PB =
2 1 35 4 6
8 7 9
 = AP
So B = P−1AP , and the columns of P is the basis we want.
Solution 2. We know B = P−1AP for some invertible matrix P and we want to find
the columns of P . Let
P =
a b cd e f
g h i

We have PB = AP , or:a b cd e f
g h i
5 4 62 1 3
8 7 9
 =
1 2 34 5 6
7 8 9
a b cd e f
g h i

This gives us a system of 9 equations in 9 variables, so we can solve for
a, b, c, d, e, f, g, h, i. The basis will be the three column vectors of P .
Page 6
Math W54
Name:
Practice Midterm 2 July 26, 2022
4. Let P3 denote the vector space of polynomials of degree at most 3, and consider the
linear map T : P3 → P3 defined by
T (p(t)) = p′(t) + p(0)t3.
Calculate the eigenvalues of T .
Proof. Observe that T (1) = 0 + t3, T (t) = 1 + 0, T (t2) = 2t+ 0, and T (t3) = 3t2 + 0.
Therefore, using the standard coordinates for P3, we see that T is represented by the
matrix

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20110 COMU7000 average acceleration BUSI4412 BFF3121 BIT233 ECMM462 ETC3430 DESN2348 INTE2584 ENME502 MA3XJ INTE2627 MA2815 MA3AM L0101 Math 3MB3 CPCCBC4004 EEEM061 0502E220 APPLIED 6SSMN310 ENV 3310 PM2PCOL3 CONS6201 MATH502 FIT5003 EEEN313 MATH3001 MECH3460 ECON30009 COSC1252 Assembler COMP 3023 ENGG7601 COMP20005 COMP SCI 1103 ICSI 402 INF2C CS320 COMPSCI7064 ECE 3220 PROGRAMMAMING MIPS Computer Science 2413 AM3611 MSOA TE2101 FFT CMPSC-121 CSE 109 COMP SCI 7064 HCI CMP-5013A CCN2042 COMP3400 EEEN30052 CO4215 EE101 KIT107 CSE 421 CSE521 MCD4720 EECS 280 MIT 403 GRPC VG101 APiE EECE 1080C FIT3143 CSE 325 ISOM 3029 CS3307 State SIT255 Engineering MQF Modelling ACX3150 SIA322 ACCT90004 COMU7201 MGMT90018 WRIT1000 COMP3702 BISM1201 COMP7505 IDEA9106 CSE3SMT BFC5926 BFC5280 BFC5935 QBUS3330 ECON3700 IFN521 ECON2121 COSC2626 MA413 DDES9903 ENGG1300 COMP9337 CO2201 ACCT2002 ICM289 MMM054 MSBA-204 DIGM707 CSC2250 Econ312 EVAT python CAN404 LUBS2840 CSCI-GA.2250 COMSM0093 COMP281 AFM 333 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LAW3016 CS371 STAT 334 FIN0008 ACC3004 STAT 371 COMP1921 MTH201 ACCT2000 UN3412 CIV6000 MATHS 3021 MATE7014 ECON1 CIBC6035 LAW121G COMM 321 CSC8204R TTM2033 CSCI 2600 MAE 115 LNG302 CENV6175W1 ECON30019 CPS2015 BST969 DDES9015 AMATH242 CS 1151 N1592 ACCT600 EG503G SMM519 EC318 Ecstatistics EEE 471 MARK2051 Topology COM00037H ACCT3000 MTHM501 EG501V COM00055H BUSN9085 GGR273 MATH 2B ITD102 MA214 ECON0060 ECE3114 ECMM447 Mechanics CCT361H5S BSAN3210 WELF2001 Fin3001 RESP5001 STA 137 Linguistics 101 CSCI368 CMPSC497 LUBS3655 OMBA 5355 BUSM2562 BFF5902 CYBR7001 CULT2005 ECON 7720 MTRX3760 FSE598 LUE1001 MGSC 7100 CS3240 COSC7500 ACCT3091 INFOGR TRAN 2080 DESIGN CIVE215001 CIVE2360 CPSC 8430 FNCE2003 Stoichiometry INFS6023 Quantitative Method Math 21D ECE5179 Design ECON8022 Chemical DMS 213 ADM1370 BFF5902 BFF2701 AGRI10051 AMN400 ECON7322 FINC5090 MATH1052 SciComp Elec4622 FINS5516 ENG 101 MMM143 CS431CA2 BSAN2201 EBSC7070 ES9v9-10 PLIN0009 DATA 8 MATH 101 BCPM0097 MAT 237 BISM2201 EEE8099 Physics 11 ECE2238B Biostat 234 ECON30009 INF10025 CHEM252 CCMA4008 CIVE 3007 STAT0031 ELEC ENG 3108 MGMT90140 CHEM 252 STAT3016 MMM095 MMGT6016 Legal STAT 311 ELEC9764 COMP 370 TCS 3053 EECS 1021 FIT1051 C/C++ EE3C COMP2003J COM398 COMP51915 Simulator FIN42330 8PRO102 PGD ELEC5160 XJCO1921 COMP5123M INFS 2042 CHE1148H ECA5313 IN3329 Microsoft Modeling B15 2TT COMP3687 COMP2432 DMS2030 KAN4TSF CFKG CSE 5322 CSE 445 STATS 3DA3 CHC5223 COMP1048 EL3105 EECS 3221 CS231 ME 4434 ITP4510 ITP4501 CSCI 5708 Controller EECE 5699 CPEN 231L CPN programs CSC8016 MATU9D2 ENGN4213 2. SVM. (10 points) Consider an SVM classifier using RBF kernel, and finish the following python code to search for the optimum kernel parameter γ. You are required to calculate the RBF function and search yourself, instead of using functions in Scikit-Learn (sklearn). def parameter search(gamma list ,Xtr,ytr,Xts,yts ,...): """ Parameters gamma list: A list of RBF kernel parameter gamma to search for Xtr: Training data, shape (ntr,nrow,ncol) ytr: Training labels, shape (ntr,) Xts: Test data, shape (nts,nrow,ncol) yts: Test labels, shape (nts,) ... Add other parameters as needed Returns: gamma optimum: The gamma with minimum error on Xts, yts """ ... return gamma optimum

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