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MATH 270 TEST 3 REVIEW1. Given subspaces H and K of a vect.docxwkyra78
MATH 270 TEST 3 REVIEW
1. Given subspaces H and K of a vector space V , the sum of H and K, written as H +K,
is the set of all vectors in V that can be written as the sum of two vectors, one in H
and the other in K; that is
H + K = {w : w = u + v}, ∃u ∈ H and ∃v ∈ K. Show that H + K is a subspace of
V .
2. Based on problem 1, show that H is a subspace of H +K and K is a subspace of H +K.
3. Find an explicit description of Nul A by listing vectors that span the null space for the
following matrix :
A =
1 −2 0 4 00 0 1 −9 0
0 0 0 0 1
4. Let A =
[
−6 12
−3 6
]
and w =
[
2
1
]
.
Determine if w ∈ Col A. Is w ∈ Nul A ?
5. Define T : P2 → R2 by T(p) =
[
p(0)
p(1)
]
.
For instance, if p(t) = 3 + 5t + 7t2, then T(p) =
[
3
15
]
.
Show that T is a linear transformation. [Hint : For arbitrary polynomials p, q ∈ P2,
compute T(p + q) and T(cp) ].
6. Find a basis for the space spanned by the given vectors v1,v2,v3,v4,v5.
1
0
−3
2
,
0
1
2
−3
,
−3
−4
1
6
,
1
−3
−8
7
,
2
1
−6
9
7. Let v1 =
4−3
7
, v2 =
19
−2
, v3 =
711
6
and H = Span{v1,v2,v3}. It can be verified that 4v1 + 5v2 − 3v3 = 0.
Use this information to find a basis for H.
8. Find the coordinate vector [x]B of x relative to the given basis B = {b1,b2,b3}.
b1 =
1−1
−3
, b2 =
−34
9
, b3 =
2−2
4
, x =
8−9
6
9. Use an inverse matrix to find [x]B for the given x and B.
B =
{[
3
−5
]
,
[
−4
6
]}
, x =
[
2
−6
]
10. The set B = {1 + t2, t + t2, 1 + 2t + t2} is a basis for P2. Find the coordinate vector of
p(t) = 1 + 4t + 7t2 relative to B.
11. Use coordinate vectors to test the linear independence of the set of polynomials.
Explain your work.
1 + 2t3, 2 + t− 3t2,−t + 2t2 − t3
12. Find the dimension of Nul A and Col A for the matrix shown below.
A =
1 −6 9 0 −2
0 1 2 −4 5
0 0 0 5 1
0 0 0 0 0
13. Assume matrix A is row equivalent to B. Find bases for Col A, Row A and Nul A of
the matrices shown below.
A =
2 −3 6 2 5
−2 3 −3 −3 −4
4 −6 9 5 9
−2 3 3 −4 1
, B =
2 −3 6 2 5
0 0 3 −1 1
0 0 0 1 3
0 0 0 0 0
14. If a 3 × 8 matrix A has rank 3, find dim(Nul A), dim(Row A) and rank(AT ).
15. Let A = {a1,a2,a3} and B = {b1,b2,b3} be bases for a vector space V and suppose
a1 = 4b1−b2, a2 = −b1 +b2 +b3 and a3 = b2−2b3. Find the change-of-coordinate
matrix from A to B. Then find [x]B for x = 3a1 + 4a2 + a3.
16. Let B = {b1,b2} and C = {c1,c2} be bases for R2. Find the change-of-coordinate
matrix from B to C and the change-of-coordinate matrix from C to B.
b1 =
[
7
5
]
, b2 =
[
−3
−1
]
, c1 =
[
1
−5
]
, c2 =
[
−2
2
]
17. In P2, find the change-of-coordinate matrix from the basis
B = {1 − 2t + t2, 3 − 5t + 4t2, 2t + 3t2} to the standard basis C = {1, t, t2}.
Then find the B-coordinate vector for −1 + 2t.
18. Find a basi.
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Matrices
CMSC 56 | Discrete Mathematical Structure for Computer Science
November 30, 2018
Instructor: Allyn Joy D. Calcaben
College of Arts & Sciences
University of the Philippines Visayas
Mathematics (from Greek μάθημα máthēma, “knowledge, study, learning”) is the study of topics such as quantity (numbers), structure, space, and change. There is a range of views among mathematicians and philosophers as to the exact scope and definition of mathematics
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MATH 270 TEST 3 REVIEW1. Given subspaces H and K of a vect.docxwkyra78
MATH 270 TEST 3 REVIEW
1. Given subspaces H and K of a vector space V , the sum of H and K, written as H +K,
is the set of all vectors in V that can be written as the sum of two vectors, one in H
and the other in K; that is
H + K = {w : w = u + v}, ∃u ∈ H and ∃v ∈ K. Show that H + K is a subspace of
V .
2. Based on problem 1, show that H is a subspace of H +K and K is a subspace of H +K.
3. Find an explicit description of Nul A by listing vectors that span the null space for the
following matrix :
A =
1 −2 0 4 00 0 1 −9 0
0 0 0 0 1
4. Let A =
[
−6 12
−3 6
]
and w =
[
2
1
]
.
Determine if w ∈ Col A. Is w ∈ Nul A ?
5. Define T : P2 → R2 by T(p) =
[
p(0)
p(1)
]
.
For instance, if p(t) = 3 + 5t + 7t2, then T(p) =
[
3
15
]
.
Show that T is a linear transformation. [Hint : For arbitrary polynomials p, q ∈ P2,
compute T(p + q) and T(cp) ].
6. Find a basis for the space spanned by the given vectors v1,v2,v3,v4,v5.
1
0
−3
2
,
0
1
2
−3
,
−3
−4
1
6
,
1
−3
−8
7
,
2
1
−6
9
7. Let v1 =
4−3
7
, v2 =
19
−2
, v3 =
711
6
and H = Span{v1,v2,v3}. It can be verified that 4v1 + 5v2 − 3v3 = 0.
Use this information to find a basis for H.
8. Find the coordinate vector [x]B of x relative to the given basis B = {b1,b2,b3}.
b1 =
1−1
−3
, b2 =
−34
9
, b3 =
2−2
4
, x =
8−9
6
9. Use an inverse matrix to find [x]B for the given x and B.
B =
{[
3
−5
]
,
[
−4
6
]}
, x =
[
2
−6
]
10. The set B = {1 + t2, t + t2, 1 + 2t + t2} is a basis for P2. Find the coordinate vector of
p(t) = 1 + 4t + 7t2 relative to B.
11. Use coordinate vectors to test the linear independence of the set of polynomials.
Explain your work.
1 + 2t3, 2 + t− 3t2,−t + 2t2 − t3
12. Find the dimension of Nul A and Col A for the matrix shown below.
A =
1 −6 9 0 −2
0 1 2 −4 5
0 0 0 5 1
0 0 0 0 0
13. Assume matrix A is row equivalent to B. Find bases for Col A, Row A and Nul A of
the matrices shown below.
A =
2 −3 6 2 5
−2 3 −3 −3 −4
4 −6 9 5 9
−2 3 3 −4 1
, B =
2 −3 6 2 5
0 0 3 −1 1
0 0 0 1 3
0 0 0 0 0
14. If a 3 × 8 matrix A has rank 3, find dim(Nul A), dim(Row A) and rank(AT ).
15. Let A = {a1,a2,a3} and B = {b1,b2,b3} be bases for a vector space V and suppose
a1 = 4b1−b2, a2 = −b1 +b2 +b3 and a3 = b2−2b3. Find the change-of-coordinate
matrix from A to B. Then find [x]B for x = 3a1 + 4a2 + a3.
16. Let B = {b1,b2} and C = {c1,c2} be bases for R2. Find the change-of-coordinate
matrix from B to C and the change-of-coordinate matrix from C to B.
b1 =
[
7
5
]
, b2 =
[
−3
−1
]
, c1 =
[
1
−5
]
, c2 =
[
−2
2
]
17. In P2, find the change-of-coordinate matrix from the basis
B = {1 − 2t + t2, 3 − 5t + 4t2, 2t + 3t2} to the standard basis C = {1, t, t2}.
Then find the B-coordinate vector for −1 + 2t.
18. Find a basi.
I am Manuela B. I am a Calculus Assignment Expert at mathsassignmenthelp.com. I hold a Master's in Mathematics from, the University of Warwick Profession. I have been helping students with their assignments for the past 8 years. I solve assignments related to Calculus.
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Matrices
CMSC 56 | Discrete Mathematical Structure for Computer Science
November 30, 2018
Instructor: Allyn Joy D. Calcaben
College of Arts & Sciences
University of the Philippines Visayas
Mathematics (from Greek μάθημα máthēma, “knowledge, study, learning”) is the study of topics such as quantity (numbers), structure, space, and change. There is a range of views among mathematicians and philosophers as to the exact scope and definition of mathematics
📚 Struggling with Math Assignments? We've got your back! 🧮
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✅ Experienced Mathematicians: Our experts have years of experience tackling math problems of all levels.
✅ Customized Solutions: We provide tailored solutions to match your unique assignment requirements.
✅ On-Time Delivery: Tight deadline? No problem! We ensure your assignments are delivered promptly.
✅ 24/7 Support: Got a question at midnight? Our support team is ready to assist you anytime.
✅ Plagiarism-Free Work: Your assignments are crafted from scratch, guaranteeing originality.
📌 Our Math Assignment Services:
🔹 Algebra, Geometry, Calculus, Statistics, and more!
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2️⃣ Fill in the assignment details and requirements.
3️⃣ Get a quote and make a secure payment.
4️⃣ Relax while our experts work on your assignment.
5️⃣ Receive your completed assignment, review it, and excel in your academics!
🌐 Don't let math assignments stress you out. Our reliable and affordable math assignment help is just a click away. Trust the experts to guide you toward academic success. Get started now at mathsassignmenthelp.com!
Unlock Your Mathematical Potential with MathAssignmentHelp.com! 🧮✨Maths Assignment Help
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🌟 **Discover Our Services:**
🔢 **1. Personalized Guidance:** Our team of experienced mathematicians is ready to provide you with personalized guidance, making even the most challenging concepts crystal clear.
📊 **2. Homework Assistance:** Don't let assignments weigh you down! We offer top-notch homework help that ensures your work is accurate and submission-ready.
🧭 **3. Exam Prep:** Tackling upcoming exams? Our comprehensive study materials and expert insights will boost your confidence and help you excel.
📈 **4. Concept Clarification:** Whether it's calculus, algebra, statistics, or any other math branch, we specialize in clarifying concepts, filling in knowledge gaps, and helping you truly grasp the subject.
🤝 **Why Choose Us?**
✅ **Qualified Experts:** Our team consists of math aficionados with profound expertise in various mathematical domains.
⏱️ **Timely Assistance:** Tight deadlines? No problem! We're equipped to handle urgent assignments without compromising on quality.
🌐 **User-Friendly Platform:** Our website's intuitive interface ensures a seamless experience from start to finish.
🔒 **100% Confidential:** Your privacy matters. We ensure all your information and interactions with us remain strictly confidential.
🌈 **Embrace the Joy of Learning Math:** Mathematics is not just about numbers; it's about problem-solving, critical thinking, and unlocking new perspectives. Let's make your math journey exciting and fulfilling together!
📣 **Special Offer:** For a limited time, new users get an exclusive discount on their first service. Don't miss out on this opportunity to experience stress-free math learning!
👉 **Visit Us Today:** [MathAssignmentHelp.com](https://www.mathsassignmenthelp.com/)
📱 **Follow Us:** Stay updated with math tips, fun facts, and more on our social media channels! 📚🎉
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This slide is special for master students (MIBS & MIFB) in UUM. Also useful for readers who are interested in the topic of contemporary Islamic banking.
Executive Directors Chat Leveraging AI for Diversity, Equity, and InclusionTechSoup
Let’s explore the intersection of technology and equity in the final session of our DEI series. Discover how AI tools, like ChatGPT, can be used to support and enhance your nonprofit's DEI initiatives. Participants will gain insights into practical AI applications and get tips for leveraging technology to advance their DEI goals.
Normal Labour/ Stages of Labour/ Mechanism of LabourWasim Ak
Normal labor is also termed spontaneous labor, defined as the natural physiological process through which the fetus, placenta, and membranes are expelled from the uterus through the birth canal at term (37 to 42 weeks
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
This presentation includes basic of PCOS their pathology and treatment and also Ayurveda correlation of PCOS and Ayurvedic line of treatment mentioned in classics.
it describes the bony anatomy including the femoral head , acetabulum, labrum . also discusses the capsule , ligaments . muscle that act on the hip joint and the range of motion are outlined. factors affecting hip joint stability and weight transmission through the joint are summarized.
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...Dr. Vinod Kumar Kanvaria
Exploiting Artificial Intelligence for Empowering Researchers and Faculty,
International FDP on Fundamentals of Research in Social Sciences
at Integral University, Lucknow, 06.06.2024
By Dr. Vinod Kumar Kanvaria
A Strategic Approach: GenAI in EducationPeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
How to Build a Module in Odoo 17 Using the Scaffold MethodCeline George
Odoo provides an option for creating a module by using a single line command. By using this command the user can make a whole structure of a module. It is very easy for a beginner to make a module. There is no need to make each file manually. This slide will show how to create a module using the scaffold method.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Pride Month Slides 2024 David Douglas School District
Linear Algebra.pptx
1. For any Assignment related queries, Call us at:- +1 (315) 557-6473
You can mail us at info@mathsassignmenthelp.com or
reach us at: https://www.mathsassignmenthelp.com/
Linear Algebra
Problem Set 4 Solutions
https://mathsassignmenthelp.com/
2. Section 4.1. Problem 7. Every system with no solution is like the one in problem
6. There are numbers y1, . . . , ym that multiply the m equations so they add up to
0 = 1. This is called Fredholm’s Alternative:
Exactly one of these problems has a solution: Ax = b OR AT y = 0 with
yT b = 1.
If b is not in the column space of A it is not orthogonal to the nullspace of AT .
Multiply the equations x1 − x2 = 1 and x2 − x3 = 1 and x1 − x3 = 1 by numbers
y1, y2, y3 chosen so that the equations add up to 0 = 1.
2
Solution Let y1 = 1, y2 = 1 and y3 = −1. Then the left-hand side of the
sum of the equations is
(x1 − x2) + (x2 − x3) − (x1 − x3) = x1 − x2 + x2 − x3 + x3 − x1 = 0
and the right-hand side is
1 + 1 − 1 = 1.
Problem 9. If AT Ax = 0 then Ax = 0. Reason: Ax is inthe nullspace of AT and
also in the of A and those spaces are . Conclusion: AT A
has the same nullspace as A. This key fact is repeated in the next section.
Solution Ax is in the nullspace of AT and also in the column space of A
and those spaces are orthogonal.
Problem 31. The command N=null(A) will produce a basis for the nullspace of
A. Then the command B=null(N’) will produce a basis for the of A.
Solution The matrix N will have as its columns a basis for the nullspace
of A. Thus if a vector is in the nullspace of N T it must have dot product 0 with
every vector in the basis of N (A), thus it must be in the row space of A. Thus the
command null(N’) will produce a basis for the row space of A.
4. 4
ans =
1.0e-13 *
Columns 1 through 6
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
Columns 6 through 12
-0.0389 0.0533 -0.0344 -0.0888 -0.2420 -0.2043
0.0047 -0.0022 0.0044 0.0089 0.0289 0.0200
-0.1710 0.1843 -0.0999 -0.3020 -0.7816 -0.6395
0.0153 -0.0089 0.0092 0.0205 0.0444 0.0333
0.1155 -0.1377 0.0688 0.2132 0.5596 0.4796
0.1488 -0.1643 0.0888 0.2709 0.6928 0.5684
Note that if B has as its columns a basis for the row space of A then the rows of BT
will form a basis for the row space of A. Since the row reduced forms of A and BT
agree (up to 13 decimal places, but the numbers up there are just rounding error)
their rows must span the same space, so the columns of B are indeed a basis for the
row space of A.
Problem 32. Suppose I give you four nonzero vectors r, n, c, l in R2 .
a. What are the conditions for those to be bases for the four fundamental sub
spaces C (AT ), N (A ), C (A ), N (AT ) of a 2 × 2 matrix?
b. What is one possible matrix A?
Solution
a. In order for r and n to be bases for N (A) and C(AT ) we must have r ·n =
0, as the row space and null space must be orthogonal. Similarly, in order
5. at least a subspace of the desired column space. On the other hand, as r1 and
5
for c and l to form bases for C(A) and N (AT ) we need c ·l = 0, as the
column space and the left nullspace are orthogonal. In addition, we need
dim N (A) +dim C(AT ) = n and dim N (AT ) +dim C(A) = m; however, in this
case n = m = 1, and as the four vectors we are given are nonzero both of these
equations reduce to 1 + 1 = 2, which is automatically satisfied.
b. One possible such matrix is A = crT . Note that each column of A will be
a multiple of c, so it will have the right column space. On the other hand,
each row of A will be a multiple of r, so A will have the right row space.
The nullspaces don’t need to be checked, as any matrix with the correct row
and column space will have the right nullspaces (as the nullspaces are just the
orthogonal complements of the row and column spaces).
Problem 33. Suppose I give you four nonzero vectors r1, r2, n1, n2, c1, c2, l1, l2 in
R2 .
a. What are the conditions for those to be bases for the four fundamental sub
spaces C (AT ), N (A ), C (A ), N (AT ) of a 2 × 2 matrix?
b. What is one possible matrix A?
Solution
a. Firstly, by the same kind of dimension considerations as before we need the four
sets {r1, r2}, {n1, n2}, {c1, c2} and {l1, l2} to each contain linearly independent
vectors. (For example, if r1 and r2 are linearly dependent the dim C(AT ) = 1
not 2, and then dim C(AT ) + dim N (A) < 4 which can’t happen.)
Secondly, for i = 1, 2 and j = 1, 2 we need ri ·nj = 0 and ci ·lj = 0. This
will imply that the specified row space and nullspace are orthogonal, and that
the specified column space and left nullpace are also orthogonal. (When we
are given subspaces in terms of bases it suffices to check orthogonality on the
basis.)
b. One possible such matrix is
A =
�
c1 c2 r
1 2
�
�
r
�
T
.
Note that every column of A is a linear combination of c1 and c2, so C(A) is
6. � �
2 2 2 2
(I − P ) = I − I P − P I + P = I − 2P + P = I − 2P + P = I − P.
6
T
r are linearly independent we know that r r
2 1 2 has full row rank, so A
will have rank 2 and thus A has the right column space.
On the other hand,
T
A = r r c c
1 2 1 2
� �
� �
T
so C(AT ) is spanned by r1 and r2, as desired. Thus A has the right row space
and column space, and thus will have the right nullspace and left nullspace.
Section 4.2. Problem 13. Suppose A is the 4 × 4 identity matrix with its last
column removed. A is 4 × 3. Project b = (1, 2, 3, 4) onto the column space of A.
What shape is the projection matrix P and what is P ?
Solution P will be 4×4 since we take a 4-dimensional vector and project
it to another 4-dimensional vector. We will have
P = ⎜
⎝
⎛ ⎞
⎜ ⎟
⎟
⎠ .
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 0
(This can be seen by direct computation, or by simply observing that the column
space of A is the wxy-space, so we just need to remove the z coordinate to project.)
The projection of b is (1, 2, 3, 0).
Problem 16. What linear combination of (1, 2, −1) and (1, 0, 1) is closest to b =
(2, 1, 1)?
Solution
Note that 1 3
2
(1, 2, −1) +
2
(1, 0, 1) = (2, 1, 1).
So this b is actually in the span of the two given vectors.
Problem 17. If P 2 = P show that (I − P )2 = I − P . When P projects onto the
column space of A, I − P projects onto the .
Solution (4 points)
7. 6
� �
� �
When P projects onto the column space of A, I −P projects onto the left nullspace
of A.
Problem 30.
a. Find the projection matrix PC ontot he column space of A.
3 6 6
4 8 8
.
b. Find the 3 × 3 projection matrix PR onto the row space of A. Multiply B =
PC APR . Your answer B should be a little suprising — can you explain it?
Solution
a. Note that as A is rank 1 its column space is spanned by the vector a =
�
3 4
�
T
. Using this matrix we can compute
C
T −1 T
P = a(a a) a =
1
25
9 12
12 16
.
�
2 2
�
T
. Then we
b. The row space of A is spanned by the vector a = 1
compute ⎛
2
⎞
PR = a(aT
a)−1
aT
= ⎝ ⎠
1 2
2 4 4 .
1
9
2 4 4
Then B = PC APR = A. First, note that PC A = A, as for every vector x,
Ax ∈ C(A), so PC Ax = Ax. Analogously, APR = A, as for every vector x we
can write it uniquely as x = n + r with n in N (A) and r in C(AT ). Then Ax =
An + Ar = Ar by the definition of nullspace. But PR x = PR n + PR r = PRr, as
the nullspace is orthogonal to the row space, so projecting onto the row space
kills the nullspace. So APR = A. Thus PC APR = (PC A)PR = APR = A, as
desired.
Problem 31. In Rm , suppose I give you b and p, and n linearly independent
vectors a1, . . . , an. How would you test to see if p is the projection of b onto the
subspace spanned by the a’s?
8. 8
Solution
The projection of b must lie in the span of the a’s, and must also be the closest
vector in this span, meaning that the error will be orthogonal to this span. Thus we
need to check (a) that p is in the span of the a’s, and (b) that b − p is orthogonal
to ai for each i = 1, . . . , n. Note that just checking (b) is not enough because if we
set p = b then (b) will be satisfied but (a) will not be if b is not in the span of the
a’s.
Problem 34. If A has r independent columns and B has r independent rows, A B
is invertible.
Proof: When A is m by r with independent columns, we know that AT A is invertible.
If B is r by n with independent rows, show that B B T is invertible. (Take A = BT .)
Now show that A B has rank r.
Solution Let A = BT . As B has independent rows, A has independent
columns, so AT A is invertible. But AT A = (BT )T BT = B B T , so B B T is invertible,
as desired.
Note that AT A is r × r and is invertible, and B B T is r × r and is invertible, so
AT A B BT is r × r and invertible, so in particular has rank r. Thus we have that
AT (AB)BT has rank r. We know that multiplying A B by any matrix on the left or
on the right cannot increase rank, but can only decrease it. Thus we see that A B
has rank at least r. However, A B is r ×r, so it has rank r and is therefore invertible.
Section 8.2. Problem 13. With conductances c1 = c2 = 2 and c3 = c4 = c5 = 3,
multiply the matrices AT CA. Find a solution to AT C A x = f = (1, 0, 0, −1). Where
these potentials x and currents y = −CAx on the nodes and edges of the square
graph.
Solution (4 points) For this graph the incidence matrix is
⎞
⎛
⎜
⎜
⎜
⎝
⎟
⎟
⎟
⎠
A = 0
−1 1 0 0
−1 0 1 0
−1 1 0 .
0
0 −1 1
0 0 −1 1
We are told that the conductance matrix has diagonal entries (2, 2, 3, 3, 3). Then
⎜
⎝
4 −2 −2 0
−2 8 −3 −3
−2 −3 8 −3
0 −3 −3 6
⎛ ⎞
⎜ ⎟
⎟
⎠
T
A C A = .
9. A solution to the given equation is x = (5/12, 1/6, 1/6, 0); then y = (1/2, 1/2, 0, 1/2, 1/2).
The picture associated to this solution is
5/12 1/2 1/6
9
1/2
0
1/2
1/2
1/6 0
Problem 17. Suppose A is a 12 × 9 incidence matrix from a connected (but
unknown) graph.
a. How many columns of A are independent?
b. What condition on f makes it possible to solve AT y = f?
c. The diagonal entries of AT A give the number of edges into each node. What
is the sum of those diagonal entries?
Solution
a. Note that as A is 12 × 9 it is a graph with 9 nodes and 12 edges. As it is
connected elimination will produce a tree with 8 edges, so the rank of A is 8,
and so it has 8 independent columns.
b. In order to solve AT y we need the entries of f to add up to 0, as f needs to
be in C(AT ), which is orthogonal to N (A) and is generated by (1, 1,. . . , 1).
c. The sum of the entries of AT A is the sum of the degrees of all of the nodes.
As each edge hits exactly two nodes it will be counted twice, so the sum of
the diagonal entries is 24.