# Mathematics II PEAK 2016 – 2017 (1)

## Atsushi Matsuo

Continue to Mathematics II PEAK 2016 – 2017 (2)
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NOTICES   CLASS SCHEDULE   COURSE INFORMATION   LECTURE PLAN   TUTORIAL   PAST NOTICES
• NOTICES

• May 2, 2017 (revised on May 3, May 5, May 8)
• On marking the make-up exam
• Problem 1 (1): The answer is ±(2/3, -2/3, 1/3)T. They are easily found by calculating the cross product (i.e. the vector product) of the given vectors and normalizing it with ± signs.
Alternatively, let (x, y, z) be a vector to be found and write down the conditions it should satisfy in terms of the inner products with the given vectors as well as the condition that the vector shoud be a unit vector. One can then find the vectors by solving the system of equations on x, y, z formed by the conditions so obtained. Thus the question is actually of high school level mathematics.
Some students erroneously applied the Gram-Schmidt process to the given vectors, which leads to some vectors w1, w2 different from what are to be found according to the requirement of the question.
One may actually apply the Gram-Schmidt process to the given vectors followed by a vector linearly independent of them. Let w1, w2, w3 be the series of vectors obtained by the process. Then normalizing the vector w3 with ± sings, one obtains the vectors to be found.
One can check his/her answer by calculating the inner products with the given vectors. Some students neglected such a check and ended up losing points by giving an obviously wrong answer with vectors not orthogonal to the given vectors.
• Problem 1 (2): The answer is given by the matrix with entries  3 2 0 -1
One can easily answer this question just by knowing what is the representation matrix of a linear transformation with respect to a basis.
Some students erroneously calculated the matrix by replacing the basis only in the first R2 (the source) or only in the second R2 (the target) of the given linear map A : R2 → R2.
• Problem 2 (1): The answer is -4.
There are many ways to calculate the determinants of 3×3 matrices. So one can check his/her result by calculating it in plural ways. Some students neglected such a check and ended up losing points by giving a wrong answer.
• Problem 2 (2): The answer is given by the matrix with entries  2 2 2 1 1 2 2 1 1 1 2 1 1 1 1 1
One can verify the result by multiplying it with the given matrix. Some students neglected such a verification and ended up losing points by giving a wrong answer.
• Problem 3 (1): The answer is given by the matrix with entries  1 1 0 3 0 0 1 -1 0 0 0 0
• Problem 3 (2): One of the bases is given by (1, -1, 0, 0)T, (3, 0, -1, -1)T, for instance. There are many other bases which fulfill the requirement of the question.
To find it, recall that a vector v = (x, y, z, w)T belongs to the kernel if and only if Av = 0. It is equivalent to Bv = 0 where B is the reduced row echelon form obtained in (1). So the condition is x + y + 3w = 0 and z - w = 0. Thus, for instance, one finds the vectors (1, -1, 0, 0)T, (3, 0, -1, -1)T form a basis of the kernel. Now it suffices to verify that they are indeed linearly independent and they indeed span the kernel.
Some students gave obviously wrong answers consisting of 3-dimensional vectors, although the kernel of a 3 × 4 matrix consists of 4-dimensional vectors.
• Problem 4 (1): The answer is 0. To see it, just use the relation among rank A, dim Ker A and n for A : Rn → Rm (sometimes called rank-nullity theorem) and the fact that rank A = rank AT.
• Problem 4 (2): The answer is (1, -1, 0)T, (1/2, 1/2, -1)T, (2/3, 2/3, 2/3)T. One may further normalize them to get an orthonormal basis.
Some students gave obviously wrong answers with vectors not mutually orthogonal.
• Problem 5: The answer is given by the matrix with entries  -1/3 2/3 2/3 2/3 -1/3 2/3 2/3 2/3 -1/3
To see it, it suffices to find the images of the fundamental unit vectors e1, e2, e3. The problem has now reduced to an easy exercise of high school level mathematics.
• April 17, 2017
• Change of the make-up exam schedule:
• The make-up exam will take place on April 26th (Wed) 10:35-12:05.
• April 4, 2017
• To those who are to take make-up exam for A semester:
• The make-up exam will take place on April 26th (Wed) 08:40-10:10.
• Review the lectures by consulting the handouts as well as your notes, solve the questions in the handouts carefully, and correct your errors by yourself. Repeat it and train yourself.
• Properly recognize the concepts in question and correctly identify what you are to calculate.
• February 7, 2017
• On marking the exam
• Problem 1 (1): The answer is (2,2,2)T. As the orthogonal projection in question is to the subspace spanned by (1,1,1)T, one can know at least that the correct answer must be parallel to (1,1,1)T.
• Problem 1 (2): One of the correct answers is given by (x,y,z)=(1-t,2-t,t) with t running over real numbers. One needs to give all the solutions by expressing them in such a way. As the question does not specify the method to find solutions, one does not need to use the elementary operations on matrices in this question.
• Problem 2 (1): The answer is 10. I recommend you to work with many ways in finding determinants in order to make sure that your answer is correct.
• Problem 2 (2): The answer is given by the matrix with entries  3 1 3 2 2 1 2 1 0 0 3 2 0 0 2 1
• Problem 3 (1): The rank is 3 and the dimension of the kernel is 1. It is required by the question that the reduced row echelon form is correctly calculated.
• Problem 3 (2): The first, the third and the fourth columns form a basis of Im A. The second, the third and the fourth columns also form a basis of Im A. It is required by the question to show that the vectors are linearly independent and span Im A. (Although Im A=R3 by (1), it is much easier to show that the vectors span Im A than to show that they span R3.) The fundamental unit vectors e1, e2, e3 form a basis of R3, of course, but they are not columns of A.
• Problem 4 (1): Some students erroneously tried to show that p' (x1+x2)=p' (x1)+p' (x2) and p' (cx)=c p' (x), but they do not hold, of course.
• Problem 4 (2): The answer is given by the matrix with entries  0 1 0 0 0 2 0 0 0
• Problem 5: Some students erroneously wrote the volume to be (a1·|a2× a3|).
• January 16, 2017
• I should have mentioned in the presetation of 2.1 that the definition of the rank of a matrix is the maximal number of linearly independent columns rather than that of rows although the two numbers agree as a matter of fact. So even the final results presented by the student was correct, the explanation given there was not quite right.
• Many students seem to be overlooking the definitions of various concepts given in the handouts and the lectures. You obviously cannot solve problems without knowing the meaning of them according to the precise definition of the terminologies. You should always check whether you correctly understand and remember the definitions or not.
• Remember I did tell you at the very beginnin of the lectures that you should listen carefully to my explanations, take your own notes on them, review the contents of the lecture, and do the homework every week before your memories fade away.
• November 21, 2016
• No class on November 28.
• Prepare for the exercise session on December 5.
• October 23, 2016
• Lecture cancelled on October 24.
As I have serious difficulty in walking caused by pulled muscle in calf, I have to cancel the lecture on October 24. Excuse me for inconvenience caused by my carelessness.
• October 3, 2016
• Corrections:
Lecture 2 Summary 2.5: ‘v + x = v = 0 + x’ should read ‘v + x = v = x + v’.
Lecture 3 Summary 2.3: ‘j-th entry yj’ should read ‘i-th entry yi’.
• September 26, 2016
• I will change the title of Lecture 1 to ‘Sets and maps’. The rest of the topics of originally planned lecture 1 will be covered in Lecture 2 under the title ‘Binary operations and vector spaces’.
• You are supposed to do homeworks, but not to submit. Instead, I will occasionally specify some of them for your reports to be checked by our TA. Note that homeworks and reports, as being purely educational, will not be considered for evaluation. The evaluation will only be based on the result of final examination in January.
• CLASS SCHEDULE

The schedule may change.

1. 2016/09/26 (Mon)  Lecture 1. Sets and maps.
2. 2016/10/03 (Mon)  Lecture 2. Binary operations and vector spaces.
3. 2016/10/17 (Mon)  Lecture 3. Matrices and linear maps.
4. 2016/10/24 (Mon)  — Cancelled —
5. 2016/10/31 (Mon)  Lecture 4. Square matrices and linear transformations.
6. 2016/11/07 (Mon)  Lecture 5. System of linear equations and matrices.
7. 2016/11/14 (Mon)  Lecture 6. Elementary operations on matrices.
8. 2016/11/21 (Mon)  Lecture 7. Bases and dimensions.
9. 2016/12/05 (Mon)  — Exercise session 1 —
10. 2016/12/12 (Mon)  Lecture 7 (continued).
11. 2016/12/19 (Mon)  Lecture 8. Ranks of matrices.
12. 2016/12/26 (Mon)  Lecture 9. Orthonormal bases.
13. 2017/01/07 (Sat)   Lecture 10. Vector products of spatial vectors.
14. 2017/01/16 (Mon)  — Exercise session 2.
15. 2017/01/23 (Mon)  Exam for A semester.
Time and Place: 13:10-14:40  in Room 121
16. 2017/04/26 (Wed)  Make-up Exam for A semester.
Time: 08:40-10:10 10:35-12:05

• LECTURE PLAN

The following topics will be covered by the course.

1. Binary operations and vector spaces
2. Matrices and linear maps
3. Systems of linear equations and matrices
4. Elementary operations on matrices
5. Bases and dimensions
6. Change of bases
7. Ranks of matrices
8. Orthonormal bases
9. Vector product of spatial vectors
10. Determinants and volumes

• TUTORIAL

Tutorial by assistant professors will take place as shown below.
The scheduled days are marked by pink.

Sepember
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567891011
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19202122 232425
26272829 30

October
MonTueWedThuFriSatSun
12
34 56789
10111213 141516
17181920 212223
24252627 282930
31

November
MonTueWedThuFriSatSun
123 4 56
78910 111213
14151617 181920
212223 24252627
282930

December
MonTueWedThuFriSatSun
1234
5678 91011
12131415 161718
19202122 232425
262728293031

January
MonTueWedThuFriSatSun
1
2345 678
9101112 131415
16171819202122
23242526272829
3031