SIT292留学生代做、LINEAR CODES代做、R编程语言调试、R代写代写

日期：2019-09-14 10:40

SIT292 ASSIGNMENT 3: ORTHOGONALISATION, SUBSPACES AND

LINEAR CODES

GUILLERMO PINEDA-VILLAVICENCIO

Instructions

This is an individual assignment. The aim of the assignment is that the student applies

concepts and methods studied in Weeks 8-10 to solve problems on the Gram-Schmidt

orthogonalisation process, subspaces and linear codes.

The assignment has a value of 30 points and is worth 15% of the unit marks. It consists

of three problems that are to be solved.

Submission

Students must submit the assignment in clear handwriting. The answers should be

provided on the assignment, after the corresponding questions. The solutions should be

clear enough so that a fellow student can understand all their steps; and they should

demonstrate the student’s understanding of all procedures used to solve the problems.

The assignment is due on Thursday September 26 2019 (Week 11) at 5pm. The student

should submit the assignment electronically through the DeakinSync unit site by the due

date. Only one pdf file must be submitted.

References

Learning materials of Weeks 8-10 of the SIT292 Unit Shell.

Problems

(1) In this problem we investigate and use the QR Decomposition of a matrix A whose

columns form a set of linearly independent vectors. The process goes as follows.

Let c1, . . . , cn be the columns of a d × n matrix A.

Step 1: Obtain an orthonormal basis q1

, . . . , qn

from c1, . . . , cn using the GramSchmidt

orthogonalisation process.

Step 2: Form a matrix Q with columns q1

, . . . , qn

.

Step 3: Find an upper triangular n × n matrix R so that A = QR.

Answer the following questions.

(a) Provide a formula to compute the (i, j)-entry ri,j , row i and column j, of the

matrix R. Prove that the matrix R is nonsingular.

Hint: For deducing a formula for ri,j use the fact that Q?1 = QT

.

(b) Let A1 be a 4×3 matrix with columns [1, 0, 0, 0], [2, 1, 0, 0], [2, 2, 0, 0], and let

A2 be a 4 × 3 matrix with columns [1, 1, 0, 1], [0, 1, 1, 0], [1, 4, 0, 1]. Determine

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whether or not the columns of each matrix form a set of linearly independent

vectors. Justify your answers.

(c) Out of the two matrices in Part (b), pick one matrix with linearly independent

columns and find its QR Decomposition (3 marks). Verify that the matrix

equals QR (1 mark).

2+4+4=10 marks

Part (a) The possible marks are 0, 1 and 2, and they have the following

meaning.

0: No formula or proof is provided.

1: Either a correct formula or a correct proof is provided, but not both.

2: Both a correct formula and a correct proof are provided.

—

Part (b) For each matrix the possible marks are 0, 1 and 2, and they have

the following meaning.

0: The linear independence of the columns of the matrix is incorrectly

determined.

1: There are mistakes in the answer, but the procedure is correct.

2: The linear independence of the columns of the matrix is correctly determined.

—

Part (c)

? The student will receive 2 marks for a correct computation of the matrix

Q using the Gram-Schmidt orthogonalisation process, 1 mark for a

correct Gram-Schmidt orthogonalisation process with significant computation

mistakes, and 0 for an incorrect procedure.

? The student will receive 1 mark for a correct computation of the matrix

R, 1 mark for a correct process with significant computation mistakes,

and 0 for an incorrect procedure.

? The student will receive 1 mark for a correct verification that the original

matrix equal QR for the computed matrices Q and R, 0.5 marks

for a correct multiplication of matrices with significant computation

mistakes, and 0 for an incorrect procedure.

Answer.

SIT292 ASSIGNMENT 3: ORTHOGONALISATION, SUBSPACES AND LINEAR CODES 3

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SIT292 ASSIGNMENT 3: ORTHOGONALISATION, SUBSPACES AND LINEAR CODES 5

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SIT292 ASSIGNMENT 3: ORTHOGONALISATION, SUBSPACES AND LINEAR CODES 7

(2) Consider the set S := {[1, 0, 0, 2], [0, 1, 0, 1], [1, 1, 0, 4], [2, 2, 0, 2]} of vectors in R

4

.

(a) Determine a basis B of the subspace W of R

4 generated by the set S (3

marks).

(b) Determine the dimension of W (1 mark). Justify your answer.

(c) Determine whether or not the vector v = [0, 0, 1, 0] is in W (2 marks). Justify

your answer.

(d) If the vector v is in W, find its coordinates with the respect the basis B (4

marks). If it is not in W, then add vectors to B to obtain a basis B0 of the

vector space R

4 and find the coordinates of v with respect to the new basis

B0

(4 marks).

3+1+2+4=10 marks

Part (a) The possible marks are 0, 1, 2 and 3.

0: The procedure to solve the question is incorrect.

1: The procedure to solve the question is roughly correct, but the final

answers are incorrect.

2: The procedure to solve the question is correct, but the final answers are

incorrect.

3: Both the procedure and the answers are correct.

—

Part (b) The possible marks are 0, 0.5 and 1.

0: The answer and the justification are incorrect.

0.5: Either the answer or the justification is correct, but not both.

1: Both the answer and the justification are correct.

—

Part (c) The possible marks are 0, 1 and 2.

0: The answer and the justification are incorrect.

1: Either the answer or the justification is correct, but not both.

2: Both the answer and the justification are correct.

—

Part (d) The possible marks are 0, 1, 2, 3 and 4.

If a correct answer and justification is provided for the relevant case, then

the mark is 4. For different degrees of correctness, the marks are 3, 2 or 1.

Otherwise the mark is 0.

Answer.

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(3) Consider the 4-dimensional cube of Fig. 1 and the highlighted code words K =

{v2, v4, v6, v8, v10, v12, v14, v16}. Answer the following questions.

(a) Verify that the code K is a linear code (2 marks).

(b) Write down a generator matrix G for this code (2 marks).

(c) Is this a (4, 3) code? (1 mark). Justify your answer.

(d) Write down the parity check matrix H for this code (1 mark).

(e) Find the error syndrome of the received words 1100 (1 mark) and 0111 (1

mark).

(f) Are the words in Part (e) code words? Justify your answer (2 marks).

2+2+1+1+2+2=10 marks

v10 v12 v13 v15 v14 v16 v7

v11 v9

v1 = 1000 v9 = 1100

v2 = 0000 v10 = 0100

v3 = 1001 v11 = 1101

v4 = 0001 v12 = 0101

v5 = 1010 v13 = 1110

v6 = 0010 v14 = 0110

v7 = 1011 v15 = 1111

v8 = 0011 v16 = 0111

Figure 1. The 4-dimensional cube with labeled vertices.

SIT292 ASSIGNMENT 3: ORTHOGONALISATION, SUBSPACES AND LINEAR CODES 11

Part (a) The possible marks are 0, 1 and 2.

0: The student has not verified that the set is closed under addition or

under scalar multiplication.

1: The student has verified that the set is closed under either addition or

scalar multiplication, but not both.

2: The student has verified that the set is closed under both addition and

scalar multiplication.

—

Part (b) The possible marks are 0, 1 and 2.

0: The student has not produced the correct matrix.

1: The student has not produced the correct matrix but the procedure is

correct.

2: The student has produced both the correct matrix and procedure.

—

Part (c) The possible marks are 0, 0.5 and 1.

0: If the answer and the justification for being a (4, 3) code is incorrect,

the student gets 0 marks.

0.5: If either the answer or the justification for being a (4, 3) code is correct,

but not both, then the student gets 0.5 marks.

1: Otherwise the students gets 1 mark.

—

Part (d) The possible marks are 0, 0.5 and 1.

0: The student has not produced the correct matrix.

0.5: The student has not produced the correct matrix but the procedure is

correct.

1: The student has produced the correct matrix and the procedure is correct.

—

Part (e) For each error syndrome the possible marks are 0, 0.5 and 1.

0: The error syndrome is not computed correctly and the procedure is

incorrect.

0.5: The procedure is correct, but the error syndrome is not correct.

1: Both the error syndrome and the procedure is correct.

—

Part (f) For each word, the possible marks are 0, 0.5 and 1.

0: If the answer and the justification for being a code word is incorrect,

the student gets 0 marks.

0.5: If either the answer or the justification for being a code word is correct,

but not both, then the student gets 0.5 marks.

1: If the answer and the justification for being a code word is correct, the

student gets 1 mark.

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Answer.

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