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
SIT292留学生作业代做、LINEAR CODES课程作业代做
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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SIT292 ASSIGNMENT 3: ORTHOGONALISATION, SUBSPACES AND LINEAR CODES 9
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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.
SIT292 ASSIGNMENT 3: ORTHOGONALISATION, SUBSPACES AND LINEAR CODES 13
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SIT292 ASSIGNMENT 3: ORTHOGONALISATION, SUBSPACES AND LINEAR CODES 15
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原文地址：https://www.cnblogs.com/toyma/p/11528993.html