Linear Algebra Assignment 3 Solution: SIT292, Matrices, and Codes

This document presents a complete solution to a Linear Algebra assignment (SIT292), addressing several key concepts. The solution begins by determining the row-rank of a given matrix and identifying the basis for its row space. It then proceeds to calculate the eigenvalues and corresponding eigenvectors of a matrix, followed by the application of the Gram-Schmidt procedure to obtain orthonormal vectors. The assignment further explores subspaces, proving that a given set forms a subspace and determining its dimension and basis. Finally, the document delves into linear binary codes, constructing a generator matrix, generating code words, and computing the parity check matrix and syndromes for received words. Each question is solved with detailed steps and explanations, providing a thorough understanding of the concepts involved.

SIT292 LINEAR ALGEBRA
ASSIGNMENT- 3
STUDENT ID
[Pick the date]

Question 1
Given matrix
(a) Row- rank of matrix
In order to determine the rank of the above matrix, it is essential to transform the matrix into
upper triangular matrix with the help of row operations.
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It can be seen from the above upper triangular matrix that there is three non-zero rows and hence,
the row- rank of matrix A is 3.
(b) Set of generators for the row space of A is highlighted below:
{ (1 2−3 0 ) ( 2 4−2 2 ) ( 36−4 3 ) }
(c) Basis for the row space of matrix A
It is apparent that all the three rows of the given matrix A are linearly independent because the
row rank of the matrix is three. Further, these three rows will make the basis for the respective
row spaces. Therefore, the basis of the row space is shown below:
{ (1 2−3 0 ) ( 2 4−2 2 ) ( 36−4 3 ) }
Therefore, the generator and the basis vectors are the same for the given problem.
Question 2
Given matrix
Let A=
[0 2 0
1 0 1
0 2 0 ]
(a) Eigenvalue
2

Eigenvalues would be the roots of the matrix which would be determined by using the
determinant of ( A−λ I )=0
( A−λ I ) =
[ 0 2 0
1 0 1
0 2 0 ] −λ [ 1 0 0
0 1 0
0 0 1 ]
( A−λ I )=
[0 2 0
1 0 1
0 2 0 ]−
[ λ 0 0
0 λ 0
0 0 λ ]
( A−λ I )=
[−λ 2 0
1 −λ 1
0 2 −λ ]
det ( A−λ I )=
|−λ 2 0
1 −λ 1
0 2 −λ|
¿−λ ( λ2−2 )−2 (−λ−0 )+ 0(2+λ)
¿−λ ( λ2−2 )+2 λ+0
¿−λ3+ 2 λ+ 2 λ
¿−λ3+ 4 λ
¿−λ ( λ2−4 )
¿−λ ( λ−2 ) ( λ+2 )
Now, put det ( A−λ I )=0
−λ ( λ−2 ) ( λ+2 ) =0
λ=0 , 2 ,−2
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Hence, the eigenvalues of the given matrix is 0 , 2 ,−2.
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(b) Eigenvector corresponding to eigenvalues
 Eigenvector corresponding to eigenvalues λ = 0
( A−λ I ) =
[ 0 2 0
1 0 1
0 2 0 ] −0 [ 1 0 0
0 1 0
0 0 1 ]
¿ [0 2 0
1 0 1
0 2 0 ]−
[0 0 0
0 0 0
0 0 0 ]
¿ [ 0 2 0
1 0 1
0 2 0 ]
Reduce the matrix into row echelon form
[ (a ⋯ b
0 ⋱ ⋮
0 0 c ) ]
¿ [ 1 0 1
0 2 0
0 0 0 ]
Reduce the matrix to reduced row echelon form
[ (1 ⋯ b
0 ⋱ ⋮
0 0 1) ]
¿ [ 1 0 1
0 1 0
0 0 0 ]
Now,
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( A−λ I ) ( x
y
z )=
(0
0
0 )
( A−0 I ) ( x
y
z ) =
( 0
0
0 )
[ 1 0 1
0 1 0
0 0 0 ] ( x
y
z )=
( 0
0
0 )
x + z=0
y=0
In isolated form
y=0
x=−z
Now, let assume z = 1
Eigenvector v=
(−z
0
z )= (−1
0
1 )
 Eigenvector corresponding to eigenvalues = 2
( A−λ I )=
[0 2 0
1 0 1
0 2 0 ]−2 [1 0 0
0 1 0
0 0 1 ]
¿ [0 2 0
1 0 1
0 2 0 ]−
[2 0 0
0 2 0
0 0 2 ] 6

¿ [−2 2 0
1 −2 1
0 2 −2 ]
Reduce the matrix into row echelon form
[ (a ⋯ b
0 ⋱ ⋮
0 0 c ) ]
¿ [
−2 2 1
0 2 −2
0 0 0 ]
Reduce the matrix to reduced row echelon form
[ (1 ⋯ b
0 ⋱ ⋮
0 0 1) ]
¿ [ 1 0 −1
0 1 −1
0 0 0 ]
Now,
( A−λ I ) ( x
y
z )=
(0
0
0 )
( A−2 I ) ( x
y
z )= (0
0
0 )
[1 0 −1
0 1 −1
0 0 0 ] ( x
y
z )=
(0
0
0)
x−z=0
7

y−z =0
In isolated form
x=z
y=z
Now, let assume z = 1
Eigenvector v=
( z
z
z )= ( 1
1
1 )
 Eigenvector corresponding to eigenvalues = -2
( A−λ I )=
[0 2 0
1 0 1
0 2 0 ]−(−2) [1 0 0
0 1 0
0 0 1 ]
¿ [0 2 0
1 0 1
0 2 0 ]−
[−2 0 0
0 −2 0
0 0 −2 ]
¿ [2 2 0
1 2 1
0 2 2 ]
Reduce the matrix into row echelon form
[ (a ⋯ b
0 ⋱ ⋮
0 0 c ) ]
¿ [ 2 2 0
0 2 2
0 0 0 ]
Reduce the matrix to reduced row echelon form
8

[ (1 ⋯ b
0 ⋱ ⋮
0 0 1) ]
¿ [ 1 0 −1
0 1 1
0 0 0 ]
Now,
( A−λ I ) ( x
y
z )=
(0
0
0 )
( A−(−2) I ) ( x
y
z )=
(0
0
0)
[1 0 −1
0 1 1
0 0 0 ] ( x
y
z )=
(0
0
0)
x−z=0
y + z=0
In isolated form
x=z
y=−z
Now, let assume z = 1
Eigenvector v=
( z
−z
z )= ( 1
−1
1 )
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Therefore, the eigenvector of the matrix is (−1
0
1 ), (1
1
1 ), ( 1
−1
1 )
(c) Set of orthonormal vectors with the help of Gram-Schmidt Procedure
Computed eigenvector of the matrix is (−1
0
1 ), (1
1
1 ), ( 1
−1
1 )
Let
x= (
−1
0
1 )
y= (1
1
1 )
z= ( 1
−1
1 )
Gram-Schmidt Procedure
u= y− ¿ x , y > x
¿ y , y >¿ ¿
u= ( 1, 1 ,1 ) − ¿ (−1,0,1 ) , ( 1 ,1 , 1 ) > x
¿ x , x>¿= ( 1,1,1 ) ¿
It is apparent that that eigenvector corresponding to distinct eigenvalues are orthogonal and
hence, {x , y , z } are orthogonal vectors.
Hence,
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{ x
||x|| , y
|| y|| , z
||z|| } are orthonormal vectors .
Thus,
Final Solution: Orthogonal vectors
Question 3
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