MAT3701/201/2/2019
Tutorial letter 201/2/2019
LINEAR ALGEBRA
MAT3701
Semester 2
Department of Mathematical Sciences
This tutorial letter contains solutions to Assignment 01.
BARCODE
Define tomorrow.
, ASSIGNMENT 01
Solution
UNIQUE ASSIGNMENT NUMBER: 656505
Please note that we will only mark a selection of the questions. It is therefore in your own best interest
to do all the questions. The fact that a question is not marked does not mean that it is less important
than one that is marked. We try to cover the whole syllabus over the two semesters (4 assignments)
and to use these assignments to help you prepare for the exam. It is therefore good practice to work
through a complete set of four assignments for a given year, and for this reason the assignments and
worked solutions of previous years are posted on myUnisa under Additional Resources – see also the
letter MAT3701 Exam Preparation under Additional Resources.
Worked solutions to all the questions for this assignment will be sent to all students and posted
on myUnisa shortly after the due date. Your answers to the assignment questions should be fully
motivated.
For this assignment, questions 2, 3, 5 and 7 will be marked.
Question 1
This question was not marked.
It is still important to work through the solution and compare it with your own attempt.
0 1 1 0
Let A = and B = , and let
1 0 0 −1
W1 = {X ∈ M2×2 (C) : AX = XB}
and
W2 = {X ∈ M2×2 (C) : AX = X}.
(1.1) Show that W1 is a subspace of M2×2 (C) .
(1.2) Given that W2 is a subspace of M2×2 (C) , find a basis for W1 ∩ W2 .
(1.3) Explain whether M2×2 (C) = W1 ⊕ W2 .
Solution
(1.1) Check the three conditions for a subspace:
ST1: W1 6= φ, since e.g. 02×2 ∈ W1 .
ST2: Suppose X1 ∈ W1 and X2 ∈ W1 . Then
A (X1 + X2 ) = AX1 + AX2
= X1 B + X2 B since X1 , X2 ∈ W1
= (X1 + X2 ) B.
Thus X1 + X2 ∈ W1 .
2
, MAT3701/201/2/2019
ST3: Suppose z ∈ C and X ∈ W1 . Then
A (zX) = z (AX)
= z (XB) since X ∈ W1
= (zX) B.
Thus zX ∈ W1 .
Since all three conditions are satisfied, W1 is a subspace of M2×2 (C) .
(1.2) Let X = [X1 X2 ] where X1 and X2 denote the first and second columns of X respectively.
Then
X ∈ W1 ∩ W2 ⇔ AX = XB and AX = X
1 0
⇔ A [X1 X2 ] = [X1 X2 ] and A [X1 X2 ] = [X1 X2 ]
0 −1
⇔ [AX1 AX2 ] = [X1 − X2 ] and [AX1 AX2 ] = [X1 X2 ]
⇔ AX1 = X1 and AX2 = −X2 = X2
⇔ AX1 = X1 and X2 = 0.
x1
Let X1 = , then
x2
0 1 x1 x1
AX1 = X1 ⇔ =
1 0 x2 x2
⇔ x2 = x1 and x1 = x2
⇔ x1 = x2 .
Thus
x1 1
X1 = = x1 , x1 ∈ C
x1 1
and therefore
1 0
X = [X1 X 2 ] = x1 , x1 ∈ C
1 0
1 0
so is a basis for W1 ∩ W2 .
1 0
(1.3) No, since W1 ∩ W2 6= {0} .
Question 2
Let V, W, U1 and U2 be subspaces of M2×2 (C) defined by
a b
V = : a + b + c + 2d = 0 ,
c d
a b
W = : a+b+c=0 ,
c 0
1 −1 2 0
U1 = span and U2 = span .
0 0 0 −1
3
, (2.1) Show that W, U1 and U2 lie in V . (6)
(2.2) Find a basis for W and extend it to a basis for V . (14)
(2.3) Explain whether V = W ⊕ U1 . (4)
(2.4) Explain whether V = W ⊕ U2 . (4)[28]
Solution
(2.1) It follows that W ⊆ V by letting d = 0 in the definition of V . Since
1 −1 1 −1 z −z
U1 = span = z :z∈C = :z∈C ,
0 0 0 0 0 0
it follows that U1 ⊆ V by letting a = z, b = −z and c = d = 0 in the definition of V.
Similarly
2 0 2 0 2z 0
U2 = span = z :z∈C = :z∈C ,
0 −1 0 −1 0 −z
hence it follows that U2 ⊆ V by letting a = 2z, b = c = 0 and d = −z in the definition of V.
(2.2)
a b −b − c b −1 1 −1 0
W = : a+b+c=0 = : b, c ∈ C = b +c : b, c ∈ C
c 0 c 0 0 0 1 0
−1 1 −1 0
= span , .
0 0 1 0
−1 1 −1 0
Since , is linearly independent, it is a basis for W. Further, since dim(V ) = 3
0 0 1 0
2 0 2 0 −1 1 −1 0
(show), ∈ V (from (2.1)) and ∈/ span , , it follows
0 −1 0 −1 0 0 1 0
that
−1 1 −1 0 2 0
B= , , (*)
0 0 1 0 0 −1
is a basis for V.
(2.3) This statement is false since U1 ⊂ W (show).
(2.4) This statement is true since B in (*) is a basis for V.
4
Tutorial letter 201/2/2019
LINEAR ALGEBRA
MAT3701
Semester 2
Department of Mathematical Sciences
This tutorial letter contains solutions to Assignment 01.
BARCODE
Define tomorrow.
, ASSIGNMENT 01
Solution
UNIQUE ASSIGNMENT NUMBER: 656505
Please note that we will only mark a selection of the questions. It is therefore in your own best interest
to do all the questions. The fact that a question is not marked does not mean that it is less important
than one that is marked. We try to cover the whole syllabus over the two semesters (4 assignments)
and to use these assignments to help you prepare for the exam. It is therefore good practice to work
through a complete set of four assignments for a given year, and for this reason the assignments and
worked solutions of previous years are posted on myUnisa under Additional Resources – see also the
letter MAT3701 Exam Preparation under Additional Resources.
Worked solutions to all the questions for this assignment will be sent to all students and posted
on myUnisa shortly after the due date. Your answers to the assignment questions should be fully
motivated.
For this assignment, questions 2, 3, 5 and 7 will be marked.
Question 1
This question was not marked.
It is still important to work through the solution and compare it with your own attempt.
0 1 1 0
Let A = and B = , and let
1 0 0 −1
W1 = {X ∈ M2×2 (C) : AX = XB}
and
W2 = {X ∈ M2×2 (C) : AX = X}.
(1.1) Show that W1 is a subspace of M2×2 (C) .
(1.2) Given that W2 is a subspace of M2×2 (C) , find a basis for W1 ∩ W2 .
(1.3) Explain whether M2×2 (C) = W1 ⊕ W2 .
Solution
(1.1) Check the three conditions for a subspace:
ST1: W1 6= φ, since e.g. 02×2 ∈ W1 .
ST2: Suppose X1 ∈ W1 and X2 ∈ W1 . Then
A (X1 + X2 ) = AX1 + AX2
= X1 B + X2 B since X1 , X2 ∈ W1
= (X1 + X2 ) B.
Thus X1 + X2 ∈ W1 .
2
, MAT3701/201/2/2019
ST3: Suppose z ∈ C and X ∈ W1 . Then
A (zX) = z (AX)
= z (XB) since X ∈ W1
= (zX) B.
Thus zX ∈ W1 .
Since all three conditions are satisfied, W1 is a subspace of M2×2 (C) .
(1.2) Let X = [X1 X2 ] where X1 and X2 denote the first and second columns of X respectively.
Then
X ∈ W1 ∩ W2 ⇔ AX = XB and AX = X
1 0
⇔ A [X1 X2 ] = [X1 X2 ] and A [X1 X2 ] = [X1 X2 ]
0 −1
⇔ [AX1 AX2 ] = [X1 − X2 ] and [AX1 AX2 ] = [X1 X2 ]
⇔ AX1 = X1 and AX2 = −X2 = X2
⇔ AX1 = X1 and X2 = 0.
x1
Let X1 = , then
x2
0 1 x1 x1
AX1 = X1 ⇔ =
1 0 x2 x2
⇔ x2 = x1 and x1 = x2
⇔ x1 = x2 .
Thus
x1 1
X1 = = x1 , x1 ∈ C
x1 1
and therefore
1 0
X = [X1 X 2 ] = x1 , x1 ∈ C
1 0
1 0
so is a basis for W1 ∩ W2 .
1 0
(1.3) No, since W1 ∩ W2 6= {0} .
Question 2
Let V, W, U1 and U2 be subspaces of M2×2 (C) defined by
a b
V = : a + b + c + 2d = 0 ,
c d
a b
W = : a+b+c=0 ,
c 0
1 −1 2 0
U1 = span and U2 = span .
0 0 0 −1
3
, (2.1) Show that W, U1 and U2 lie in V . (6)
(2.2) Find a basis for W and extend it to a basis for V . (14)
(2.3) Explain whether V = W ⊕ U1 . (4)
(2.4) Explain whether V = W ⊕ U2 . (4)[28]
Solution
(2.1) It follows that W ⊆ V by letting d = 0 in the definition of V . Since
1 −1 1 −1 z −z
U1 = span = z :z∈C = :z∈C ,
0 0 0 0 0 0
it follows that U1 ⊆ V by letting a = z, b = −z and c = d = 0 in the definition of V.
Similarly
2 0 2 0 2z 0
U2 = span = z :z∈C = :z∈C ,
0 −1 0 −1 0 −z
hence it follows that U2 ⊆ V by letting a = 2z, b = c = 0 and d = −z in the definition of V.
(2.2)
a b −b − c b −1 1 −1 0
W = : a+b+c=0 = : b, c ∈ C = b +c : b, c ∈ C
c 0 c 0 0 0 1 0
−1 1 −1 0
= span , .
0 0 1 0
−1 1 −1 0
Since , is linearly independent, it is a basis for W. Further, since dim(V ) = 3
0 0 1 0
2 0 2 0 −1 1 −1 0
(show), ∈ V (from (2.1)) and ∈/ span , , it follows
0 −1 0 −1 0 0 1 0
that
−1 1 −1 0 2 0
B= , , (*)
0 0 1 0 0 −1
is a basis for V.
(2.3) This statement is false since U1 ⊂ W (show).
(2.4) This statement is true since B in (*) is a basis for V.
4
