MAT2611 LINEAR ALGEBRA REVISION
ASSIGNMENTS AND EXAMS
Question 1: 10 Marks
Show that the set X with the given operations fails to be a vector space
... [Show More] by identifying all axioms that
hold and fail to hold:
The set X = R 3 with vector addition ⊕ defined by
(a, b, c) ⊕ (x, y, z) = (1, y, c + z)
and scalar multiplication defined by k (a, b, c) = (ka, kb, kc).
In the following let u = (u 1
, u2
, u3) ∈ R3, v = (v 1
, v2
, v3) ∈ R 3, w = (w 1
, w2
, w3) ∈ R 3 be arbitrary
elements of R3 and k, m ∈ R . We have
1. u ⊕ v = (1, v2
, u3 + v3) ∈ R3 holds. X
2. u ⊕ v = (1, v2
, u3 + v3).
v ⊕ u = (1, u2
, v3 + u3).
Choosing u = (1, 0, 0) and v = (0, 1, 0) we see that u ⊕ v = v ⊕ u does not hold in general. X
3. u ⊕ (v ⊕ w) = u ⊕ (1, w2
, v3 + w3) = (1, w2
, u3 + v 3 + w3).
(u ⊕ v) ⊕ w = (1, v2
, u3 + v3) ⊕ w = (1, w2
, u3 + v3 + w3).
Thus u ⊕ (v ⊕ w) = (u ⊕ v) ⊕ w holds.X
4. Suppose the zero vector 0 = (a, b, c) ∈ R3 exists. Then
u ⊕ 0 = (u1
, u2
, u3) ⊕ (a, b, c) = (1, b, u3 + c) = (u 1
, u2
, u3) ⇔ u 1 = 1, u 2 = b, c = 0.
The zero vector property obviously does not hold, in particular the equation cannot be satisfied
for u = (0, 0, 0).X
5. Since the zero vector does not exist, the negative is undefined.Thus the existence of negatives
does not hold. X
6. k u = (ku 1
, ku2
, ku3) ∈ R3 holds. X
7. k (u + v) = k (1, v 2
, u3 + v3) = (k, kv 2
, ku3 + kv 3).
k u k ⊕ v = (1, kv 2
, ku3 + kv 3).
Choosing k = 2, for example, we find k (u + v) = k u + k v does not hold in general. X
8. (k + m) u = ((k + m)u 1, (k + m)u2, (k + m)u3).
k u m ⊕ u = (1, mu 2
, ku1 + mu2).
Choosing k = m = 0,
for example, we find (k + m) u = k u m ⊕ u does not hold in
general.X
4
MAT2611/106
9. k(m u) = k (mu 1
, mu2
, mu3) = (kmu 1
, kmu2
, kmu3).
(km) u = ((km)u 1, (km)u2, (km)u3).
Thus k(m u) = (km) u holds. X
10. 1 u = (1 · u 1, 1 · u2, 1 · u3) = u holds. X
Question 2: 16 Marks
Consider the vector space M22 of all 2 × 2 matrices.
(2.1)Show that B = {A 1
, A2
, A3
, A4} is a basis for M22 where (6)
A1 =
3 6
3 −6
; A2 =
0 −1
−1 0
; A3 =
0 −8
−12 −4
; and A4 =
1 0
−1 2
.
First we show linear independence of the vectors.Let c1
, c2
, c3
, c4 ∈ R determined by
c1A1 + c2A2 + c3A3 + c4A4 = 0 X
i.e.
3c1 6c1
3c1 −6c1
+
0 −c 2
−c 2 0
+
0 −8c3
−12c3 −4c3
+
c4 0
−c 4 2c4
=
3c1 + c4 6c1
− c2 − 8c3
3c1
− c2 − 12c3
− c4 −6c1 − 4c3 + 2c4
=
0 0
0 0
Thus we obtain the four equations
3c1 + c4 = 0
6c1
− c2 − 8c3 = 0
3c1
− c2 − 12c3
− c4 = 0
−6c1 − 4c3 + 2c4 = 0
X
or in matrix form
3 0 0 1
6 −1 −8 0
3 −1 −12 −1
−6 0 −4 2
c1
c2
c3
c4
=
0
0
0
0
.
5
Row reduction of the augmented matrix yields
3 0 0 1 0
−2R1 6 −1 −8 0 0
−R1 3 −1 −12 −1 0
+2R1 −6 0 −4 2 0
→
3 0 0 1 0
0 −1 −8 −2 0
−R2 0 −1 −12 −2 0
0 0 −4 4 0
→
3 0 0 1 0
0 −1 −8 −2 0
0 0 −4 0 0
−R3 0 0 −4 4 0
→
3 0 0 1 0
0 −1 −8 −2 0
0 0 −4 0 0
0 0 0 4 0
Here −2R1 means subtract twice the first row from the second row (appearing right
of −2R1). Thus c4 = 0, c3 = 0, c2 = −8c 3 − 2c4 = 0 and c 1 = −c 4/3 = 0. Since
c1 = c 2 = c 3 = c4 = 0 X is the only solution, the matrices A1
, A2
, A3
and A4
are linearly
independent.
Next we show that any element of M22 can be expressed as a linear combination of A1
,
A2
, A3
and A4
. Let a, b, c, d ∈ R and
a b
c d
= a1A1 + a2A2 + a3A3 + a4A4 =
3a1 + a4 6a1
− a2 − 8a3
3a1
− a2 − 12a3
− a4 −6a1 − 4a3 + 2a4
Thus we obtain the four equations
3a1 + a4 = a
6a1
− a2 − 8a3 = b
3a1
− a2 − 12a3
− c4 = c
−6a1 − 4a3 + 2a4 = d
X
or in matrix form
3 0 0 1
6 −1 −8 0
3 −1 −12 −1
−6 0 −4 2
a1
a2
a3
a4
=
a
b
c
d
.
Row reduction of the augmented matrix yields
3 0 0 1 a
−2R1 6 −1 −8 0 b
−R1 3 −1 −12 −1 c
+2R1 −6 0 −4 2 d
→
3 0 0 1 a
0 −1 −8 −2 b − 2a
−R2 0 −1 −12 −2 c − a
0 0 −4 4 d + 2a
→
3 0 0 1 a
0 −1 −8 −2 b − 2a
0 0 −4 0 a − b + c
−R3 0 0 −4 4 d + 2a
6
MAT2611/106
→
3 0 0 1 a
0 −1 −8 −2 b − 2a
0 0 −4 0 a − b + c
0 0 0 4 a + b + d − c
X
2
Thus we find
a4 =
a + b + d − c
4
a3 =
b − a − c
4
a2 = −8a 3 − 2a4 − b + 2a = −2b + 2a + 2c −
a + b + d − c
2
− b + 2a =
7a − 7b + 5c − d
2
a1 =
a − a4
3
=
3a − b + c − d
12
Since a solution exists, we have span(B) = M22. Thus B is a basis for M22.
(2.2)Write A = (4) 6 2
5 3
as a linear combination of vectors from B.
From 2.1 we have (for a = 6, b = 2, c = 5 and d = 3)
a4 =
3
2
, X a3 = −
9
4
, X a2 = 25, X a1 =
3
2
X
so that
6 2
5 3
=
3
2
3 6
3 −6
+ 25 0 −1
−1 0
−
9
4
0 −8
−12 −4
+
3
2
1 0
−1 2
.
(2.3)Prove that the subset (6)
D = {A M ∈ 22 : A
T
+ A = 0}
forms a subspace of M22.
First we must show that D is not empty. Note that if D is a subspace of M22 then it is a
vector space with the same zero vector as for M22 (i.e. the 2 × 2 zero matrix). It suffices
to check whether the zero vector is in D:
0 0
0 0
T
+
0 0
0 0
=
0 0
0 0
+
0 0
0 0
=
0 0
0 0
X
2
so that 0 0
0 0
∈ D.
Next we prove the closure under vector addition and under scalar multiplication.
Let A D and k ∈ ∈ R. Then, using the properties of the transpose and vector space, we
have
(kA)
T
+ (kA) = kA T
+ kA = k(A
T
+ A) = k0 = 0 X
2
7
where 0 on the right hand side denotes the 2 × 2 zero matrix. Thus D is closed under
scalar multiplication.
Let A, B D. ∈ Then, using the properties of the transpose and vector space, we find
(A + B)
T
+ (A + B) = (A
T
+ B T
) + (A + B) = (A
T
+ A) + (B
T
+ B) = 0 + 0 = 0 X
2
where 0 on the right hand side denotes the 2 × 2 zero matrix. Thus D is closed under
vector addition. [Show Less]