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Title: TMC1874 Mathematics For Computing tutorial 6 solution
Description: Tutorial Solution Chapter 6: Linear Transformations
Description: Tutorial Solution Chapter 6: Linear Transformations
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TMC1874 Mathematics For Computing
Tutorial Solution
Chapter 6: Linear Transformations
1
...
Then
L([u 1
u 2 ] + [v1
L([u 1 + v1
=
u1 + u2]
v2 ]),
u 2 + v2 ]),
= [u 1 + v1
u 2 + v2
= [u 1
u 1 + u 2 ] + [v1
u2
u 1 + u 2 + v1 + v2 ],
v2
v1 + v2 ],
L(u) + L(v)
=
and let c be any scalar
...
=
cL(u)
...
(b) L : R2 → R3 defined by L([u 1
Solution
Let u = [u 1
u 2 ]) = [u 1 + u 2
u 2 ] and v = [v1
L(u + v) =
v2 ]
...
and
L(u) + L(v) = [u 1 + u 2
u2
u 2 − 1] + [v1 + v2
= [u 1 + u 2 + v1 + v2
=
u 2 + v2
v2
v2 − 1],
u 2 + v2 − 2],
L(u + v)
...
2
...
]
W
...
Tiong
Page 1 of 7
TMC1874 Mathematics For Computing
(a) L : P1 → P2 defined by L(p(t)) = tp(t) + p(0)
Solution
Let p(t), q(t) be polynomials in P1 and let c be any scalar
...
and
L(c p(t)) = t[c p(t)] + c p(0) = c[tp(t) + p(0)] = cL(p(t))
...
(b) L : P1 → P2 defined by L(p(t)) = tp(t) + 1
Solution
Let p(t), q(t) be polynomials in P1 and let c be any scalar
...
Therefore, L is not a linear transformation
...
Find the standard matrix representing each given linear transformation
...
Compute L(e j ) for j = 1, 2 as follows:
L(e1 ) = L
1
0
L(e2 ) = L
Hence the matrix A = [L(e1 )
W
...
Tiong
L(e2 )] =
−1
0
=
0
1
,
0
1
...
Page 2 of 7
TMC1874 Mathematics For Computing
u1
u2
(b) L : R2 → R2 defined by L
=
u1 − u2
u1
Solution
Let {e1 , e2 } be the natural basis for R2
...
=
=
1
1
L(e2 )] =
−1
0
...
Compute L(e j ) for j = 1, 2, 3 as follows:
1
L(e1 ) = L 0 =
0
0
L(e2 ) = L 1 =
0
0
L(e3 ) = L 0
1
Hence the matrix A = [L(e1 )
L(e2 )
1
0 ,
0
0
0 ,
0
0
= 0
...
Consider the function L : M34 → M24 defined by L(A) =
1
(a) Find L 3
4
W
...
Tiong
2
0
1
0
2
−2
0
0
0
2
0
0
2
0
0
...
−1
3
1
Page 3 of 7
TMC1874 Mathematics For Computing
Solution
1
3
L
4
2
0
1
0
2
−2
−1
3 =
1
2
0
0
2
6
−6
=
1
3
4
1
−3
5
−3
−2
10
−1
3
2
0
1
0
2
−2
−1
3
1
...
Solution
Let M =
2
0
0
2
1
−3
...
Therefore
L(A + B) =
L(c A) =
M(A + B) = M A + MB = L(A) + L(B),
M(c A) = c(M A) = cL(A)
...
5
...
1
(a) What is L −2 ?
3
Solution
1
L −2 =
3
1
0
0
L 0 − 2L 1 + 3L 0
0
0
1
=
1
2
=
2
−3
5
1
12
11
−2
+3
u1
(b) What is L u 2 ?
u3
W
...
Tiong
Page 4 of 7
TMC1874 Mathematics For Computing
Solution
u1
L u 2 =
u3
=
1
0
0
u 1 L 0 + u 2 L 1 + u 3 L 0
0
0
1
u1
1
2
2
−3
+ u2
u 1 + 2u 2 + 5u 3
2u 1 − 3u 2 + u 3
=
u1
u2
6
...
in range L?
(e) Find ker L
...
Solution
(a) L
0
2
=
0
2
...
(b) L
2
0
=
0
0
...
(c) No because
3
0
does not have the form
0
u2
...
(e) The kernel of L consists of all vectors
0
0
u1
u2
u1
u2
such that L
=
0
u2
=
...
Therefore ker L
consists of all vectors
a
0
, where a is any real number
...
W
...
Tiong
Page 5 of 7
TMC1874 Mathematics For Computing
7
...
2] in range L?
(d) Is [0
u2
0] in range L?
−2
− 1] is in ker L?
1] is in ker L?
(e) Find ker L
...
Solution
(a) Yes since L([1
−1
(b) No since L([0 2
1
− 1]) = [0
0]
...
(c) Yes since [1
2] = L([1
2 0
0])
...
This leads to the system of equations u 1 + u 4 = u 2 + u 3 = 0
...
Therefore the kernel
consists of all vectors of the form [r s − s − r], where r and s are real numbers
...
Since u 1 + u 4
and u 2 + u 3 can assume any real value, the range is all vectors [a b] = a[1 0] +
b[0 1]
...
8
...
(a) Find a basis for ker L
...
(d) What is dim range L?
Solution
(a) [u 1 u 2 u 3 u 4] is in the kernel if and only if u 1 + u 3 = u 2 + u 4 = 0
...
Therefore the kernel consists of all vectors of the form [−a −
W
...
Tiong
Page 6 of 7
TMC1874 Mathematics For Computing
b
a
b], where a and b are any real numbers
...
1]}
...
(c) The range consists of all vectors of the form
[u 1 + u 3
u2 + u4
u1 + u3] =
u 1 [1 0
+ u 4 [0
1] + u 2 [0
1
= (u 1 + u 3 )[1
A basis is {[1
0 1], [0
1
0] + u 3 [1
0 1]
0],
0 1] + (u 2 + u 4 )[0
1
0]
1 0]}
(d) dim range L = 2
9
...
(a) Find a basis for ker L
...
Solution
(a) The kernel consists of all polynomials at2 + bt + c such that L(at2 + bt + c) = [a b] =
[0 0]
...
Hence ker L consists of the polynomials c, c any
number
...
(b) The range consists of matrices [a
W
...
Tiong
b] = a[1 0] + b[0 1]
Title: TMC1874 Mathematics For Computing tutorial 6 solution
Description: Tutorial Solution Chapter 6: Linear Transformations
Description: Tutorial Solution Chapter 6: Linear Transformations