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CH-6 Linear Transformation

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Lecture Notes: Compiled by Maqsood Ahmad (A.P. Maths.) for students
of CUI, Lahore. (SP21-BSE-A, B & C).
Chapter 6: Linear Transformations (function) and Matrices
Exercise 6.1 : 1-16, 20-23; Exercise 6.2 : 1-11,16,17; Exercise 6.3 : 1-8,13,14,22.
Important Note:(1) some books combine these two conditions into one.
𝐿(π‘Žπ’– + 𝑏𝒗) = π‘ŽπΏ(𝒖) + 𝑏𝐿(𝒗) ∢ π‘Ž, 𝑏 π‘Žπ‘Ÿπ‘’ π‘Ÿπ‘’π‘Žπ‘™π‘ 
(2)𝐿(0𝑉 ) = 0π‘Š i.e., Zero vector of vector space 𝑉 always transform on zero of vector
space π‘Š.
(3) Every Linear Transformation have the form 𝐿(𝒖) = 𝑨𝒖
(4) (a) No product of components (e.g. 𝑒1 𝑒2 ) appear in LT.
(b) No power of component is allowed in LT. (e.g.𝑒1 2 )
(c) No constant will be added into any component in LT (e.g. 𝑒1 + 2)
Examples of LT:(1) π‘…π‘’π‘“π‘™π‘’π‘π‘‘π‘–π‘œπ‘› π‘Žπ‘π‘‘ 𝒀 π‘Žπ‘₯𝑖𝑠: 𝐿: 𝑅 2 → 𝑅 2 defined by
π‘₯
−π‘₯
𝑒1
−𝑒1
𝐿(π‘₯, 𝑦) = (−π‘₯, 𝑦) 𝑂𝑅 𝐿 [𝑦] = [ 𝑦 ] 𝑂𝑅 𝐿(𝒖) = 𝐿 [𝑒 ] = [ 𝑒 ]
2
2
(2) π‘…π‘’π‘“π‘™π‘’π‘π‘‘π‘–π‘œπ‘› π‘Žπ‘π‘‘ π’š = 𝒙 𝑙𝑖𝑛𝑒: 𝐿: 𝑅 2 → 𝑅 2 defined by
π‘₯
𝑒1
𝑒2
𝑦
𝐿(π‘₯, 𝑦) = (𝑦, π‘₯) 𝑂𝑅 𝐿 [𝑦] = [ ] 𝑂𝑅 𝐿(𝒖) = 𝐿 [𝑒 ] = [𝑒 ]
π‘₯
2
1
(3) π‘…π‘’π‘“π‘™π‘’π‘π‘‘π‘–π‘œπ‘› π‘Žπ‘π‘‘ π‘‹π‘Œ π‘π‘™π‘Žπ‘›π‘’; 𝐿: 𝑅 3 → 𝑅 3 ;
𝑒1
𝑒1
π‘₯
π‘₯
𝐿(π‘₯, 𝑦, 𝑧) = (π‘₯, 𝑦, −𝑧),
𝑂𝑅 𝐿 [𝑦] = [ 𝑦 ] ; 𝐿(𝒖) = 𝐿 [𝑒2 ] = [ 𝑒2 ]
𝑒3
−𝑒3
𝑧
−𝑧
𝑒1
𝑒1
𝑒
𝑒
(4)π·π‘–π‘™π‘Žπ‘‘π‘–π‘œπ‘› 𝐿(𝑒) = π‘Ÿπ‘’ , π‘Ÿ > 1 𝑂𝑅 𝐿(𝒖) = 𝐿 [ 2 ] = π‘Ÿ [ 2 ]
𝑒3
𝑒3
𝑒1
𝑒1
(5)πΆπ‘œπ‘›π‘‘π‘Ÿπ‘Žπ‘π‘‘π‘–π‘œπ‘› 𝐿(𝑒) = π‘Ÿπ‘’ , 0 < π‘Ÿ < 1 𝑂𝑅 𝐿(𝒖) = 𝐿 [𝑒2 ] = π‘Ÿ [𝑒2 ]
𝑒3
𝑒3
(1) π‘…π‘’π‘“π‘™π‘’π‘π‘‘π‘–π‘œπ‘› π‘Žπ‘π‘‘ 𝑋 π‘Žπ‘₯𝑖𝑠: 𝐿: 𝑅 2 → 𝑅 2
𝑒1
𝑒1
𝐿(𝒖) = 𝐿 [𝑒 ] = [−𝑒 ]
2
2
1
1
𝐿(𝑒1 ) = 𝐿 [ ] = [ ]
0
−0
0
0
𝐿(𝑒2 ) = 𝐿 [ ] = [ ]
1
−1
𝑨 = [𝐿(𝑒1 ) 𝐿(𝑒2 )]
𝐿(𝒖) = [
𝑒1
1 0 𝑒1
] [𝑒 ] = 𝑨𝒖 = [−𝑒 ]
0−1 2
2
(2) π‘ƒπ‘Ÿπ‘œπ‘—π‘’π‘π‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘£π‘’π‘π‘‘π‘œπ‘Ÿ 𝑖𝑛 π‘‹π‘Œ − π‘π‘™π‘Žπ‘›π‘’; 𝐿: π‘ΉπŸ‘ → 𝑅 2
𝑒1
𝑒1
𝐿(𝒖) = 𝐿 [𝑒2 ] = [𝑒2 ]
𝑒3
0
1
1
0
0
0
0
𝐿(𝑒1 ) = 𝐿 [0] = [0]; 𝐿(𝑒2 ) = 𝐿 [1] = [1]; 𝐿(𝑒3 ) = 𝐿 [0] = [0]
0
0
0
0
1
0
𝑨 = [𝑳(π’†πŸ ) 𝑳(π’†πŸ )𝑳(π’†πŸ‘ )]
100 𝑒1
𝐿(𝒖) = [010] [𝑒2 ] = 𝑨𝒖
000 𝑒3
𝑒1
𝑒1
(3) π‘…π‘’π‘“π‘™π‘’π‘π‘‘π‘–π‘œπ‘› π‘Žπ‘π‘‘ π‘‹π‘Œ π‘π‘™π‘Žπ‘›π‘’; 𝐿: 𝑅 → 𝑅 ; 𝐿(𝒖) = 𝐿 [𝑒2 ] = [ 𝑒2 ]
𝑒3
−𝑒3
3
3
1
1
0
0
0
0
𝐿(𝑒1 ) = 𝐿 ([0]) = [ 0 ]; 𝐿(𝑒2 ) = 𝐿 ([1]) = [ 1 ] ; 𝐿(𝑒3 ) = 𝐿 ([0]) = [ 0 ]
0
−0
0
−0
1
−1
𝐴 = [𝐿(𝑒1 ) 𝐿(𝑒2 ) 𝐿(𝑒3 )]
𝑒1
𝑒1
10 0 𝑒1
𝑒
𝑒
𝑒
𝐿(𝒖) = 𝐿 [ 2 ] = [01 0 ] [ 2 ] = 𝑨𝒖 = [ 2 ]
𝑒3
−𝑒3
00−1 𝑒3
𝒗 = π’„πŸ π’—πŸ + π’„πŸ π’—πŸ + π’„πŸ‘ π’—πŸ‘ + π’„πŸ’ π’—πŸ’ − − − (𝟏) 𝑡𝑯𝑺
1 0 01| 3
1000| 2
|
0 1 20 −5
0100| 1
[
] 𝑅𝑅𝐸𝐹 [
]
1−120|−5
0010|−3
0 2 11| 0
0001| 1
𝒗 = πŸπ’—πŸ + π’—πŸ − πŸ‘π’—πŸ‘ + π’—πŸ’
Now apply Linear transformation on Both Sides
𝑳(𝒗) = 𝑳(πŸπ’—πŸ + π’—πŸ − πŸ‘π’—πŸ‘ + π’—πŸ’ )
π‘π‘Žπ‘Ÿπ‘‘ (π‘Ž)π‘œπ‘“ 𝐿𝑇: 𝐿(𝑒 + 𝑣) = 𝐿(𝑒) + 𝐿(𝑣)
𝑳(𝒗) = 𝑳(πŸπ’—πŸ ) + 𝑳(π’—πŸ ) + 𝑳(−πŸ‘π’—πŸ‘ ) + 𝑳(π’—πŸ’ )
π‘π‘Žπ‘Ÿπ‘‘ (𝑏)π‘œπ‘“ 𝐿𝑇: 𝐿(𝒄𝑒) = 𝒄𝐿(𝑒)
𝑳(𝒗) = πŸπ‘³(π’—πŸ ) + 𝑳(π’—πŸ ) − πŸ‘π‘³(π’—πŸ‘ ) + 𝑳(π’—πŸ’ )
𝑳(𝒗) = 𝟐[𝟏
𝟐] + [𝟎
πŸ‘] − πŸ‘[𝟎
Exercise 6.1
𝟎] + [𝟐
𝟎] = [πŸ’
πŸ•]
Solution: Method 1: (a) Let 𝑣1 = [1
1]and 𝑣2 = [−1
1], 𝑣 = [−1
5], 𝑀 = [𝑒1 𝑒2 ]
𝑣 = π’„πŸ 𝑣1 + π’„πŸ 𝑣2
[−1
5] = π’„πŸ [1
[−1
1] + π’„πŸ [−1
1]
5] = [π’„πŸ π’„πŸ ] + [−π’„πŸ π’„πŸ ]
[−1
5] = [π’„πŸ −π’„πŸ π’„πŸ +π’„πŸ ]
Equating components of equal vectors
𝑐1 −𝑐2 = −1
𝑐1 +𝑐2 = 5
2𝑐1 = 4 𝑔𝑖𝑣𝑒𝑠 π’„πŸ = 𝟐 𝒕𝒉𝒆𝒏 π’„πŸ = πŸ‘
[−1
1] + 3 [−1
5] = 2[1
1]
Now apply transformation on Both Sides
𝐿[−1
1] + 3 [−1
5] = 𝐿(2[1
1])
𝐿[−1
5] = 𝐿(2[1
1]) + 𝐿(3 [−1
1]); π‘π‘Žπ‘Ÿπ‘‘ (π‘Ž)π‘œπ‘“ 𝐿𝑇: 𝐿(𝑒 + 𝑣) = 𝐿(𝑒) + 𝐿(𝑣)
𝐿[−1
5] = 2𝐿([1
1]) + 3𝐿( [−1
1]); π‘π‘Žπ‘Ÿπ‘‘ (𝑏)π‘œπ‘“ 𝐿𝑇: 𝐿(𝒄𝑒) = 𝒄𝐿(𝑒)
𝐿[−1
− 2] + 3[2
3] = [2 + 6
− 4 + 9] = [8
[𝑒1 𝑒2 ] = 𝑐1 [1
1] + 𝑐2 [−1
1]
5] = 2[1
(b)𝑀 = 𝑐1 𝑣1 + 𝑐2 𝑣2
[𝑒1 𝑒2 ] = [𝑐1 𝑐1 ] + [−𝑐2 𝑐2 ]
[𝑒1 𝑒2 ] = [𝑐1 −𝑐2 𝑐1 +𝑐2 ]
𝑐1 −𝑐2 = 𝑒1
𝑐1 +𝑐2 = 𝑒2
Adding both equations 2𝑐1 = 𝑒1 + 𝑒2
𝑒1 + 𝑒2
[𝑒1 𝑒2 ] = (
) [1
2
𝑒2 − 𝑒1
1] + (
2
) [−1
1]
Now apply transformation on BS;
𝐿[𝑒1 𝑒2 ] = 𝐿 ((
𝑒1 +𝑒2
𝐿[𝑒1 𝑒2 ] = 𝐿 ((
2
) [1
𝑒1 + 𝑒2
2
𝑒2 −𝑒1
1]) + 𝐿 ((
2
) [1
) [−1
𝑒2 − 𝑒1
1] + (
1]);
2
) [−1
1])
5]
π‘π‘Žπ‘Ÿπ‘‘ (π‘Ž)π‘œπ‘“ 𝐿𝑇: 𝐿(𝑒 + 𝑣) = 𝐿(𝑒) + 𝐿(𝑣)
𝑒1 +𝑒2
𝐿[𝑒1 𝑒2 ] = (
𝐿[𝑒1 𝑒2 ] = (
2
𝑒1 + 𝑒2
) [1
2
𝐿[𝑒1 𝑒2 ] = [(
𝐿[𝑒1 𝑒2 ] = [
) 𝐿[1
𝑒1 + 𝑒2
)
2
𝑒2 −𝑒1
1] + (
2
) 𝐿 [−1
𝑒2 − 𝑒1
) [2
2
− 2] + (
− 2(
1]; π‘π‘Žπ‘Ÿπ‘‘ (𝑏)π‘œπ‘“ 𝐿𝑇: 𝐿(π‘˜π‘’) = π‘˜πΏ(𝑒)
3]
𝑒1 + 𝑒2
𝑒2 − 𝑒1
)] + [2 (
)
2
2
𝑒1 + 𝑒2
+ 𝑒2 − 𝑒1
2
− (𝑒1 + 𝑒2 ) + 3 (
1
3
𝐿[𝑒1 𝑒2 ] = [− 𝑒1 + 𝑒2
2
2
𝑒2 − 𝑒1
)]
2
3(
𝑒2 − 𝑒1
)]
2
5
1
− 𝑒1 + 𝑒2 ]
2
2
𝑴𝒆𝒕𝒉𝒐𝒅 𝟐: See Slader.com
Definition1:
Note: Problem of finding 𝑲𝒆𝒓 𝑳 is same as solving homogeneous system (Null or solution
space).
Definition2:
Theorem1 :
Definition3:
Note: Problem of finding π‘…π‘Žπ‘›π‘”π‘’ 𝐿(onto) is same as of finding spanning set of vector space.
Theorem 6.6: π‘‘π‘–π‘š(πΎπ‘’π‘Ÿ(𝐿)) + π‘‘π‘–π‘š(π‘…π‘Žπ‘›π‘”π‘’(𝐿)) = π‘‘π‘–π‘š(𝑉)
Example:
𝑒1
1 0
𝑒
Solution (a):- 𝐿 [ 2 ] = [1 1
𝑒3
2 1
𝑒1 + 0𝑒2 + 𝑒3
1 𝑒1
𝑒
2] [ 2 ] = [ 𝑒1 + 𝑒2 + 2𝑒3 ]
2𝑒1 + 𝑒2 + 3𝑒3
3 𝑒3
1
1
0
0
0
1
Extra 𝐿(𝑒1 ) = 𝐿 [0] = [1] ; 𝐿(𝑒2 ) = 𝐿 [1] = [1] ; 𝐿(𝑒3 ) = 𝐿 [0] = [2]
0
2
0
1
1
3
(a) Is L is onto → To find Range(L)= 𝑅 3
𝑒1
π‘Ž
π‘Ž
101 𝑒1
𝑒
𝑒
𝐿 [ 2 ] = [𝑏 ] → [112] [ 2 ] = [𝑏 ] − −(1)
𝑒3
𝑐
𝑐
213 𝑒3
π‘Ž
101|π‘Ž 𝑅 − 𝑅 101| π‘Ž
101|
2
1
| 𝑏 − π‘Ž ] 𝑅3 − 𝑅2 [011| 𝑏 − π‘Ž ]
[112|𝑏 ]
[
𝑅3 − 2𝑅1 011
𝑐
213|
011|𝑐 − 2π‘Ž
000|𝑐 − π‘Ž − 𝑏
No Solution to NHS (1) therefore L is not onto( Range(L)≠ 𝑅 3 ).
(b) Find a basis for Range of L
𝑒1
𝑒1 + 0𝑒2 + 𝑒3
101 𝑒1
1
0
1
𝑒
𝑒
Consider 𝐿(𝒖) = 𝐿 [ 2 ] = [112] [ 2 ] = [ 𝑒1 + 𝑒2 + 2𝑒3 ] = 𝑒1 [1] + 𝑒2 [1] + 𝑒3 [2]
𝑒3
2𝑒1 + 𝑒2 + 3𝑒3
213 𝑒3
2
1
3
1
0
1
𝑆 = {𝑣1 = [1] , 𝑣2 = [1] , 𝑣3 = [2]}is spanning set for Range of L.
2
1
3
Take LI vectors from S.
101|0
101|0
[112|0] 𝑅𝐸𝐹 [011|0]
213|0
000|0
𝟏
𝟎
Basis for Range of L =𝑻 = {π’—πŸ = [𝟏] , π’—πŸ = [𝟏]}
𝟐
𝟏
𝑒1
0
101 𝑒1
0
(c) To find Ker(L); 𝑳(𝒖) = 𝟎 → 𝐿 [𝑒2 ] = [0] → [112] [𝑒2 ] = [0]
𝑒3
0
213 𝑒3
0
101|0
101|0
[112|0] 𝑅𝑅𝐸𝐹 [011|0]
213|0
000|0
π’–πŸ + π’–πŸ‘ = 0, π’–πŸ +π’–πŸ‘ = 0 ; π’–πŸ‘ = π‘Ž 𝒕𝒉𝒆𝒏 π’–πŸ = −π’–πŸ‘ = −π‘Ž; π’–πŸ = −π’–πŸ‘ = −π‘Ž
−π‘Ž
−1
−π‘Ž
(𝑳)
𝑲𝒆𝒓
= {[ ] = π‘Ž [−1] ∢ π‘Ž ∈ 𝑹}
π‘Ž
1
−π‘Ž + 0 + π‘Ž
101 −π‘Ž
0
Verify [112] [−π‘Ž] = [ −π‘Ž − π‘Ž + 2π‘Ž ] = [0]
−2π‘Ž − π‘Ž + 3π‘Ž
213 π‘Ž
0
(d) Since 𝑲𝒆𝒓 (𝑳) ≠ {𝟎} hence 𝑳 is not one to one.
----------------------------------------------------------------------------------------------------------Note: Additional question: find basis for πΎπ‘’π‘Ÿ 𝐿and π‘‘π‘–π‘š(πΎπ‘’π‘Ÿ 𝐿);π‘‘π‘–π‘š(π‘…π‘Žπ‘›π‘”π‘’ 𝐿) and
proveπ‘‘π‘–π‘š(πΎπ‘’π‘Ÿ(𝐿)) + π‘‘π‘–π‘š(π‘…π‘Žπ‘›π‘”π‘’(𝐿)) = π‘‘π‘–π‘š(𝑅 3 )
−1
Basis for πΎπ‘’π‘Ÿ 𝐿 = {[−1]} and π‘‘π‘–π‘š(πΎπ‘’π‘Ÿ 𝐿) = π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘£π‘’π‘π‘‘π‘œπ‘Ÿπ‘  𝑖𝑛 π‘π‘Žπ‘ π‘–π‘  π‘œπ‘“ πΎπ‘’π‘Ÿ 𝐿 = 1
1
π‘‘π‘–π‘š(π‘…π‘Žπ‘›π‘”π‘’ 𝐿) = π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘£π‘’π‘π‘‘π‘œπ‘Ÿπ‘  𝑖𝑛 π‘π‘Žπ‘ π‘–π‘  π‘œπ‘“ π‘…π‘Žπ‘›π‘”π‘’ 𝐿 = 2
1+2=3
----------------------------------------------------------------------------------------------------------
Exercise 6.2
Solution:(a)𝐿([2 3−2 3]) = [0 6] ⟹ [2 3−2 3] πΎπ‘’π‘Ÿ 𝐿
(b) 𝐿([4 −2−4 2]) = [0
(c) [π’–πŸ + π’–πŸ‘
0] ⟹ [4
π’–πŸ + π’–πŸ’ ] = [𝟏
−2−4 2] ∈ πΎπ‘’π‘Ÿ 𝐿
𝟐] ⟹ π’–πŸ + π’–πŸ‘ = 𝟏 &π’–πŸ + π’–πŸ’ = 𝟐.
We have 2 equations and 4 unknowns and have infinite many solution.
One solution may be, 𝐼𝑓 𝑒1 = 1 π‘‘β„Žπ‘’π‘› 𝑒3 = 0 & 𝑖𝑓 𝑒2 = 1 π‘‘β„Žπ‘’π‘› 𝑒4 = 1
𝐿([1 10 1]) = [1
(d) [𝑒1 + 𝑒3
𝑒2 + 𝑒4 ] = [0
2] ∈ π‘…π‘Žπ‘›π‘”π‘’ 𝐿
0] ⟹ 𝑒1 + 𝑒3 = 0 &𝑒2 + 𝑒4 = 0.
We have 2 equations and 4 unknowns and have infinite many solution.
One solution may be, 𝐼𝑓 𝑒1 = 1 π‘‘β„Žπ‘’π‘› 𝑒3 = −1 & 𝑖𝑓 𝑒2 = 1 π‘‘β„Žπ‘’π‘› 𝑒4 = −1
𝐿([1
1−1 −1]) = [0 0] ∈ π‘…π‘Žπ‘›π‘”π‘’ 𝐿
(e) To obtainπΎπ‘’π‘Ÿ 𝐿,
find[𝑒1
𝑒2 𝑒3
𝑒4 ] ∈ 𝑅4 such that𝐿(𝒖) = 𝟎 π‘–π‘šπ‘π‘™π‘–π‘’π‘  𝐿([𝑒1
⟹ [𝑒1 + 𝑒3
𝑒2 𝑒3
𝑒4 ]) = πŸŽπ‘ΉπŸ
𝑒2 + 𝑒4 ] = [0 0] ⟹ 𝑒1 + 𝑒3 = 0 &𝑒2 + 𝑒4 = 0 (𝐻𝑆)
⟹ 𝑒1 = −𝑒3 &𝑒2 = −𝑒4 ;
We have 2 equations and 4 unknowns and have infinite many solution. Suppose
𝑒3 = π‘Ÿ π‘‘β„Žπ‘’π‘› 𝑒1 = −π‘Ÿ&𝑒4 = 𝑠 π‘‘β„Žπ‘’π‘› 𝑒2 = −𝑠
πΎπ‘’π‘Ÿ 𝐿{[𝑒1
𝑒2 𝑒3
𝑒4 ] = [−π‘Ÿ
−π‘ π‘Ÿ
𝑠] ∢ π‘Ÿ, 𝑠 ∈ 𝑅}
(f) To find spanning set for π‘…π‘Žπ‘›π‘”π‘’ 𝐿, consider the element of range
𝐿(𝒖) = 𝐿([𝑒1
𝑒2 𝑒3 𝑒4 ]) = [𝑒1 + 𝑒3 𝑒2 + 𝑒4 ]
= 𝑒1 [1 0] + 𝑒2 [0 1] + 𝑒3 [1 0] + 𝑒4 [0 1]
Spanning set for π‘…π‘Žπ‘›π‘”π‘’ 𝐿 = {[1
0], [0
1]}
----------------------------------------------------------------------------------------------------------------Note: Additional question: find basis for πΎπ‘’π‘Ÿ 𝐿 and π‘‘π‘–π‘š(πΎπ‘’π‘Ÿ 𝐿).
Answer: consider[−π‘Ÿ
−π‘ π‘Ÿ
𝑠] = π‘Ÿ[−1 01 0] + 𝑠[0 −10
1]
Hence basis for πΎπ‘’π‘Ÿ 𝐿 = {[−1 01 0], [0 −10 1]}.
π‘‘π‘–π‘š(πΎπ‘’π‘Ÿ 𝐿) = 2
Note: Additional question: find basis for π‘…π‘Žπ‘›π‘”π‘’ 𝐿 and π‘‘π‘–π‘š(π‘…π‘Žπ‘›π‘”π‘’ 𝐿).
Basis for π‘…π‘Žπ‘›π‘”π‘’ 𝐿 = {[1 0], [0 1]}
π‘‘π‘–π‘š(π‘…π‘Žπ‘›π‘”π‘’ 𝐿) = 2
*** π‘‘π‘–π‘š(πΎπ‘’π‘Ÿ(𝐿)) + π‘‘π‘–π‘š(π‘…π‘Žπ‘›π‘”π‘’(𝐿)) = π‘‘π‘–π‘š(𝑅4 )
2+2=4
Solution:(a) To obtain𝑲𝒆𝒓 𝑳, find 𝒑(𝒕) = π’‚π’•πŸ + 𝒃𝒕 + 𝒄 ∈ π‘·πŸ such that
𝐿(𝑝(𝑑)) = 0𝑃3
⟹ π’•πŸ 𝒑 (𝒕) = 0𝑑3 + 0𝑑2 + 0𝑑 + 0
⟹ π’•πŸ (πŸπ’‚π’• + 𝒃) = 0𝑑 3 + 0𝑑 2 + 0𝑑 + 0 ⟹ πŸπ’‚π’•πŸ‘ + π’ƒπ’•πŸ = 0𝑑 3 + 0𝑑 2 + 0𝑑 + 0
Equating coefficients of like powers
2π‘Ž = 0, 𝑏 = 0, 𝒄 ∈ 𝑹 ⟹ 𝒑(𝒕) = 𝒄
𝑲𝒆𝒓 𝑳 = {π‘Žπ‘‘ 2 + 𝑏𝑑 + 𝑐 ∢ π‘Ž = 0, 𝑏 = 0} = {𝒄} = {𝒄(1)}
Basis of 𝑲𝒆𝒓𝑳 = {1} and π’…π’Šπ’Ž (𝑲𝒆𝒓 𝑳) = 1
(b) To obtainπ‘Ήπ’‚π’π’ˆπ’† 𝑳, Let 𝒑(𝒕) = π’‚π’•πŸ + 𝒃𝒕 + 𝒄 ∈ π‘·πŸ
Consider𝐿 (𝑝(𝑑))
= π’•πŸ 𝒑 (𝒕) = π’•πŸ (πŸπ’‚π’• + 𝒃) = πŸπ’‚(π’•πŸ‘ ) + 𝒃(π’•πŸ )
π‘Ίπ’‘π’‚π’π’π’Šπ’π’ˆ 𝒔𝒆𝒕 𝒇𝒐𝒓 π‘Ήπ’‚π’π’ˆπ’† 𝑳 = {𝑣1 = π’•πŸ‘ , 𝑣2 = π’•πŸ }
Basis of π‘Ήπ’‚π’π’ˆπ’†π‘³ = {π’•πŸ , π’•πŸ‘ } and π’…π’Šπ’Ž (π‘Ήπ’‚π’π’ˆπ’† 𝑳) = 𝟐
*** π‘‘π‘–π‘š(πΎπ‘’π‘Ÿ(𝐿)) + π‘‘π‘–π‘š(π‘…π‘Žπ‘›π‘”π‘’(𝐿)) = π‘‘π‘–π‘š(𝑃2 )
1+2=3
Solution: πΊπ‘’π‘›π‘’π‘Ÿπ‘Žπ‘™π‘™π‘¦ 𝐿: 𝑉 → π‘Š ; π»π‘’π‘Ÿπ‘’ 𝐿: 𝑀22 → 𝑀22
π‘Žπ‘
π‘Ž + 𝑏𝑏 + 𝑐
𝐿[ ] = [
]
𝑐𝑑
π‘Ž + 𝑑𝑏 + 𝑑
(a)
𝐿(𝑣) = 0π‘Š , 𝐿 [
π‘Žπ‘
] = 0𝑀22
𝑐𝑑
π‘Ž + 𝑏𝑏 + 𝑐
00
[
]=[ ]
π‘Ž + 𝑑𝑏 + 𝑑
00
π‘Ž+𝑏 =0;
𝑏+𝑐 =0 ; π‘Ž+𝑑 =0;
𝑏+𝑑 = 0
𝐴𝑋 = 0
1100
0110
𝐴=[
]
1001
0101
, |𝐴| ≠ 0; π‘’π‘›π‘–π‘žπ‘’π‘’ π‘ π‘œπ‘™π‘’π‘‘π‘–π‘œπ‘›, π‘‘π‘Ÿπ‘–π‘£π‘–π‘Žπ‘™ π‘ π‘œπ‘™π‘›
(Homogeneous System of 4 linear eqs in 4 unknowns)
π‘Ž=𝑏=𝑐=𝑑=0
πΎπ‘’π‘Ÿ 𝐿 = {[
00
]}(Transformation is one to one)
00
π΅π‘Žπ‘ π‘–π‘  π‘œπ‘“ πΎπ‘’π‘Ÿ 𝐿 = {[
00
]} ; dim πΎπ‘’π‘Ÿ (𝐿) = 0
00
(b) For Range 𝐿 Consider;
𝐿[
π‘Žπ‘
π‘Ž + 𝑏𝑏 + 𝑐
10
11
01
00
]=[
] = π‘Ž[ ]+𝑏[ ]+𝑐[ ]+𝑑[ ]
𝑐𝑑
π‘Ž + 𝑑𝑏 + 𝑑
10
01
00
11
10 11 01 00
π‘†π‘π‘Žπ‘›(π‘…π‘Žπ‘›π‘”π‘’ 𝐿) = {[ ] , [ ] , [ ] , [ ]}
10 01 00 11
***** π‘‘π‘–π‘š(πΎπ‘’π‘Ÿ(𝐿)) + dim(Range(𝐿)) = dim(𝑉)
π‘‘π‘–π‘š(πΎπ‘’π‘Ÿ(𝐿)) + dim(Range(𝐿)) = dim(𝑀22 )
0 + dim(Range(𝐿)) = 4
10 11 01 00
π΅π‘Žπ‘ π‘–π‘  π‘œπ‘“ π‘…π‘Žπ‘›π‘”π‘’ 𝐿 = {[ ] , [ ] , [ ] , [ ]}
10 01 00 11
Working in Exercise 6.1, 6.2:
Theorem: : 𝑉 → π‘Š; π‘†π‘‘π‘Žπ‘›π‘‘π‘Žπ‘Ÿπ‘‘ π΅π‘Žπ‘ π‘–π‘  π‘œπ‘“ 𝑉 = 𝑆 = {𝑒1 , 𝑒2 , 𝑒3 }
π‘†π‘‘π‘Žπ‘›π‘‘π‘Žπ‘Ÿπ‘‘ π΅π‘Žπ‘ π‘–π‘  π‘œπ‘“ π‘Š = 𝑇 = {𝐸1 , 𝐸2 , 𝐸3 }
Then 𝐿[𝑿] = 𝐴 𝑿 π‘“π‘œπ‘Ÿ π‘Žπ‘™π‘™ 𝑋 ∈ 𝑉
𝑨 = [𝑳(π’†πŸ ) 𝑳(π’†πŸ ) 𝑳(π’†πŸ‘ )]
Note: But Now to deal with “Ordered Basis”
Theorem (1):: 𝑉 → π‘Š; π‘Άπ’“π’…π’†π’“π’†π’…π΅π‘Žπ‘ π‘–π‘  π‘œπ‘“ 𝑉 = 𝑆 = {𝑒1 = 𝑣1 , 𝑒2 = 𝑣2 , 𝑒3 = 𝑣3 }
π‘Άπ’“π’…π’†π’“π’†π’…π΅π‘Žπ‘ π‘–π‘  π‘œπ‘“ π‘Š = 𝑇 = {𝑀1 , 𝑀2 , 𝑀3 }
Then [𝐿(𝑋)] 𝑇 = 𝐴[𝑋]𝑆 π‘“π‘œπ‘Ÿ π‘Žπ‘™π‘™ 𝑋 ∈ 𝑉
𝐴 = [[𝐿(𝑣1 )] 𝑇 [𝐿(𝑣2 )] 𝑇 [𝐿(𝑣3 )]𝑇 ]with respect to S and T.
[𝑀1 𝑀2 𝑀3 | 𝐿(𝑣1 )| 𝐿(𝑣2 ) | 𝐿(𝑣3 )]
[𝐼 | 𝐴]
Theorem (2): : π‘Š → 𝑉; π‘‚π‘Ÿπ‘‘π‘’π‘Ÿπ‘’π‘‘ π΅π‘Žπ‘ π‘–π‘  π‘œπ‘“ 𝑉 = 𝑆 = {𝑣1 , 𝑣2 , 𝑣3 }
π‘‚π‘Ÿπ‘‘π‘’π‘Ÿπ‘’π‘‘ π΅π‘Žπ‘ π‘–π‘  π‘œπ‘“ π‘Š = 𝑇 = {𝑀1 , 𝑀2 , 𝑀3 }
Then [𝐿(𝑋)]𝑆 = 𝐴 [𝑋] 𝑇 π‘“π‘œπ‘Ÿ π‘Žπ‘™π‘™ 𝑋 ∈ π‘Š
𝐴 = [[𝐿(𝑀1 )]𝑆 [𝐿(𝑀2 )]𝑆 [𝐿(𝑀3 )]𝑆 ]with respect to T and S.
[𝑣1 𝑣2 𝑣3 | 𝐿(𝑀1 )| 𝐿(𝑀2 ) | 𝐿(𝑀3 )]
[𝐼 | 𝐴]
Theorem (3):: 𝑉 → π‘Š; π‘‚π‘Ÿπ‘‘π‘’π‘Ÿπ‘’π‘‘ π΅π‘Žπ‘ π‘–π‘  π‘œπ‘“ 𝑉 = 𝑆 = {𝑣1 , 𝑣2 , 𝑣3 }
Then 𝐿[𝑋] = 𝐴 [𝑋]𝑆 π‘“π‘œπ‘Ÿ π‘Žπ‘™π‘™ 𝑋 ∈ 𝑉
𝐴 = [𝐿[𝑣1 ]
𝐿[𝑣2 ]
𝐿[𝑣3 ]]with respect to S.
[𝑣1 𝑣2 𝑣3 | 𝐿(𝑣1 )| 𝐿(𝑣2 ) | 𝐿(𝑣3 )]
[𝐼 | 𝐴]
Theorem (4): : 𝑉 → π‘Š; π‘‚π‘Ÿπ‘‘π‘’π‘Ÿπ‘’π‘‘ π΅π‘Žπ‘ π‘–π‘  π‘œπ‘“ 𝑉 = 𝑇 = {𝑀1 , 𝑀2 , 𝑀3 }
Then [𝐿(𝑋)] 𝑇 = 𝐴 𝑋 π‘“π‘œπ‘Ÿ π‘Žπ‘™π‘™ 𝑋 ∈ 𝑉
𝐴 = [𝐿[𝑀1 ]
𝐿[𝑀2 ]
𝐿[𝑀3 ]]with respect to T.
[𝑀1 𝑀2 𝑀3 | 𝐿(𝑀1 )| 𝐿(𝑀2 ) | 𝐿(𝑀3 )]
[𝐼 | 𝐴]
Solution to some important problem Exercise 6.3
Solution:Given
1
2
0
−1
𝑆 = {𝑣1 = [ ] , 𝑣2 = [ ]} π‘Žπ‘›π‘‘ 𝑇 = {𝑀1 = [ ] , 𝑀2 = [ ]}
0
0
1
2
(a)
Matrix of linear transformation with respect to 𝑆 =?
[𝑣1 𝑣2 | 𝐿(𝑣1 )| 𝐿(𝑣2 )] … … … … … … . . (1)
1
1
𝐿(𝑣1 ) = 𝐿 [ ] = [ ]
0
2
0
2
𝐿(𝑣2 ) = 𝐿 [ ] = [ ]
1
−1
Put these values in (1) we have
[
10|1| 2
]
01|2|−1
Matrix of linear transformation with respect to 𝑆 is
𝐴=[
12
]
2−1
(b) Matrix of linear transformation with respect to 𝑆 π‘Žπ‘›π‘‘ 𝑇=?
[𝑀1 𝑀2 | 𝐿(𝑣1 )| 𝐿(𝑣2 )] … … … … … … . . (2)
Put values of 𝐿(𝑣1 )and 𝐿(𝑣2 )from part (a) in (2) we have
[
−12|1| 2
]
2 0|2|−1
π·π‘œ π‘¦π‘œπ‘’π‘Ÿπ‘ π‘’π‘™π‘“
𝑅𝑅𝐸𝐹~ [
10|1|−1/2
]
01|1| 3/4
Matrix of linear transformation with respect to 𝑆 π‘Žπ‘›π‘‘ 𝑇 is
𝐴=[
1−1/2
]
1 3/4
(c) Matrix of linear transformation with respect to 𝑇 π‘Žπ‘›π‘‘ 𝑆=?
𝑒1
𝑒 + 2𝑒2
𝐿 [𝑒 ] = [ 1
]
2𝑒1 − 𝑒2
2
[𝑣1 𝑣2 | 𝐿(𝑀1 )| 𝐿(𝑀2 )] … … … … … … . . (3)
−1
3
𝐿(𝑀1 ) = 𝐿 [ ] = [ ]
2
−4
2
2
𝐿(𝑀2 ) = 𝐿 [ ] = [ ]
0
4
Put values of 𝐿(𝑀1 )and 𝐿(𝑀2 ) in (3) we have
[
10| 3 |2
]
01|−4|4
Matrix of linear transformation with respect to 𝑇 π‘Žπ‘›π‘‘ 𝑆 is
𝐴=[
32
]
−44
(d) Matrix of linear transformation with respect to 𝑇=?
𝑒1
𝑒 + 2𝑒2
𝐿 [𝑒 ] = [ 1
]
2𝑒1 − 𝑒2
2
[𝑀1 𝑀2 | 𝐿(𝑀1 )| 𝐿(𝑀2 )] … … … … … … . . (4)
Putvalues of 𝐿(𝑀1 )and 𝐿(𝑀2 )from part (c) in (4) we have
[
−12| 3 |2
]
2 0|−4|4
π·π‘œ π‘¦π‘œπ‘’π‘Ÿπ‘ π‘’π‘™π‘“ 𝑅𝑅𝐸𝐹~ [
10| −2 |2
]
01|1/2|2
Matrix of linear transformation with respect to 𝑇 is
𝐴=[
−2 2
]
1/22
Home Work: Question 13 is similar to Question 1
Solution:(a) Matrix of linear transformation with respect to S and T natural/standard basis is
same as given in the definition of linear transformation, i.e.,
1
0
0
0
1
0
0
0
1
0
0
𝑆 = {𝑣1 = [ ] , 𝑣2 = [ ] , 𝑣3 = [ ] , 𝑣4 = [ ]} π‘Žπ‘›π‘‘ 𝑇 = {𝑀1 = [0] , 𝑀2 = [1] , 𝑀3 = [0]}
0
0
1
0
0
0
1
0
0
0
1
(a)
Matrix of linear transformation with respect to 𝑆and 𝑇?
Recall Theorem1:- 𝐴 = [[𝐿(𝑣1 )] 𝑇 [𝐿(𝑣2 )] 𝑇 [𝐿(𝑣3 )] 𝑇 ] with respect to S and T.
[𝑀1 𝑀2
𝑀3 | 𝐿(𝑣1 )| 𝐿(𝑣2 ) | 𝐿(𝑣3 )| 𝐿(𝑣4 )] − − − −(1)
[𝐼 | 𝐴]
1
1 0 1
0
𝐿(𝑣1 ) = 𝐿 ([ ]) = [ 0 1 2
0
−1 −2 1
0
1
1
1
0
1] [ ] = [ 0 ]
0
0
−1
0
0
1 0 1
1
𝐿(𝑣2 ) = 𝐿 ([ ]) = [ 0 1 2
0
−1 −2 1
0
0
1
0
1
1] [ ] = [ 1 ]
0
0
−2
0
0
1 0 1
0
𝐿(𝑣3 ) = 𝐿 ([ ]) = [ 0 1 2
1
−1 −2 1
0
0
1
1
0
1] [ ] = [2]
1
0
1
0
0
1 0 1
0
)
([
𝐿(𝑣4 = 𝐿
]) = [ 0 1 2
0
−1 −2 1
1
0
1
1
0
1] [ ] = [1]
0
0
0
1
Put these values in (1) we have
1 0 0| 1 | 0 | 1 | 1
[0 1 0| 0 | 1 | 2 | 1]
0 0 1|−1 | −2 | 1 | 0
Matrix of linear transformation with respect to 𝑆 and 𝑇 is
1 0 1
𝐴=[ 0 1 2
−1 −2 1
1
1]
0
(b) Matrix of linear transformation with respect to 𝑆 ′ and 𝑇 ′ =?
1
1 0 1
1
𝐿(𝑣1 ) = 𝐿 [ ] = [ 0 1 2
0
−1 −2 1
0
1
1
1
1
]
[
]
=
[
1
1]
0
0
−3
0
0
1 0 1
1
𝐿(𝑣2 ) = 𝐿 [ ] = [ 0 1 2
0
−1 −2 1
0
0
1
0
1
1] [ ] = [ 1 ]
0
0
−2
0
0
1 0 1
0
𝐿(𝑣3 ) = 𝐿 [ ] = [ 0 1 2
1
−1 −2 1
1
0
1
2
0
]
[
]
=
[
1
3]
1
0
1
1
0
1 0 1
1
𝐿(𝑣4 ) = 𝐿 [ ] = [ 0 1 2
1
−1 −2 1
0
0
1
1
1
1] [ ] = [ 3 ]
1
0
−1
0
[𝑀1 𝑀2 𝑀3| 𝐿(𝑣1 )| 𝐿(𝑣2 ) | 𝐿(𝑣3 ) | 𝐿(𝑣4 )] − − − −(1)
1 0
[0 1
1 1
0| 1 | 0 |2| 1
0| 1 | 1 |3| 3 ]
1|−3|−2|1|−1
1 0
π·π‘œ π‘¦π‘œπ‘’π‘Ÿ 𝑠𝑒𝑙𝑓 𝑅𝑅𝐸𝐹~ [0 1
0 0
0| 1 | 0 | 2 | 1
0| 1 | 1 | 3 | 3 ]
1|−5|−3|−4|−5
Hence Matrix representing L with respect to 𝑆 ′ and 𝑇 ′ is
1 0 2 1
𝐴=[ 1 1 3 3 ]
−5−3−4−5
Solution:Given 𝑆 = {𝑣1 , 𝑣2 , 𝑣3 , 𝑣4 } π‘Žπ‘›π‘‘ 𝑇 = {𝑀1 , 𝑀2 , 𝑀3 , 𝑀4 }
(a)
Matrix of linear transformation with respect to 𝑆 =?
[𝑣1 𝑣2 𝑣3 𝑣4 | 𝐿(𝑣1 )| 𝐿(𝑣2 )| 𝐿(𝑣3 )| 𝐿(𝑣4 )] … . . (1)
𝐿(𝑣1 ) = 𝐿 [
1 0
1 2 1 0
1 0
]=[
][
]=[
]
0 0
3 4 0 0
3 0
𝐿(𝑣2 ) = 𝐿 [
0 1
1 2 0 1
0 1
]=[
][
]=[
]
0 0
3 4 0 0
0 3
𝐿(𝑣3 ) = 𝐿 [
1 2 0 0
0 0
2 0
]=[
][
]=[
]
3 4 1 0
1 0
4 0
𝐿(𝑣4 ) = 𝐿 [
0 0
1 2 0 0
0 2
]=[
][
]=[
]
0 1
3 4 0 1
0 4
Put these values in (1) we have
[𝑣1 𝑣2 𝑣3 𝑣4 | 𝐿(𝑣1 )| 𝐿(𝑣2 )| 𝐿(𝑣3 )| 𝐿(𝑣4 )] … . . (1)
1
0
[
0
0
0
1
0
0
0
0
1
0
0|1
0|0
0|3
1|0
|0
|1
|0
|3
|2
|0
|4
|0
|0
|2
]
|0
|4
[𝐼 | 𝐴]
1
0
Matrix of linear transformation with respect to 𝑆 is 𝐴 = [
3
0
0
1
0
3
2
0
4
0
0
2
]
0
4
(c) Matrix of linear transformation with respect to 𝑆 π‘Žπ‘›π‘‘ 𝑇=?
[𝑀1 𝑀2 𝑀3 𝑀4 | 𝐿(𝑣1 )| 𝐿(𝑣2 )| 𝐿(𝑣3 )| 𝐿(𝑣4 )] … . . (2)
Put values of 𝐿(𝑣1 ), 𝐿(𝑣2 ), 𝐿(𝑣3 )and 𝐿(𝑣4 )from part (a) in (2) we have
1110|1|0|2|0
0101|0|1|0|2
[
]
0010|3|0|4|0
1000|0|3|0|4
1000| 0 | 3 | 0 | 4
0100|−2|−3|−2|−4
𝑅𝑅𝐸𝐹~ [
]
0010| 3 | 0 | 4 | 0
0001| 2 | 4 | 2 | 6
Matrix of linear transformation with respect to 𝑆 π‘Žπ‘›π‘‘ 𝑇 is
0 3 0 4
−2−3−2−4
𝐴=[
]
3 0 4 0
2 4 2 6
Solution of part (b) and (d) given below
Question 22: First time in our exercise, he gave “Matrix w.r.t. Ordered basis”
1 21
Solution: (a) Given 𝐴 = [[𝐿(𝑣1 )] 𝑇 [𝐿(𝑣2 )] 𝑇 [𝐿(𝑣3 )] 𝑇 ] = [
]
−110
[𝐿(𝑣1 )] 𝑇 = [ 1 ]; [𝐿(𝑣2 )] 𝑇 = [2]; [𝐿(𝑣3 )] 𝑇 = [1]
0
−1
1
𝑐1
(Recall:-[𝑣] 𝑇 =? ; [𝑣] 𝑇 = [𝑐 ] π‘ π‘’π‘β„Ž π‘‘β„Žπ‘Žπ‘‘ 𝑣 = 𝑐1 𝑀1 + 𝑐2 𝑀2
2
𝑐
[𝐿(𝑣)]𝑇 =? ; [𝐿(𝑣)]𝑇 = [𝑐1 ] π‘ π‘’π‘β„Ž π‘‘β„Žπ‘Žπ‘‘ 𝐿(𝑣) = 𝑐1 𝑀1 + 𝑐2 𝑀2)
2
(b)
0
1
1
𝐿(𝑣1 ) = 1𝑀1 − 1𝑀2 = [ ] − [ ] = [ ];
3
2
−1
3
1
1
𝐿(𝑣2 ) = 2𝑀1 + 1𝑀2 = 2 [ ] + [ ] = [ ]
3
2
−1
1
1
1
𝐿(𝑣3 ) = 1𝑀1 + 0𝑀2 = [ ] + 0 [ ] = [ ]
2
−1
2
(c) [𝐿(𝑋)]𝑇 = 𝐴[𝑋]𝑆
;
2
𝑋=[ 1 ]
−1
;
2
𝐿(𝑋) = 𝐿 [ 1 ] =?
−1
[𝑋]𝑆 =? 𝑐1 𝑣1 + 𝑐2 𝑣2 + 𝑐3 𝑣3 = 𝑋 − −(1)π‘π‘œπ‘› β„Žπ‘œπ‘šπ‘œπ‘”π‘’π‘›π‘’π‘œπ‘’π‘  π‘ π‘¦π‘ π‘‘π‘’π‘š
𝑐1
[𝑋]𝑆 = [𝑐2 ]
𝑐3
−101| 2
[𝐴|𝑋] = [ 1 10| 1 ]
0 10|−1
100| 2
𝑅𝑅𝐸𝐹 [010|−1]
001| 4
𝑐1
2
[𝑋]𝑆 = [𝑐2 ] = [−1]
𝑐3
4
[𝐿(𝑋)] 𝑇 = 𝐴[𝑋]𝑆
2
4
1 21
𝑅𝐻𝑆 = 𝐴[𝑋]𝑆 = [
] [−1] = [ ] = 𝐿𝐻𝑆 = [𝐿(𝑋)] 𝑇
−3
−110
4
[𝐿(𝑋)] 𝑇 = [ 4 ]
−3
𝐿(𝑋) = 4𝑀1 − 3𝑀2 = [
1
]
11
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