Transcript Vector Spaces, Linear Transformations and Matrices

Vector Spaces, Linear Transformations
and Matrices
Md. Rabiul Haque
Lecturer
Department of Mathematics
University of Rajshahi, Rajshahi.
Email: [email protected]
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Vector Spaces, Linear Transformations and
Matrices
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Vector Spaces, Linear Transformations and Matrices
A non-empty set 𝑭 with two
operations
addition and multiplication
Fields
(A) Axioms for addition
(A1)
(A2)
(A3)
(A4)
(A5)
If 𝑥 ∈ 𝐹 and 𝑦 ∈ 𝐹, then their sum 𝑥 + у is in 𝐹.
Addition is commutative: 𝑥 + у = у + 𝑥 for all x, у ∈ 𝐹.
Addition is associative: (𝑥 + 𝑦) + 𝑧 = 𝑥 + (𝑦 + 𝑧) for all 𝑥, 𝑦, 𝑧 ∈ 𝐹.
𝐹 contains an element 0 such that 0 + x = x for every 𝑥 ∈ 𝐹.
To every 𝑥 ∈ 𝐹 corresponds an element −𝑥 ∈ 𝐹 such that 𝑥 + −𝑥 = 0.
(M) Axioms for multiplication
(M1) If 𝑥 ∈ 𝐹 and 𝑥 ∈ 𝐹, then their product 𝑥𝑦 is in 𝐹.
(M2) Multiplication is commutative: 𝑥𝑦 = 𝑦𝑥 for all 𝑥, 𝑦 ∈ 𝐹 .
(M3) Multiplication is associative: (𝑥𝑦)𝑧 = 𝑥(𝑦𝑧) for all 𝑥, 𝑦, 𝑧 ∈ 𝐹 .
(M4) 𝐹 contains an element 1 ≠ 0 such that 1𝑥 = 𝑥 for every 𝑥 ∈ 𝐹 .
(M5) If 𝑥 ∈ 𝐹 and 𝑥 ≠ 0 then there exists an element
1
𝑥
(D) The distributive law
𝑥(𝑦 + 𝑧) = 𝑥𝑦 + 𝑥𝑧
1
𝑥
∈ 𝐹 such that 𝑥. = 1.
holds for all
𝑥, 𝑦, 𝑧 ∈ 𝐹.
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Vector Spaces, Linear Transformations and
Matrices
Example
ℚ = the set of rational numbers
ℝ = the set of real numbers
ℂ = the set
of complex numbers
1
are fields.
ℕ =the set of natural numbers
ℤ =the set of integers
not fields ( why?? )
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Vector
𝒗
𝑩
𝒖
𝒌𝒖
𝒖
𝐴
𝒖
𝒌𝒖
𝒀
(𝒂 + 𝒄, 𝒃 + 𝒅 )
𝒀
(𝒌𝐚, 𝐤𝐛)
𝒗
𝒐
(𝒂, 𝒃)
𝒖 (𝒂, 𝒃)
𝑿
𝒐
𝑿
Marriage between Geometry and Algebra
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𝒖 = (𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 )
𝒗 = (𝑦1 , 𝑦2 , 𝑦3 , 𝑦4 )
𝒖+𝒗
= (𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 ) + 𝑦1 , 𝑦2 , 𝑦3 , 𝑦4
= (𝑥1 + 𝑦1 , 𝑥2 + 𝑦2 , 𝑥3 + 𝑦3 , 𝑥4 + 𝑦4 )
𝑘𝒖 = 𝑘 𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 = (𝑘𝑥1 , 𝑘𝑥2 , 𝑘𝑥3 , 𝑘𝑥4 )
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Let
𝑉 = ℝ𝑛 = { 𝑎1 , 𝑎2 , … … , 𝑎𝑛 ∶ 𝑎𝑖 ∈ ℝ } and 𝐹 = ℝ.
Define two operations vector addition and scalar multiplication as follows
𝒂𝟏 , 𝒂𝟐 , … … , 𝒂𝒏 + 𝒃𝟏 , 𝒃𝟐 , … … , 𝒃𝒏 = 𝒂𝟏 + 𝒃𝟏 , 𝒂𝟐 + 𝒃𝟐 , … … , 𝒂𝒏 + 𝒃𝒏
𝒌 𝒂𝟏 , 𝒂𝟐 , … … , 𝒂𝒏 = 𝒌𝒂𝟏 , 𝒌𝒂𝟐 , … … , 𝒌𝒂𝒏 where 𝒌 ∈ 𝑭
𝑛 is the n-tuple of zeros
The
zero
element
in
ℝ
− 𝒂𝟏 , 𝒂𝟐 , … , 𝒂𝒏 is the negative of 𝒂𝟏 , 𝒂𝟐 , … , 𝒂𝒏
− 𝒂𝟏 , 𝒂𝟐 , …
−𝒂, 0)
0 =, 𝒂(0,0,0,
𝒏 = …
𝟏 , −𝒂𝟐 , … , −𝒂𝒏
Let 𝒖 = 𝑎1 , 𝑎2 , … … , 𝑎𝑛 , 𝒗 = 𝑏1 , 𝑏2 , … … , 𝑏𝑛 and 𝒘 = 𝑐1 , 𝑐2 , … … , 𝑐𝑛
𝐴1 𝒖 + 𝒗 + 𝒘
= 𝑎1 , 𝑎2 , … … , 𝑎𝑛 + 𝑏1 , 𝑏2 , … … , 𝑏𝑛 + 𝑐1 , 𝑐2 , … … , 𝑐𝑛
= 𝑎1 + 𝑏1 , 𝑎2 + 𝑏2 , … … , 𝑎𝑛 + 𝑏𝑛 + 𝑐1 , 𝑐2 , … … , 𝑐𝑛
= 𝑎1 + 𝑏1 + 𝑐1 , 𝑎2 + 𝑏2 + 𝑐2 , … … , 𝑎𝑛 + 𝑏𝑛 + 𝑐𝑛
= 𝑎1 + (𝑏1 + 𝑐1 ), 𝑎2 + (𝑏2 +𝑐2 ), … … , 𝑎𝑛 + (𝑏𝑛 +𝑐𝑛 )
= 𝑎1 , 𝑎2 , … … , 𝑎𝑛 + 𝑏1 + 𝑐1 , 𝑏2 + 𝑐2 , … … , 𝑏𝑛 + 𝑐𝑛
=𝒖+ 𝒗+𝒘
𝑨𝟐 𝒖 + 𝒐
= 𝑎1 , 𝑎2 , … … , 𝑎𝑛 + (0, 0, … , 0) = 𝑎1 + 0, 𝑎2 + 0, … … , 𝑎𝑛 + 0 = 𝑎1 , 𝑎2 , … … , 𝑎𝑛
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=𝒖.
𝑨𝟑 𝒖 + −𝒖
= 𝑎1 , 𝑎2 , … … , 𝑎𝑛 + −𝑎1 , −𝑎2 , … , −𝑎𝑛
= 𝑎1 + −𝑎1 , 𝑎2 + (−𝑎2 ), … … , 𝑎𝑛 + (−𝑎𝑛 )
= (0, 0, … , 0)
=𝟎
𝑨𝟒 𝒖 + 𝒗
= 𝑎1 , 𝑎2 , … … , 𝑎𝑛 + 𝑏1 , 𝑏2 , … … , 𝑏𝑛
= 𝑎1 + 𝑏1 , 𝑎2 + 𝑏2 , … … , 𝑎𝑛 + 𝑏𝑛
= 𝑏1 + 𝑎1 , 𝑏2 + 𝑎2 , … … , 𝑏𝑛 + 𝑎𝑛
=𝒗+𝒖
𝑴𝟏 For any scalar 𝑘 ∈ 𝐹
𝒌 𝒖+𝒗
= 𝑘 𝑎1 + 𝑏1 , 𝑎2 + 𝑏2 , … … , 𝑎𝑛 + 𝑏𝑛
= 𝑘𝑎1 + 𝑘𝑏1 , 𝑘𝑎2 + 𝑘𝑏2 , … … , 𝑘𝑎𝑛 + 𝑘𝑏𝑛
= 𝑘𝑎1 , 𝑘𝑎2 , … … , 𝑘𝑎𝑛 + 𝑘𝑏1 , 𝑘𝑏2 , … … , 𝑘𝑏𝑛
= 𝑘 𝑎1 , 𝑎2 , … … , 𝑎𝑛 + 𝑘 𝑏1 , 𝑏2 , … … , 𝑏𝑛
= 𝒌𝒖 + 𝒌𝒗.
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𝑴𝟐 For any scalar 𝒌, 𝒍 ∈ 𝑭
𝒌+𝒍 𝒖
= 𝑘 + 𝑙 𝑎1 , 𝑎2 , … … , 𝑎𝑛
= (𝑘 + 𝑙)𝑎1 , (𝑘 + 𝑙)𝑎2 , … … , (𝑘 + 𝑙)𝑎𝑛
= 𝑘𝑎1 + 𝑙𝑎1 , 𝑘𝑎2 + 𝑙𝑎2 , … … , 𝑘𝑎𝑛 + 𝑙𝑎𝑛
= 𝑘𝑎1 , 𝑘𝑎2 , … … , 𝑘𝑎𝑛 + 𝑙𝑎1 , 𝑙𝑎2 , … … , 𝑙𝑎𝑛
= 𝑘 𝑎1 , 𝑎2 , … , 𝑎𝑛 + 𝑙 𝑎1 , 𝑎2 , … … , 𝑎𝑛 = 𝒌𝒖 + 𝒍𝒖
𝑴𝟑 For any scalar 𝒌, 𝒍 ∈ 𝑭
𝒌𝒍 𝒖
= 𝑘𝑙 𝑎1 , 𝑎2 , … … , 𝑎𝑛
= 𝑘𝑙𝑎1 , 𝑘𝑙𝑎2 , … … , 𝑘𝑙𝑎𝑛 = 𝑘(𝑙𝑎1 ), 𝑘(𝑙𝑎2 ), … … , 𝑘(𝑙𝑎𝑛 )
= 𝑘 𝑙𝑎1 , 𝑙𝑎2 , … … , 𝑙𝑎𝑛 = 𝑘 𝑙 𝑎1 , 𝑎2 , … … , 𝑎𝑛
= 𝒌 𝒍𝒖
𝑴𝟒 For unit scalar 𝟏 ∈ 𝑭
𝟏𝒖
= 𝟏 𝒂𝟏 , 𝒂𝟐 , … … , 𝒂𝒏 = 𝟏𝒂𝟏 , 𝟏𝒂𝟐 , … … , 𝟏𝒂𝒏 = 𝒂𝟏 , 𝒂𝟐 , … … , 𝒂𝒏
=𝒖
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𝑭(𝑿) =
𝒇𝜶 , 𝒇𝜷 , 𝒇𝜸
𝒇
,𝒇 𝜶
,………
𝜹
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𝑭 𝑿 = 𝒇: 𝑿 → 𝑲 = ℝ
𝑭(𝑿) denote the set of all functions of 𝑋 into 𝐾.
Can define
Addition and Scalar
Scalar multiplication
Multiplication
Addition
(𝒇 + 𝒈)(𝒙) = 𝒇(𝒙) + 𝒈(𝒙) ∀𝒙
in∈ 𝑿𝐹
𝑋
?
(𝒌𝒇)(𝒙) = 𝒌𝒇(𝒙) ∀𝒙 ∈ 𝑿
Zero and negative element in
𝐹 𝑋 ?
Zero element in 𝐹(𝑋)
Zero function 0
𝟎 𝒙 = 0 ∀𝒙 ∈ 𝑿
−𝒇 𝒊𝒔 𝒕𝒉𝒆 negative of the function 𝒇
−𝒇 𝒙 = −𝒇 𝒙 ∀ 𝒙 ∈ 𝑿
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Let 𝒇, 𝒈, 𝒉 ∈ 𝑭(𝑿)
𝑨𝟏
𝒇+𝒈 +𝒉 𝑥 = 𝑓+𝑔 𝑥 +ℎ 𝑥 = 𝑓 𝑥 +𝑔 𝑥
=𝑓 𝑥 + 𝑔 𝑥 +ℎ 𝑥 = 𝒇+ 𝒈+𝒉 𝒙
∀𝑥 ∈ 𝑋
+ℎ 𝑥
𝒇+𝒈 +𝒉=𝒇+ 𝒈+𝒉
𝑨𝟐
𝒇+𝟎 𝒙 =𝒇 𝒙 +𝟎 𝒙 =𝒇 𝒙 +𝟎= 𝒇
∀𝑥 ∈ 𝑋
𝒇+𝟎=𝒇
𝑨𝟑
𝒇 + −𝒇
𝒙 = 𝒇 𝒙 + −𝒇 𝒙 = 𝒇 𝒙 − 𝒇 𝒙 = 𝟎 = 𝟎 𝒙 ∀𝑥 ∈ 𝑋
𝒇 + −𝒇 = 𝟎
𝑨𝟒
𝒇 + 𝒈 𝒙 = 𝒇 𝒙 + 𝒈 𝒙 = 𝒈 𝒙 + 𝒇 𝒙 = 𝒈 + 𝒇 𝒙 ∀𝑥 ∈ 𝑋
𝒇+𝒈=𝒈+𝒇
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𝑴𝟏 For any scalar 𝑘 ∈ 𝐾
𝑘 𝑓+𝑔 𝑥 =𝑘 𝑓+𝑔 𝑥 =𝑘 𝑓 𝑥 +𝑔 𝑥
= 𝑘𝑓 𝑥 + 𝑘𝑔 𝑥 = 𝑘𝑓 + 𝑘𝑔 𝑥
= 𝑘𝑓 𝑥 + 𝑘𝑔 𝑥
𝒌 𝒇 + 𝒈 = 𝒌𝒇 + 𝒌𝒈
𝑴𝟐 For any scalar 𝒌, 𝒍 ∈ 𝑲
𝒌 + 𝒍 𝒇 𝒙 = 𝒌 + 𝒍 𝒇 𝒙 = 𝒌𝒇 𝒙 + 𝒍𝒇 𝒙 = 𝒌𝒇 𝒙 + 𝒍𝒇 𝒙
𝒌 + 𝒍 𝒇 = 𝒌𝒇 + 𝒍𝒇
𝑴𝟑 For any scalar 𝒌, 𝒍 ∈ 𝑲 𝒌𝒍𝒇 𝒙 = 𝒌𝒍𝒇 𝒙 = 𝒌 𝒍𝒇 𝒙
= 𝒌 𝒍𝒇
𝒙
(𝒌𝒍)𝒇 = 𝒌(𝒍𝒇)
𝑴𝟒 For unit scalar 𝟏 ∈ 𝑲
𝟏𝒇 𝒙 = 𝟏𝒇 𝒙 = 𝒇 𝒙
𝟏𝒇 = 𝒇
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Let ℝ be field and let
𝑷 𝒕 = { 𝑝 𝑡 = 𝑎0 + 𝑎1 𝑡 1 + 𝑎2 𝑡 2 + ⋯ + 𝑎𝑠 𝑡 𝑠 ,
𝑠 = 1,2,3, … 𝑎𝑛𝑑 𝑎𝑖 ∈ ℝ }
𝑷 𝒕 denote the set of all real polynomials 𝒑(𝒕) . 𝒑 𝒕 =
𝟑 + 𝟖𝒕𝟏 − 𝟕𝒕𝟐 + ⋯ + 𝟒𝟓𝒕𝟑𝟏 + ⋯
Let ℝ be field and 𝐥𝐞𝐭
𝑷𝒏 𝒕 = { 𝒑 𝒕 = 𝒂𝟎 + 𝒂𝟏 𝒕𝟏 + 𝒂𝟐 𝒕𝟐 + ⋯ + 𝒂𝒔 𝒕𝒔 ,
𝑠 ≤ 𝑛 𝑎𝑛𝑑 𝑎𝑖 ∈ ℝ }
𝑷𝒏 𝒕 denote the set of all polynomials 𝒑(𝒕) over the
field ℝ ,
where the degree of 𝒑(𝒕) is less than or equal to 𝒏 .
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Usual addition and scalar multiplication, we can see that
𝑃 𝑡 and 𝑃𝑛 𝑡
satisfies
𝑨𝟏 𝒑 + 𝒒 + 𝒓 = 𝒑 + 𝒒 + 𝒓
𝑨𝟐 𝒑 + 𝟎 = 𝒑
𝑨𝟑
𝑨𝟒
𝑴𝟏
𝑴𝟐
𝑴𝟑
𝑴𝟒
𝒑 + −𝒑 = 𝟎
𝒑+𝒒=𝒒+𝒑
𝒌 𝒑 + 𝒒 = 𝒌𝒑 + 𝒌𝒒
𝒌 + 𝒍 𝒑 = 𝒌𝒑 + 𝒍𝒑
𝒌𝒍 𝒑 = 𝒌 𝒍𝒑
𝟏𝒑 = 𝒑
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𝑛
ℝ
ℝ
2
ℝ
𝑀𝑚,𝑛
3
Vector
Spaces over
the field 𝑭
???????
𝑭(𝑿)
𝑷𝒏 (𝒕)
𝑃(𝑡)
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Vector Spaces
Abstract Concept
Let 𝑽 Vector
be a nonempty
set involves
with two operations:
Space
four things
(i) Vector Addition
: This assigns to any
𝒖, 𝒗(𝑽
∈ 𝑽, a𝑭)
sumand
𝒖 + 𝒗 in 𝑉.
Two non-empty
sets
two
algebraic
operations
( addition,
(ii)
Scalar
Multiplication:
This assigns
to any 𝒖 ∈scalar
𝑽, 𝒌 ∈ multiplication)
𝑲 a product 𝒌𝒖 ∈ 𝑉.
Then 𝑽 is called a vector space over the field 𝐾
𝑨𝟏 𝒖 + 𝒗 + 𝒘 = 𝒖 + 𝒗 + 𝒘
Abelian Group under
𝑨𝟐 𝒖 + 𝟎 = 𝟎 + 𝒖 = 𝒖
Addition
𝑨𝟑 𝒖 + −𝒖 = −𝒖 + 𝒖 = 𝟎.
𝑨𝟒 𝒖 + 𝒗 = 𝒗 + 𝒖
+: 𝑉×𝑉 →𝑉
⋅ ∶𝐹×𝑉 →𝑉
𝑴𝟏 𝒄 𝒖 + 𝒗 = 𝒄𝒗 + 𝒄𝒘
𝑴𝟐 𝒄 + 𝒌 𝒖 = 𝒄𝒖 + 𝒌𝒖
𝑴𝟑 𝒄𝒌 𝒖 = 𝒄 𝒌𝒖
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3
ℝ
𝑛
ℝ
𝑀𝑚,𝑛
2
ℝ
ℝ
Vector
Spaces over
the field ℝ
Integrable
functions on the
same interval
𝑭(𝑿)
𝑷𝒏 (𝒕)
𝑃(𝑡)
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𝟐
(𝟎, 𝟎)
vector space ?
𝒀
ℝ
(𝟐, 𝟏)
vector space ?
𝟐, 𝟏 , (𝟒, 𝟐)
vector space ?
𝒙, 𝒚 ∶ 𝒚 = 𝒎𝒙
vector space ?
𝟐
ℝ
●
●
●𝒐
𝑿
Subset
Vector space
Not Vector space
Subspace
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Subspaces
V
W
is vector space over a field 𝑲
𝑾 ⊂ 𝑽 , 𝑾 is subspace of 𝑽
Let 𝑽 be a vector space over a field 𝑲 and let 𝑾 be a subset of V.
Then 𝑾 is a subspace of 𝑽 if 𝑾 is itself a vector space over 𝑲 with
respect to the operations of vector addition and scalar
multiplication on 𝑽.
if
𝟎∈𝑾
and 𝒂𝒖 + 𝒃𝒗 ∈ 𝑾
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Subspaces of ℝ𝟐
𝒀
∎ {(𝟎, 𝟎)}
∎ Any line through (𝟎, 𝟎)
∎ The whole Space (ℝ𝟐 )
⦁
𝒐
⦁
⦁
𝑿
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Subspaces of ℝ𝟑
𝒀
∎ {(𝟎, 𝟎, 𝟎)}
∎ Any line through (𝟎, 𝟎, 𝟎)
∎ Any Plane through (𝟎, 𝟎, 𝟎)
∎ The whole Space
(ℝ𝟑 )
𝒐
⦁
𝑿
𝒁
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𝑺𝒖𝒃𝒔𝒑𝒂𝒄𝒆𝒔 𝒐𝒇 𝑭(𝑿)
All real valued function
on
𝑿
All bounded/odd/even/
analytic functions on 𝑿
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𝑺𝒖𝒃𝒔𝒑𝒂𝒄𝒆𝒔 𝒐𝒇 𝑷(𝒕)
The set of all real
polynomials
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𝑺𝒖𝒃𝒔𝒑𝒂𝒄𝒆𝒔 𝒐𝒇 𝑴𝒎,𝒏
All matrices of order
𝒎×𝒏
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𝑈∩ 𝑉
V
𝑈
𝑠𝑢𝑏𝑠𝑝𝑎𝑐𝑒
subspace
subspace
The intersection of any number of subspaces
of a vector space 𝑽 is a subspace of 𝑽.
26
Linear combinations
𝒗𝟏 , 𝒗 𝟐 , … , 𝒗𝒎 ∈ 𝑽
𝒗 = 𝒂 𝟏 𝒗𝟏 + 𝒂 𝟐 𝒗𝟐 + ⋯ + 𝒂 𝒎 𝒗𝒎
𝒘𝒉𝒆𝒓𝒆 𝒂𝟏 , 𝒂𝟐 , … , 𝒂𝒎 ∈ 𝑲
2
(𝟏, 𝟎), (𝟎, 𝟏) ∈ ℝ
2 1, 0 + 3 0, 1 = (2, 3)
−4 1, 0 + 2 0, 1 = −4, 2
0 1, 0 + 0 0, 1 = (0, 0)
𝟏, 𝟏, 𝟐 , 𝟑, 𝟎, 𝟏 , (𝟐, 𝟐, 𝟒) ∈ ℝ3
2 1, 1, 2 + 3 3,0, 1 + 4 2, 2, 4 = 17, 10,23
−3 1, 1, 2 + 3 3,0, 1 + 5(2, 2, 4) = (10, 7,17)
0 1, 1, 2 + 0 3,0, 1 + 0(2, 2, 4) = (0, 0,0)
27
Spanning sets
𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒎 ∈ 𝑽 span 𝑽 if every 𝒗 in 𝑽 is
𝒗 = 𝒂 𝟏 𝒗𝟏 + 𝒂𝟐 𝒗𝟐 + ⋯ + 𝒂𝒎 𝒗𝒎
Are 𝒆𝟏 = 𝟏, 𝟎 , 𝒆𝟐 = 𝟎, 𝟏 span ℝ𝟐 ?
𝒂, 𝒃 = 𝒙𝒆𝟏 + 𝒚𝒆𝟐
= 𝒙 𝟏, 𝟎 + 𝒚 𝟎, 𝟏
= 𝒙, 𝟎 + 𝟎, 𝒚 = 𝒙, 𝒚
⇒𝒙=𝒂,
𝒂, 𝒃 = 𝒂𝒆𝟏 + 𝒃𝒆𝟐
𝒚=𝒃
𝟐
𝒆𝟏 , 𝒆𝟐 span ℝ
𝟓, −𝟔 = 𝟓𝒆𝟏 − 𝟔𝒆𝟐
28
Are 𝒘𝟏 = 𝟐, 𝟏 , 𝒘𝟐 = 𝟑, 𝟐 span ℝ𝟐 ?
𝒂, 𝒃 = 𝒙𝒘𝟏 + 𝒚𝒘𝟐
= 𝒙 𝟐, 𝟏 + 𝒚 𝟑, 𝟐
= 𝟐𝒙, 𝒙 + 𝟑𝒚, 𝟐𝒚 = 𝟐𝒙 + 𝟑𝒚, 𝒙 + 𝟐𝒚
⇒ 𝟐𝒙 + 𝟑𝒚 = 𝒂,
𝒙 + 𝟐𝒚 = 𝒃
⇒ 𝒙 = 𝟐𝒂 − 𝟑𝒃 ,
𝒚 = 𝟐𝒃 − 𝒂,
𝒂, 𝒃 = (𝟐𝒂 − 𝟑𝒃)𝒘𝟏 + (𝟐𝒃 −𝒂)𝒘𝟐
𝒘𝟏 , 𝒘𝟐 span ℝ𝟐
𝟏, −𝟐 = 𝟕𝒘𝟏 − 𝟓𝒘𝟐
29
𝑨𝒓𝒆 𝒆𝟏 = 𝟏, 𝟎 𝒔𝒑𝒂𝒏 ℝ𝟐 ?
𝒂, 𝒃 = 𝒙𝒆𝟏 = 𝒙 𝟏, 𝟎 = 𝒙, 𝟎
⇒ 𝒙 = 𝒂,
𝒃=𝟎
𝒆𝟏 do not span ℝ𝟐
Are 𝒖𝟏 = 𝟏, 𝟎 , 𝒖𝟐 = 𝟎, 𝟏 , 𝒖𝟑 = 𝟏, 𝟏 s𝐩𝐚𝐧 ℝ𝟐 ?
𝒂, 𝒃 = 𝒂𝒖𝟏 + 𝒃𝒖𝟐 + 𝟎𝒖𝟑
𝒖𝟏 , 𝒖𝟐 , 𝒖𝟑 span ℝ𝟐 .
Are 𝒖𝟏 = 𝟏, 𝟎 , 𝒖𝟐 = 𝟎, 𝟏 , 𝒖𝟑 = 𝟏, 𝟏 , 𝒖𝟒 = (𝟐, 𝟐) s𝐩𝐚𝐧 ℝ𝟐 ?
𝒂, 𝒃 = 𝒂𝒖𝟏 + 𝒃𝒖𝟐 − 𝟐𝒖𝟑 + 𝒖𝟒
𝒖𝟏 , 𝒖𝟐 , 𝒖𝟑 , 𝒖𝟒 span ℝ𝟐 .
S𝒖𝒑𝒑𝒐𝒔𝒆 𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒎 span 𝑽,
the set 𝒘, 𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒎 also span 𝑽 .
𝒗 = 𝒂𝟏 𝒗𝟏 + 𝒂𝟐 𝒗𝟐 + ⋯ + 𝒂𝒎 𝒗𝒎
𝒗 = 𝒂𝟏 𝒗𝟏 + 𝒂𝟐 𝒗𝟐 + ⋯ + 𝒂𝒎 𝒗𝒎 + 𝟎𝒘
30
Linear Spans
𝒖𝟏 , 𝒖𝟐 , … , 𝒖𝒎 ∈ 𝑽
𝒔𝒑𝒂𝒏 𝒖𝒊 =
𝒎
𝒊=𝟏
𝒂𝒊 𝒖𝒊 | 𝒂𝒊 ∈ 𝑲
𝒔𝒑𝒂𝒏(𝒖) consists of all
scalar multiples 𝑜𝑓 𝒖.
𝒖
𝒐
𝒀
𝒖
𝒗
Geometrically, 𝒔𝒑𝒂𝒏(𝒖) is the
line through the origin 𝑶 and the
endpoint of 𝒖.
Then 𝒔𝒑𝒂𝒏 𝒖, 𝒗 is the plane
through the origin O and the
endpoints of 𝑢 𝑎𝑛𝑑 𝑣 .
𝟎
𝑿
𝒁
31
𝑺 = ∅ 𝒔𝒑𝒂𝒏(𝑺) = {𝟎}.
𝑺 is a spanning set of 𝒔𝒑𝒂𝒏(𝑺).
𝒆𝟏 = 𝟏, 𝟎, 𝟎 ,
𝒆𝟐 = 𝟎, 𝟏, 𝟎 , 𝒆𝟑 = 𝟎, 𝟎, 𝟏
𝒂, 𝒃, 𝒄 = 𝒂𝒆𝟏 + 𝒃𝒆𝟐 + 𝒄𝒆𝟑
s𝒑𝒂𝒏(𝒆𝟏 , 𝒆𝟐 , 𝒆𝟑 ) = ℝ𝟑 .
𝑺𝒑𝒂𝒏(𝒖𝒊 ) is a subspace of 𝑽.
𝟎 ∈ 𝒔𝒑𝒂𝒏 𝒖𝒊 , since 𝟎 = 𝟎𝒖𝟏 + 𝟎𝒖𝟐 + … + 𝟎𝒖𝒎
Let 𝒖 , 𝒗 ∈ 𝒔𝒑𝒂𝒏 𝒖𝒊
⟹ 𝒖 = 𝒏𝒊=𝟏 𝒂𝒊 𝒖𝒊 and 𝒗 = 𝒏𝒊=𝟏 𝒃𝒊 𝒖𝒊
⇒ 𝒂𝒖 + 𝒃𝒗 = 𝒂
=
𝒏
𝒊=𝟏
𝒏
𝒊=𝟏
𝒖𝒊 + 𝒃
𝒏
𝒊=𝟏
𝒃𝒊 𝒖𝒊
𝒂𝒂𝒊 + 𝒃𝒂𝒊 𝒖𝒊
⇒ 𝒂𝒖 + 𝒃𝒗 ∈ 𝑺𝒑𝒂𝒏(𝒖 )
32
Theorem
Let 𝑺 be a subset of a vector space 𝑽.
(i) Then 𝒔𝒑𝒂𝒏(𝑺) is a subspace of 𝑉 that contains 𝑆.
(ii) If 𝑾 is a subspace of 𝑽 containing 𝑺, then 𝒔𝒑𝒂𝒏(𝑺) ⊂ 𝑾.
𝒔𝒑𝒂𝒏(𝑺) is the "smallest" subspace of 𝑽 containing 𝑺.
𝑆
𝑽
𝑺𝒑𝒂𝒏(𝒔)
33
Linear Dependence and Independence
{ 𝟐, 𝟏 , (𝟑, 𝟐)}
↓
ℝ
{ 𝟏, 𝟎 }
{ 𝟏, 𝟎 , 𝟎, 𝟏 , (𝟏, 𝟏)}
↓
{ 𝟏, 𝟎 , (𝟎, 𝟏)}
↓𝟐
{ 𝟏, 𝟎 , 𝟎, 𝟏 , 𝟏, 𝟏 , (𝟐, 𝟐)}
𝒖𝟏 , 𝒖𝟐 , … , 𝒖𝒎 ∈ 𝑽 are linearly dependent
if ∃ scalars 𝒂𝟏 , 𝒂𝟐 , … , 𝒂𝒎 in 𝑲, not all of them 𝟎, such that
𝒂𝟏 𝒖𝟏 + 𝒂𝟐 𝒖𝟐 + ⋯ + 𝒂𝒎 𝒖𝒎 = 𝟎.
Otherwise, we say that the vectors
𝒖𝟏 , 𝒖𝟐 , … , 𝒖𝒎 are linearly independent.
34
Linear Dependence and Independence
𝑆 = 𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒎 one vector is 0 , 𝑠𝑎𝑦 𝒗𝟏 =0.
𝑺 linearly independent or not ?
𝟏𝒗𝟏 + 𝟎𝒗𝟐 + ⋯ + 𝟎𝒗𝒎
𝑺 dependent.
= 𝟏. 𝟎 + 𝟎 + ⋯ + 𝟎 = 𝟎
𝒗, 𝒌𝒗 two vectors
−𝒌𝒗 + 𝟏 𝒌𝒗 = 𝟎
𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒎
one vector is multiple of others , 𝑠𝑎𝑦 𝒗𝟏 = 𝒌𝒗𝟑 .
𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒎 linearly independent or not ?
−𝒌𝒗𝟏 + 𝟎𝒗𝟐 + 𝒌𝒗𝟑 + ⋯ + 𝟎𝒗𝒎
= 𝟏. 𝟎 + 𝟎 + ⋯ + 𝟎
=𝟎
𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒎 linearly dependent.
35
{ 𝟏, 𝟎 , 𝟎, 𝟏 } 𝒍𝒊𝒏𝒆𝒂𝒓𝒍𝒚
𝒊𝒏𝒅𝒆𝒑𝒆𝒏𝒅𝒆𝒏𝒕 𝒊𝒏 ℝ𝟐 ?
{ 𝟐, 𝟏 , 𝟑, 𝟐 } 𝒍𝒊𝒏𝒆𝒂𝒓𝒍𝒚
𝒊𝒏𝒅𝒆𝒑𝒆𝒏𝒅𝒆𝒏𝒕 𝒊𝒏 ℝ𝟐 ?
𝒙 𝟏, 𝟎 + 𝒚 𝟎, 𝟏 = 𝟎, 𝟎
𝒙 𝟐, 𝟏 + 𝒚 𝟑, 𝟐 = 𝟎, 𝟎
⇒ 𝑥, 0 + 0, 𝑦 = 0, 0
⇒ 2𝑥, 𝑥 + 3𝑦, 2𝑦 = 0, 0
⇒ 𝑥, 𝑦 = 0, 0
⇒ 2𝑥 + 3𝑦, 𝑥 + 2𝑦 = 0, 0
⇒ 𝒙 = 𝟎, 𝒚 = 𝟎
⇒ 𝒙 = 𝟎, 𝒚 = 𝟎
𝟏, 𝟎 , 𝟎, 𝟏 𝒂𝒏𝒅 𝟐, 𝟏 , 𝟑, 𝟐 𝒍𝒊𝒏𝒆𝒂𝒓𝒍𝒚 𝒊𝒏𝒅𝒆𝒑𝒆𝒏𝒅𝒆𝒏𝒕 .
{ 𝟏, 𝟎 , 𝟎, 𝟏 , 𝟏, 𝟏 } 𝒍𝒊𝒏𝒆𝒂𝒓𝒍𝒚 𝒊𝒏𝒅𝒆𝒑𝒆𝒏𝒅𝒆𝒏𝒕 𝒊𝒏 ℝ2 ?
𝒙 𝟏, 𝟎 + 𝒚 𝟎, 𝟏 + 𝒛 𝟏, 𝟏 = 𝟎, 𝟎
⇒ 𝑥, 0 + 0, 𝑦 + (𝑧, 𝑧) = 0, 0
⇒ 𝑥 + 𝑧, 𝑦 + 𝑧 = 0, 0 ⇒ 𝒙 + 𝒛 = 𝟎,
𝒚+𝒛=𝟎
Let z =1 ⇒ 𝑥 = −1, 𝑦 = −1
−𝟏 𝟏, 𝟎 + 𝟏 𝟎, 𝟏 + 𝟏 −𝟏, 𝟏 = 𝟎, 𝟎
{ 𝟏, 𝟐 , 𝟏, 𝟏 , 𝟎, 𝟏 } 𝒂𝒓𝒆 𝒏𝒐𝒕 𝒍𝒊𝒏𝒆𝒂𝒓𝒍𝒚 𝒊𝒏𝒅𝒆𝒑𝒆𝒏𝒅𝒆𝒏𝒕 𝒊𝒏 ℝ2 .
{ 𝟏, 𝟎 , 𝟎, 𝟏 , 𝟏, 𝟏 , (𝟐, 𝟐)} are linearly dependent.
Since −𝟏 𝟏, 𝟎 + 𝟏 𝟎, 𝟏 − 𝟐 𝟏, 𝟏 + 𝟏(𝟐, 𝟐) = 𝟎, 𝟎
36
Basis
Span + Linearly Independent ⟹ Basis
𝑽
𝑺 span 𝑽
𝑺 linearly independent
𝑺 = { 𝒖𝟏 , 𝒖𝟐 , 𝒖𝟑 , … , 𝒖𝒏 }
𝟐
𝟏, 𝟎 , (𝟎, 𝟏) is a basis of ℝ
and called the usual or standard basis .
𝟐, 𝟏 , (𝟑, 𝟐) is a basis of ℝ𝟐 .
37
𝟏, 𝟏 = 𝟏 𝟕 𝟏, 𝟑 + 𝟐 𝟕 (𝟑, 𝟐)
𝑺=
𝟏, 𝟑 , (𝟑, 𝟐) in ℝ𝟐 .
𝒙 𝟏, 𝟑 + 𝒚 𝟑, 𝟐 = 𝟎, 𝟎
⇒ 𝒙, 𝟑𝒙 + 𝟑𝒚, 𝟐𝒚 = 𝟎, 𝟎
⇒ 𝒙 + 𝟑𝒚, 𝟑𝒙 + 𝟐𝒚 = 𝟎, 𝟎
⇒ 𝒙 + 𝟑𝒚 = 𝟎 ,
𝟑𝒙 + 𝟐𝒚 = 𝟎
⇒ 𝒙 = 𝟎, 𝒚 = 𝟎
⇒ 𝟎 𝟏, 𝟑 + 𝟎 𝟑, 𝟐 = 𝟎, 𝟎
𝒂, 𝒃 = 𝒙 𝟏, 𝟑 + 𝒚 𝟑, 𝟐 = 𝒙, 𝟑𝒙 + 𝟑𝒚, 𝟐𝒚
= 𝒙 + 𝟑𝒚, 𝟑𝒙 + 𝟐𝒚
⇒ 𝒙 + 𝟑𝒚 = 𝒂 ,
𝟑𝒙 + 𝟐𝒚 = 𝒃
⇒ 𝒙 = (𝟑𝒃 − 𝟐𝒂) 𝟕 ,
𝒚 = (𝟑𝒂 − 𝒃) 𝟕
𝒂, 𝒃 = (𝟑𝒃 − 𝟐𝒂) 𝟕 𝟏, 𝟑 + (𝟑𝒂 − 𝒃) 𝟕 𝟑, 𝟐
𝟐
𝑺 is a basis of ℝ .
38
{ 𝟏, 𝟎 }
{ 𝟏, 𝟎 , 𝟎, 𝟏 , (𝟏, 𝟏)}
Not a basis
of ℝ𝟐
{ 𝟏, 𝟎 , 𝟎, 𝟏 , 𝟏, 𝟏 , (𝟐, 𝟐)}
39
𝟏, 𝟎 , (𝟎, 𝟏) ,
basis for ℝ𝟐 ??
𝟏, 𝟑 , (𝟑, 𝟐) , 𝟐, 𝟏 , (𝟑, 𝟐) only
𝒀
𝟏, 𝟐 , 𝟏, 𝟎
{ 𝟏, 𝟐 , (𝟎, 𝟏)}
(𝟏, 𝟐)
(−𝟒, 𝟐)
(𝟎, 𝟏)
{ −𝟒, 𝟐 , (𝟏, 𝟎)} { 𝟏, 𝟐 , (−𝟒, 𝟐)}
𝒐
(𝟏, 𝟎)
(−𝟐, −𝟏)
(𝟐, −𝟐)
𝑿
𝟑
ℝ
𝒏
,ℝ
Orthogonal, orthonormal basis
Any three vectors in ℝ𝟐 are linearly dependent.
ℝ2 (a vector space ) has infinitely many different bases.
40
Vector space 𝑷 𝒕 of all polynomials
Consider any finite set 𝑺 = 𝒇𝟏 𝒕 , 𝒇𝟐 𝒕 , … , 𝒇𝒎 𝒕 of
polynomials in 𝑷 𝒕 , and let 𝒎 denote the largest of the degrees
of the polynomials.
Then any polynomial 𝒈(𝒕) of degree exceeding 𝒎 cannot be expressed as a
linear combination of the elements of 𝑺. Thus 𝑺 cannot be a basis of 𝑷 𝒕 .
We note that the infinite set 𝑺′ = {𝟏, 𝒕, 𝒕𝟐 , 𝒕𝟑 , … … }, consisting of
all the powers of 𝑡, spans 𝑃 𝑡 and is linearly independent.
Accordingly, 𝑆 ′ is an infinite basis of 𝑃 𝑡 .
41
Dimension
Basis ⟹ Dimension
Theorem
𝑽
𝑺 = {𝒖𝟏 , 𝒖𝟐 , … , 𝒖𝒎 }
𝑺′ = {𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒏 }
𝒎=𝒏
Every basis of a vector space 𝑽 has the same
number of elements..
dim 𝑽 = number of vectors in any basis for 𝑉.
42
𝑽
finite-dimensional spaces.
infinite−dimensional spaces
𝒅𝒊𝒎 ℝ = 𝟏, standard basis 𝟏
𝒅𝒊𝒎 ℝ𝟐 = 𝟐, standard basis 𝟏, 𝟎 , (𝟎, 𝟏)
𝒅𝒊𝒎 ℝ𝒏 = 𝒏, standard basis
𝟏, 𝟎, 𝟎, … , 𝟎 , 𝟎, 𝟏, 𝟎, … , 𝟎 , … … , (𝟎, 𝟎, 𝟎, … , 𝟏)
𝐝𝐢𝐦 𝑷𝒏 = 𝒏 + 𝟏,
standard basis {𝟏, 𝒕, 𝒕𝟐 , 𝒕𝟑 , … , 𝒕𝒏 }
𝐝𝐢𝐦 𝑴𝒎,𝒏 = 𝒎𝒏
𝐝𝐢𝐦 𝑷 = 𝐝𝐢𝐦 𝑭 𝑿 = 𝐝𝐢𝐦 𝑪 𝒂, 𝒃 = ∞
𝑺′ = {𝟏, 𝒕, 𝒕𝟐 , 𝒕𝟑 , … … }
43
Theorem: Let 𝑽 be a vector space of finite
dimension 𝒏. Then any 𝒏 + 𝟏 or more vectors
in 𝑽 are linearly dependent.
Since 𝑽 is vector space of dimension 𝑛 .
Then {𝒖𝟏 , 𝒖𝟐 , … , 𝒖𝒏 } span 𝑽.
Let 𝒗 ∈ 𝑽, then 𝑣 = 𝑎1 𝑢1 + 𝑎2 𝑢2 + ⋯ + 𝑎𝑛 𝑢𝑛
⟹ 1𝑣 − 𝑎1 𝑢1 − 𝑎2 𝑢2 − ⋯ − 𝑎𝑛 𝑢𝑛 = 0
⇒ {𝒗, 𝒖𝟏 , 𝒖𝟐 , … , 𝒖𝒏 } are linearly dependent
⇒ 𝑛 + 1 vectors are linearly dependent.
44
Sums and Direct sums
𝑈
𝑊
𝑽
𝑼 + 𝑾 = {𝒗 ∶ 𝒗 = 𝒖 + 𝒘, 𝒘𝒉𝒆𝒓𝒆 𝒖 ∈ 𝑼 𝒂𝒏𝒅 𝒘 ∈ 𝑾 }
𝒅𝒊𝒎(𝑼 + 𝑾) = 𝒅𝒊𝒎𝑼 + 𝒅𝒊𝒎 𝑾 − 𝒅𝒊𝒎(𝑼 ∩ 𝑾)
45
𝑉 = 𝑈⨁𝑊
if every 𝒗 ∈ 𝑽 can be written in one and only one way as
𝑣 = 𝑢 + 𝑤 where 𝑢 ∈ 𝑈 𝑎𝑛𝑑 𝑤 ∈ 𝑊.
The vector space 𝑽 is the direct sum of its subspaces 𝑼 and 𝑾 if and only if:
𝑖 𝑉 = 𝑈 + 𝑊, 𝑖𝑖 𝑈 ∩ 𝑊 = {0}.
𝒀
𝑽 = ℝ𝟑
𝑼 =
𝒂, 𝒃, 𝟎
𝒂, 𝒃 ∈ ℝ}
𝑉 ≠ 𝑈⨁𝑊
𝟎
𝒁
𝑿
𝑾 = {(𝟎, 𝒃, 𝒄)| 𝒃, 𝒄 ∈ ℝ}
𝟑, 𝟓, 𝟕 = 𝟑, 𝟏, 𝟎 + 𝟎, 𝟒, 𝟕 𝒂𝒏𝒅 𝒂𝒍𝒔𝒐
(𝟑, 𝟓, 𝟕) = (𝟑, −𝟒, 𝟎) + (𝟎, 𝟗, 𝟕)
46
𝑽 = ℝ𝟑
𝑼 =
𝑽 = 𝑼⨁𝑾
𝒂, 𝒃, 𝟎
𝒂, 𝒃 ∈ ℝ}
𝒀
𝒂, 𝒃, 𝒄 = 𝒂, 𝒃 , 𝟎 + (𝟎, 𝟎, 𝒄)
𝟎
𝒁
𝑿
𝑾 = {(𝟎, 𝟎, 𝒄)| 𝒄 ∈ ℝ}
𝒀
𝟎
𝑿
ℝ𝟐 = 𝑳𝟏 ⊕ 𝑳𝟐
47
Coordinates
K
n-dimension
𝑣
𝑽
𝑠
= 𝑎1 , 𝑎2 , … , 𝑎𝑛
𝒗 ∈ 𝑽, 𝒗 = 𝒂𝟏 𝒖𝟏 + 𝒂𝟐 𝒖𝟐 + ⋯ + 𝒂𝒏 𝒖𝒏
𝒗 = 𝒃𝟏 𝒖𝟏 + 𝒃𝟐 𝒖𝟐 + ⋯ + 𝒃𝒏 𝒖𝒏
(𝒂𝟏 −𝒃𝟏 )𝒖𝟏 + (𝒂𝟐 −𝒃𝟐 )𝒖𝟐 + ⋯ + (𝒂𝒏 −𝒃𝒏 )𝒖𝒏 = 𝟎
Since 𝒖𝟏 , 𝒖𝟐 , … , 𝒖𝒏 are linearly independent
𝒂𝟏 − 𝒃𝟏 = 𝟎 , 𝒂𝟐 − 𝒃𝟐 = 𝟎 , … … ,𝒂𝒏 − 𝒃𝒏 = 𝟎
𝒂𝟏 = 𝒃𝟏 , 𝒂𝟐 = 𝒃𝟐 , … … , 𝒂𝒏 = 𝒃𝒏
48
Coordinates
ℝ𝟐
basis 𝐒 = 𝒖𝟏 = 𝟏, 𝟏 , 𝒖𝟐 = (−𝟏, 𝟏)
standard basis 𝐄 =
𝟏, 𝟎 , (𝟎, 𝟏)
𝒙−𝒚
𝟏
−𝟏
𝟓
=𝒙
+𝒚
= 𝒙+𝒚
𝟏
𝟏
𝟑
𝒙−𝒚=𝟓 𝒙=𝟒
𝒙 + 𝒚 = 𝟑 𝒚 = −𝟏
𝟓, 𝟑 = 𝟓 𝟏, 𝟎 + 𝟑(𝟎, 𝟏)
𝒗
′
𝒀
𝑬
= [𝟓, 𝟑]
𝑿′
𝒀
𝒗
𝑷 𝟓, 𝟑
⦁
[2, 3] [𝟒, −𝟏]
𝟎
𝑺 = [𝟒, −𝟏]
𝑿′ along 𝒖𝟏
with unit length 𝒖𝟏
𝒀′ along 𝒖𝟐
with unit length 𝒖𝟐
𝑿
basis 𝑺′ = 𝟏, 𝟎 , (𝟏, 𝟏)
𝒗 𝑺′ = [2, 3]
49
Linear Mappings (Linear Transformations
𝑉
𝑭 𝒗+𝒘 =𝑭 𝒗 +𝑭 𝒘
∀ 𝑣, 𝑤 ∈ 𝑉
𝑭 𝒄𝒗 = 𝒄𝑭 𝒗
∀𝑐 ∈𝐾
𝐾
𝑈
Target space
Domain space
𝑭
𝐾
A linear transformation is said to be operation preserving
( because the same result occurs whether the operations of addition and
scalar multiplication are performed before or after the linear transformation is applied )
𝑭 𝒖+𝒘 =𝑭 𝒖 +𝑭 𝒘
Addition
in 𝑽n
Addition
in 𝑼n
𝑭 𝒄𝒗 = 𝒄𝑭(𝒗)
Scalar
multiplication
in 𝑽n
Scalar
multiplication
in 𝑼n
50
Linear Mappings (Linear Transformations
𝑉
𝐾
𝑈
𝑭 𝒗+𝒘 =𝑭 𝒗 +𝑭 𝒘
∀ 𝑣, 𝑤 ∈ 𝑉
𝑭 𝒄𝒗 = 𝒄𝑭 𝒗
∀𝑐 ∈𝐾
Target space
Domain space
𝑭
𝐾
𝑭 𝟎 =𝟎
𝟎 𝒂𝒏𝒅 𝟎 𝒔𝒂𝒎𝒆 ? ?
𝐹
𝐹
51
Linear Mappings (Linear Transformations
𝑉
𝐾
𝑭 𝒗+𝒘 =𝑭 𝒗 +𝑭 𝒘
∀ 𝑣, 𝑤 ∈ 𝑉
𝑭 𝒄𝒗 = 𝒄𝑭 𝒗
∀𝑐 ∈𝐾
𝑈
Target space
Domain space
𝑭
𝐾
𝑭(𝒂𝒗 + 𝒃𝒘) = 𝑭(𝒂𝒗) + 𝑭(𝒃𝒘) = 𝒂𝑭(𝒗) + 𝒃𝑭(𝒘)
𝐒 = 𝒖𝟏 , 𝒖𝟐 , … , 𝒖𝒏 , 𝒗 ∈ 𝑽, 𝒗 = 𝒂𝟏 𝒖𝟏 + 𝒂𝟐 𝒖𝟐 + ⋯ + 𝒂𝒏 𝒖𝒏
𝑭 𝒗 = 𝑭 𝒂𝟏 𝒖𝟏 + 𝒂𝟐 𝒖𝟐 + ⋯ + 𝒂𝒏 𝒖𝒏
= 𝒂𝟏 𝑭(𝒖𝟏 ) + 𝒂𝟐 𝑭(𝒖𝟐 ) + ⋯ + 𝒂𝒏 𝑭(𝒖𝒏 )
A linear transformation from 𝑽 𝒕𝒐 𝑽 is called a linear operator.
52
Example of Linear Transformations
0
𝑽
0 (𝒗) = 𝟎 ∀𝒗 ∈ 𝑽
Only Constant linear Mapping
𝑼
𝑰 𝒗 = 𝒗 ∀𝒗 ∈ 𝑽
𝑣
𝑣
𝑽
𝑽
53
𝑽 = ℝ 𝒂𝒏𝒅 𝑼 = ℝ 𝑭: ℝ → ℝ
𝒚 = 𝒂𝒙
𝒀
𝒚 = 𝒂𝒙 + 𝒃
𝒐
𝑿
54
𝑽 = ℝ𝟑 𝒂𝒏𝒅 𝑼 = ℝ𝟑 𝑭: ℝ𝟑 → ℝ𝟑
𝑍
⋅
𝑣 = (𝑎, 𝑏, 𝑐)
Projection Mapping
𝑭 𝒙, 𝒚, 𝒛 = (𝒙, 𝒚, 𝟎)
𝑭
⋅
0
𝑌
𝐹 𝑣 = (𝑎, 𝑏, 0)
𝑋
Rotation Mapping
𝑭(𝒙, 𝒚) = (𝒙𝒄𝒐𝒔𝜽 − 𝒚𝒔𝒊𝒏𝜽, 𝒙𝒔𝒊𝒏𝜽 + 𝒚𝒄𝒐𝒔𝜽)
𝜽
55
𝑍
𝑽 = ℝ𝟑 𝒂𝒏𝒅 𝑼 = ℝ𝟑 𝑹 ∶ ℝ𝟑 → ℝ𝟑
𝑹𝒆𝒇𝒍𝒆𝒄𝒕𝒐𝒓
𝑌
𝑹 𝒙, 𝒚, 𝒛 = (𝒙, 𝒚, −𝒛)
𝑋
⦁
𝑻𝒓𝒂𝒏𝒔𝒍𝒂𝒕𝒊𝒐𝒏
𝑹 𝒙, 𝒚 = (𝒙 + 𝟏, 𝒚 + 𝟏)
⦁
56
Derivative Mapping
Let 𝑽 be the vector space of polynomials over ℝ.
𝑫: 𝑽 → 𝑽
𝒅(𝒖+𝒗)
𝒅𝒕
𝒅𝒖
𝒅𝒕
define 𝑫 𝒑 𝒕
𝒅𝒗
𝒅𝒕
𝒅(𝒌𝒕)
𝒅𝒕
𝒅𝒑
=
𝒅𝒕
𝒅𝒖
𝒌
𝒅𝒕
= +
and
=
𝑫 𝒖 + 𝒗 = 𝑫 𝒖 + 𝑫 𝒗 and 𝑫 𝒌𝒗 = 𝒌𝑫(𝒗)
Derivative Mapping is Linear.
Integral Mapping
Let 𝑽 be the vector space of polynomials over ℝ.
𝑱∶𝑽→ℝ
𝑱∶𝑽→𝑽
define 𝐉 𝒇 =
𝟏
𝒇
𝟎
define 𝐉 𝒇 =
𝑱 𝒖+𝒗 =𝑱 𝒖 +𝑱 𝒗
𝒕 𝒅𝒕
𝒇 𝒕 𝒅𝒕
and 𝑱 𝒌𝒗 = 𝒌𝑱(𝒗)
57
𝒏
𝑚
Let 𝑽 = ℝ and 𝑼 = ℝ .
Let 𝑨 𝑏𝑒 𝒎 × 𝒏 𝑚𝑎𝑡𝑟𝑖𝑥
𝑳 ∶ 𝑽 → 𝑼 𝒃𝒚
𝑳 𝒗 = 𝑨𝒗
𝐴 𝑣 + 𝑤 = 𝐴𝑣 + 𝐴𝑤 and 𝐴 𝑐𝑣 = 𝑐𝐴𝑣
for all 𝑣, 𝑤 ∈ ℝ 𝑛 , 𝑐 ∈ ℝ
Any m x n matrix A over a field ℝ
𝒏
𝒎
viewed as a linear map 𝑨 ∶ ℝ → ℝ .
58
Kernel and Image of a Linear Mapping
𝑭
𝟎
𝑽 𝑲𝒆𝒓 𝑭 = 𝒗 ∈ 𝑽 𝑭 𝒗 = 𝟎} 𝑼
𝑰𝒎 𝑭 = 𝒖 ∈ 𝑼
∃ 𝒗 ∈ 𝑽 𝒔𝒖𝒄𝒉 𝒕𝒉𝒂𝒕 𝑭 𝒗 = 𝒖}
𝑲𝒆𝒓 𝑭 is a subspace of 𝑽
𝑰𝒎 𝑭 is a subspace of 𝑼
59
𝑽 = ℝ𝟑 𝒂𝒏𝒅 𝑼 = ℝ𝟑 𝑭: ℝ𝟑 → ℝ𝟑
𝑍
⋅
⋅ ⋅
Projection Mapping
𝑭 𝒙, 𝒚, 𝒛 = (𝒙, 𝒚, 𝟎)
𝑣 = (𝑎, 𝑏, 𝑐)
0
𝑋
𝐹 𝑣 = (𝑎, 𝑏, 0)
𝐾𝑒𝑟 𝐹 =
𝑌
𝐼𝑚 𝐹 =
0, 0, 𝑐
𝑎, 𝑏, 0
= 𝑧 − 𝑎𝑥𝑖𝑠
= 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒
Rotation Mapping
𝑭(𝒙, 𝒚) = (𝒙𝒄𝒐𝒔𝜽 − 𝒚𝒔𝒊𝒏𝜽, 𝒙𝒔𝒊𝒏𝜽 + 𝒚𝒄𝒐𝒔𝜽)
𝜽
𝐾𝑒𝑟 𝐹 = 0
𝐼𝑚 𝐹 = ℝ2 , the entire space
60
Let
𝑽 = 𝑷(𝒕)
𝑯: 𝑽 → 𝑽
be the vector space of polynomials over ℝ.
define 𝑯 𝒇 𝒕
=
𝒅𝟑 𝒇
𝒅𝒕𝟑
𝐾𝑒𝑟 𝐹 = 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙𝑠 𝑜𝑓 𝑑𝑒𝑔𝑟𝑒𝑒 ≤ 2 = 𝑃2 (𝑡)
𝐼𝑚 𝐹 = 𝑉, the entire space
61
Rank and Nullity of a Linear Mapping
𝑭
𝑽
𝒓𝒂𝒏𝒌 𝐹 = dim 𝐼𝑚 𝐹 𝑎𝑛𝑑
𝒏𝒖𝒍𝒍𝒊𝒕𝒚(𝐹) = dim(𝐾𝑒𝑟 𝐹)
𝑼
𝐋𝐞𝐭 𝐕 be of finite dimension, 𝐚𝐧𝐝 𝒍𝒆𝒕 𝑭: 𝑽 → 𝑼 𝒃𝒆 𝒍𝒊𝒏𝒆𝒂𝒓.
𝑻𝒉𝒆𝒏
𝒅𝒊𝒎 𝑽 = 𝒅𝒊𝒎(𝑲𝒆𝒓 𝑭) + 𝒅𝒊𝒎(𝑰𝒎 𝑭)
= 𝒏𝒖𝒍𝒍𝒊𝒕𝒚(𝑭) + 𝒓𝒂𝒏𝒌(𝑭)
62
∃𝒗≠𝟎
𝑭
singular
∃𝒗≠𝟎∈𝑽
such that 𝑭 𝒗 = 𝟎
𝑽
𝑼
𝑭
0
𝒐𝒏𝒍𝒚
𝑭
nonsingular
𝑽
0
𝑲𝒆𝒓 𝑭 = 𝟎
0
𝑼
Rotation Mapping
𝑭
Projection Mapping
Singular and Nonsingular Linear Mappings
63
Nonsingular
linear Mapping
𝑽 finite dimension
𝒅𝒊𝒎(𝑽) = 𝒅𝒊𝒎(𝒊𝒎𝒂𝒈𝒆 𝒐𝒇 𝑭 )
𝒅𝒊 𝒎 𝑽 = 𝒅𝒊 𝒎 𝑲𝒆𝒓 𝑭 + 𝒅𝒊 𝒎 𝑰𝒎 𝑭
= 𝒅𝒊 𝒎 𝟎 + 𝒅𝒊 𝒎 𝑰𝒎 𝑭
= 𝒅𝒊 𝒎 𝑰𝒎 𝑭
independent
𝑭
{𝑭(𝒖𝟏 ), 𝑭(𝒖𝟐 , ) … , 𝑭(𝒖𝒏 )}
{𝒖𝟏 , 𝒖𝟐 , … , 𝒖𝒏 }
independent
𝑽
Let 𝒖𝟏 , 𝒖𝟐 , … , 𝒖𝒏 independent in 𝑽.
𝒂𝟏 𝑭(𝒖𝟏 ) + 𝒂𝟐 𝑭 𝒖𝟐 + ⋯ + 𝒂𝒏 𝑭 𝒖𝒏 = 0
𝑭 𝒂𝟏 𝒖𝟏 + 𝒂𝟐 𝒖𝟐 + ⋯ + 𝒂𝒏 𝒖𝒏 = 𝟎
𝒂𝟏 𝒖𝟏 + 𝒂𝟐 𝒖𝟐 + ⋯ + 𝒂𝒏 𝒖𝒏 = 𝟎
⟹ 𝒂𝟏 = 𝟎, 𝒂𝟐 = 𝟎 … , 𝒂𝒏 = 𝟎
𝑼
64
Vector Space Isomorphism
𝑭
bijective
(one-to-one and onto)
Isomorphic
𝑽≅𝑼
𝑭 isomorphism
𝑽
Isomorphism
“ dimension preserving”
𝑼
Suppose 𝑽 finite dimensional and 𝒅𝒊𝒎 𝑽 = 𝒅𝒊𝒎 𝑼.
Isomorphism ⇔nonsingular.
65
Theorem : Every vector space 𝑽 of dimension
𝒏
n over ℝ is isomorphic to ℝ .
𝟔
𝑴𝟑,𝟐 ≅ ℝ
𝟏𝟎
𝑷𝟗 𝒕 ≅ ℝ
𝑻
𝑻 𝒗 = (𝒂𝟏 , 𝒂𝟐 , … , 𝒂𝒏 )
ℝ
𝒏
∀ 𝒂𝟏 , 𝒂𝟐 , … , 𝒂𝒏
∃ unique vector
𝑽
𝒏
𝒄𝟏 𝒖𝟏 + 𝒄𝟐 𝒖𝟐 + ⋯ + 𝒄𝒏 𝒖𝒏
𝒏
𝑽≅ℝ
66
𝑻(𝒗)
𝑻 𝒗𝟏 = 𝒂𝟏𝟏 𝒖𝟏 + 𝒂𝟏𝟐 𝒖𝟐 + ⋯ + 𝒂𝟏𝒎 𝒖𝒎
𝑻 𝒗𝟐 = 𝒂𝟐𝟏 𝒖𝟏 + 𝒂𝟐𝟐 𝒖𝟐 + ⋯ + 𝒂𝟐𝒎 𝒖𝒎
………………………………………………………………
𝑻 𝒗𝒏 = 𝒂𝒏𝟏 𝒖𝟏 + 𝒂𝒏𝟐 𝒖𝟐 + ⋯ + 𝒂𝒏𝒎 𝒖𝒎
𝒂𝒏𝟏
𝒂𝟏𝟏
𝒂𝟐𝟏
𝒂𝟏𝟐
𝒂𝟐𝟐 ⋯ 𝒂𝒏𝟐
𝑻 𝑺,𝑺′ = 𝑨 =
⋮
⋱
⋮
𝒂𝟏𝒎 𝒂𝟐𝒎
⋯ 𝒂𝒏𝒎
The transpose of the above matrix of coefficients, denoted by 𝑻
matrix representation of 𝑻 relative to the bases 𝑺 and 𝑺′ .
𝑻 𝒗 = 𝑨𝒗
𝒔
=𝑨𝒗
𝑼
𝑺′
𝑻
𝑺′ = {𝒖𝟏 , 𝒖𝟐 , … , 𝒖𝒎 }
𝑽
𝑺 = {𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒏 }
Matrix Representation of a Linear Transformation
𝑺,𝑺′
is called the
67
Let 𝑭: ℝ𝟐 → ℝ𝟐 be linear operator
by 𝑭 𝒙, 𝒚 = (𝟐𝒙 + 𝟑𝒚, 𝟒𝒙 − 𝟓𝒚)
Find the matrix representation of 𝑭 relative to the (usual) basis
{𝒆𝟏 = (𝟏, 𝟎), 𝒆𝟐 = (𝟎, 𝟏)}.
𝑨=
𝟐
𝟒
𝟑
−𝟓
𝑭 𝒗 = 𝑨𝒗
2 3
8
1
1
𝐹
=
=
4 −5 2
−6
2
Find the matrix representation of 𝑭
relative to the basis 𝐒 = {𝒖𝟏 = (𝟏, 𝟐), 𝒖𝟐 = (𝟐, 𝟓)}.
𝟓𝟐 𝟏𝟐𝟗
𝑨=
−𝟐𝟐 −𝟓𝟓
𝐹(𝑣)
𝑠
=𝐴 𝑣
𝑠
Every linear transformation 𝑭: ℝ𝒏 →
ℝ𝒎 is given by matrix multiplication
𝑭 𝒗 = 𝑨𝒗,
68
where 𝑨 is 𝒎 × 𝒏 matrix.
Let 𝑅𝜃 : ℝ2 → ℝ2 be linear operator
that rotates the vector in the plane around the origin by a specified angle 𝜃.
Find the matrix representation of 𝑅𝜃 relative to the (usual) basis
{𝑒1 = (1, 0), 𝑒2 = (0, 1)}.
𝒆𝟐
𝜽
𝑅𝜃 (𝑣) = 𝐴𝜃 𝑣
𝜽
𝒆𝟏
𝑹𝜽
𝟏
𝒄𝒐𝒔𝜽
=
𝟑
𝒔𝒊𝒏𝜽
𝟏
𝟑
−𝒔𝒊𝒏𝜽
𝒄𝒐𝒔𝜽
𝑅𝜃 𝑒1 = 𝑅𝜃 1, 0 = 𝑐𝑜𝑠𝜃, 𝑠𝑖𝑛𝜃 = 𝑒1 𝑐𝑜𝑠𝜃 + 𝑒2 𝑠𝑖𝑛𝜃
𝑅𝜃 𝑒2 = 𝑅𝜃 0, 1 = − 𝑠𝑖𝑛𝜃, 𝑐𝑜𝑠𝜃 = −𝑒1 𝑠𝑖𝑛𝜃 + 𝑒2 𝑐𝑜𝑠𝜃
𝑐𝑜𝑠𝜃
𝐴𝜃 =
𝑠𝑖𝑛𝜃
−𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃
69
Let 𝐹: ℝ3 → ℝ3 be linear operator
Projection Mapping 𝑭 𝒙, 𝒚, 𝒛 = (𝒙, 𝒚, 𝟎)
𝐹 𝑒1 = 𝐹 1, 0, 0 = 1, 0, 0 = 1𝑒1 + 0𝑒2 + 0𝑒3
𝐹 𝑒2 = 𝐹 0, 1,0 = 0, 1,0 = 0𝑒1 + 1𝑒2 + 0𝑒3
𝐹 𝑒3 = 𝐹 0, 0,1 = 0, 0,0 = 0𝑒1 + 0𝑒2 + 0𝑒3
𝟏
𝑨= 𝟎
𝟎
𝟎
𝟏
𝟎
𝟎
𝟎
𝟎
𝑭 𝒗 = 𝑨𝒗
Let 𝑅: ℝ3 → ℝ3 be linear operator
𝑹𝒆𝒇𝒍𝒆𝒄𝒕𝒐𝒓 𝑹 𝒙, 𝒚, 𝒛 = (𝒙, 𝒚, −𝒛)
𝐹 𝑒1 = 𝐹 1, 0, 0 = 1, 0, 0 = 1𝑒1 + 0𝑒2 + 0𝑒3
𝐹 𝑒2 = 𝐹 0, 1,0 = 0, 1,0 = 0𝑒1 + 1𝑒2 + 0𝑒3
𝐹 𝑒3 = 𝐹 0, 0,1 = 0, 0, −1 = 0𝑒1 + 0𝑒2 − 1𝑒3
𝟏
𝑨= 𝟎
𝟎
𝟎
𝟏
𝟎
𝟎
𝟎
−𝟏
𝑹 𝒗 = 𝑨𝒗
70
𝑽
be the space of all polynomial functions from ℝ into ℝ of
the form 𝒇(𝒙) = 𝒄𝟎 + 𝒄𝟏 𝒙 + 𝒄𝟐 𝒙𝟐 +𝒄𝟑 𝒙𝟑
𝒅𝒇(𝒙)
𝑫 ∶ 𝑽 → 𝑽 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒃𝒚
𝑫 𝒇 𝒙 =
𝒅𝒙
𝟐
𝟑
𝕭 = {𝒇𝟏 = 𝟏, 𝒇𝟐 = 𝒙, 𝒇𝟑 = 𝒙 , 𝒇𝟒 = 𝒙 } be a ordered basis of 𝑽 .
Let
1
So that the matrix of D in ordered basis 𝕭 is
𝑫
𝕭
𝟎
= 𝟎
𝟎
𝟎
𝟏
𝟎
𝟎
𝟎
𝟎
𝟐
𝟎
𝟎
𝟎
𝟎
𝟑
𝟎
𝑫 𝒇 𝒙
= 𝑫 𝓑 𝒇(𝒙)
71
It would be wrong to infer from
that all linear transformations
can be represented by matrices
(of finite size).
For example, the differential and integral
operators do not have matrix
representations because they are
defined on infinite-dimensional spaces.
But linear transformations on finitedimensional spaces will always have
matrix representations.
72
Change of Basis
How do our representations change if we select another basis
𝑽
𝑺′ = {𝒗𝟏 , 𝒗𝟐 , … , 𝒗𝒏 }
𝑺 = {𝒖𝟏 , 𝒖𝟐 , … , 𝒖𝒏 }
𝒗𝟏 = 𝒂𝟏𝟏 𝒖𝟏 + 𝒂𝟏𝟐 𝒖𝟐 + ⋯ + 𝒂𝟏𝒏 𝒖𝒏
𝒗𝟐 = 𝒂𝟐𝟏 𝒖𝟏 + 𝒂𝟐𝟐 𝒖𝟐 + ⋯ + 𝒂𝟐𝒏 𝒖𝒏
……………………………………………………
𝒗𝒏 = 𝒂𝒏𝟏 𝒖𝟏 + 𝒂𝒏𝟐 𝒖𝟐 + ⋯ + 𝒂𝒏𝒏 𝒖𝒏 𝑷 =
𝑷 is the change-of-basis matrix
(or transition matrix) from the
"old" basis 𝑺 to the "new"
basis 𝑺′.
?
𝒂𝟏𝟏
𝒂𝟏𝟐
𝒂𝟏𝒏
⋮
𝒂𝟐𝟏
𝒂𝟐𝟐
𝒂𝟐𝒏
𝒂𝒏𝟏
⋯ 𝒂
𝒏𝟐
⋱
⋮
⋯ 𝒂𝒏𝒏
𝑸 is the change-of-basis matrix
(or transition matrix) from the
𝑺′ to 𝑺.
Matrix 𝑷 and 𝑸 are invertible and 𝑸 = 𝑷−𝟏 .
73
Change of Basis
Applications of Change-of-Basis Matrix
How a change of basis affects the coordinates of a
vector in a vector space 𝑉 ?
Let 𝑷 be the change-of-basis matrix from a basis 𝑺
to a basis 𝑺′ in a vector space 𝑽. Then, for any vector
𝒗 ∈ 𝑽, we have :
𝑷𝒗
hence
−𝟏
𝑷
𝑺′
𝒗
𝒔
= 𝒗
𝑺
= 𝒗
𝑺′
and
𝑷−𝟏 transforms the coordinates of 𝒗 in the original basis 𝑺
into the coordinates of 𝒗 in the new basis 𝑺′.
74
Change of Basis
Consider vector space ℝ𝟐
𝑺 = {𝒆𝟏 , 𝒆𝟐 } =
𝟏, 𝟎 , 𝟎, 𝟏
and 𝑺′ = {𝒖𝟏 , 𝒖𝟐 } = {(𝟏, 𝟑), (𝟏, 𝟒)}
𝒖𝟏 = 𝟏, 𝟑 = 𝟏 𝟏, 𝟎 + 𝟑(𝟎, 𝟏)
𝒖𝟐 = 𝟏, 𝟒 = 𝟏 𝟏, 𝟎 + 𝟒(𝟎, 𝟏)
𝟏 𝟏
𝑷=
𝟑 𝟒
𝒗 = (𝟓, 𝟑)
𝒗
𝑺
𝑸=
𝑺′
=
𝟐𝟑
−𝟏𝟖
𝒗
𝟒
=
−𝟑
−𝟏
𝟏
𝒗 = 𝟓, −𝟑 = 𝟐𝟑 𝟏, 𝟑 − 𝟏𝟖(𝟏, 𝟒)
𝟓
=
−𝟑
𝒗
𝑷−𝟏
𝑺′
𝟒
=
−𝟑
−𝟏
𝟏
𝟐𝟑
𝟓
=
−𝟏𝟖
−𝟑
75
How a change of basis affects the matrix
representation of a linear operator ?
Let 𝑷 be the change-of-basis matrix from a basis
𝑺 to a basis 𝑺′ in a vector space 𝑽. Then, for any
linear operator 𝑻 on 𝑽,
−𝟏
𝑻 𝑺′ = 𝑷 𝑻 𝒔 𝑷
That is, if 𝑨 and 𝑩 are the matrix representations
of 𝑻 relative, respectively, to 𝑺 and 𝑺′ then
𝑩 =
−𝟏
𝑷
𝑨𝑷
76
𝑭(𝒙, 𝒚) = (𝟐𝒙 + 𝟑𝒚 , 𝟒𝒙 − 𝟓𝒚)
Find the matrix representation of 𝑭 relative to the bases
𝑺 = {𝒆𝟏 , 𝒆𝟐 } = 𝟏, 𝟎 , 𝟎, 𝟏
and 𝑺′ = {𝒖𝟏 , 𝒖𝟐 } = {(𝟏, 𝟑), (𝟏, 𝟒)}
𝑭 𝒖𝟏 = 𝑭
𝟏
𝟑
𝒙 + 𝒚 = 𝟏𝟏
𝟏
𝟏𝟏
𝟏
=
=𝒙
+𝒚
and
𝟑𝒙 + 𝟒𝒚 = −𝟏𝟏
𝟑
−𝟏𝟏
𝟒
𝟏
𝟒
𝒙 + 𝒚 = 𝟏𝟒
𝟏𝟒
𝟏
𝟏
=
=𝒙
+𝒚
and
𝟑𝒙 + 𝟒𝒚 = −𝟏𝟔
−𝟏𝟔
𝟑
𝟒
Solving the system 𝑥 = 55 , 𝑦 = −44 .
Hence 𝑭 𝒖𝟏 = 𝟓𝟓𝒖𝟏 − 𝟒𝟒𝒖𝟐
𝑭 𝒖𝟐 = 𝑭
Solving the system 𝑥 = 72 , 𝑦 = −58 .
Hence 𝑭 𝒖𝟐 = 𝟕𝟐𝒖𝟏 − 𝟓𝟖𝒖𝟐
𝑭
𝑺′
𝟓𝟓
=
−𝟒𝟒
𝟕𝟐
−𝟓𝟖
𝑭
𝒔
𝟐 𝟑
=
𝟒 −𝟓
77
𝟏
𝑷=
𝟑
𝟏
𝟒
−𝟏
𝑸=𝑷
𝟒 −𝟏
=
−𝟑 𝟏
𝑷−𝟏 𝑭 𝑺 𝑷
𝟒 −𝟏 𝟐 𝟑
𝟏 𝟏
=
−𝟑 𝟏
𝟒 −𝟓 𝟑 𝟒
𝟒 −𝟏
𝟏𝟏
𝟏𝟒
=
−𝟑 𝟏
−𝟏𝟏 −𝟏𝟔
𝟓𝟓 𝟕𝟐
=
= 𝑭
𝟒𝟒 −𝟓𝟖
𝑺′
𝐝𝐞𝐭 𝒐𝒇 𝑭 𝒔 = −𝟏𝟎 − 𝟏𝟐 = −𝟐𝟐
𝑻𝒓𝒂𝒄𝒆 𝒐𝒇 𝑭 𝒔 = 𝟐 − 𝟓 = −𝟑
𝐝𝐞𝐭 𝒐𝒇 𝑭
= −𝟑𝟏𝟗𝟎 + 𝟑𝟏𝟔𝟖 = −𝟐𝟐
𝑻𝒓𝒂𝒄𝒆 𝒐𝒇 𝑭 𝒔 = 𝟓𝟓 − 𝟓𝟖 = −𝟑
78
𝑺′
Similarity
𝑨 𝒂𝒏𝒅 𝑩 square matrices
∃ an invertible matrix 𝑷
𝑩 = 𝑷−𝟏 𝑨𝑷
𝑩≈𝑨
Similarity of matrices is an equivalence relation.
Two matrices represent the same
linear operator if and only if the
matrices are similar.
That is, all the matrix representations of a linear operator
𝑻 form an equivalence class of similar matrices.
79
References
1. Lipschutz S. Schaum's Outline of Theory and
Problems of Linear Algebra 3rd Edition
(Schaum,2004)
2. Peter J. O. Shakiban C. Applied Linear Algebra 1st
Edition
3. Carl D. Meyer Matrix Analysis And Applied Linear
Algebra
80