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Math 121A Linear Algebra (Friday, Week 0)
 5 to 7 quizzes (homework problems)
 2 midterms
 1 final exam
 Chapters 1 to 4
 Solutions, practice exams, and schedule on website.
 Website: http://math.uci.edu/~swise/classes/math_121a.html
 Email: swise(at)math(dot)uci(dot)edu
 Office Hours: Monday 3pm – 4pm, Wednesday 1pm – 2pm.
 TA: Andrew-David Bjork
 Office: RH 420B
 Office Hours: Tu 12pm – 2pm, W 10am – 12pm.
 Email: abjork(at)math(dot)uci(dot)edu
Linear Algebra
 About vector space, denoted V and W, and about linear transformation (operation) on vector spaces.
Suppose V and W are vector spaces.
A transformation (operator) L: V  W is linear if and only if
L(u + v) = L(u) + L(v) u, v  V and “scalars”  and .
Example: Let , g  C[a, b] and ,   .
ba ( + g) dx =  ba  dx +  ba g dx
 A linear transformation
Arrows in Real Two-Space
x = (a1, a2)  2
y = (b1, b2)  2
 Addition of arrows at the origin
 Parallelogram Law (textbook 1.1)
 Addition of arrows at a point P  O
 Apply Parallelogram Law at P.
(a1, a2) + (b1, b2) = (a1 + b1, a2 + b2)
 Scalar Vector Multiplication
t   and x  2
t(a1, a2) = (ta1, ta2)
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Example: A(–2, 0, 1) and B(4, 5, 3)
C is a vector emanating from the origin and having the same direction from A to B.
(4, 5, 3) – (–2, 0, 1) = (6, 5, 2)
 The equation of line through A and B is x = (–2, 0, 1) + t(6, 5, 2).
Example: A(1, 0, 2), B(–3, –2, 4), and C(1, 8, –5).
A vector from A to B: (–3, –2, 4) – (1, 0, 2) = (–4, –2, 2)
A vector from A to C: (1, 8, –5) – (1, 0, 2) = (0, 8, –7)
 The equation of plane containing A, B, and C is x = (1, 0, 2) + s(–4, –2, 2) + t(0, 8, –7)
x = (a1, a2)  2, ai  
y = (b1, b2)  2, bi  
x + y = (a1 + b1, a2 + b2)
tx = (ta1, ta2)
–x = (–1)x = (–a1, –a2)
x – y = x + (–1)y = (a1 – b1, a2 – b2)
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Lecture (Monday, Week 1)
Properties of Arrows
A1. x, y  2 x + y = y + x  commutativity of addition
A2. x, y, z  2 (x + y) + z = x + (y + z)  associativity of addition
A3. x  2 0  2 s. t. x + 0 = x, where 0 = (0, 0)
A4. x  2 y  2 s. t. x + y = 0  If x = (a1, a2), then y = (–a1, –a2).
A5. x  2 1x = x
A6. a, b   x  2 (ab)x = a(bx).
A7. a   x, y  2 a(x + y) = ax + ay.
A8. a, b   x  2 (a + b)x = ax + bx.
Fields
Examples: , Q, and C.
x2 + 1 = 0  x =  i   is not algebraically closed.
Properties of Fields
F1. a + b = b + a and ab = ba  commutativity of addition and multiplication
F2. (a + b) + c = a + (b + c) and (ab)c = a(bc)  associativity of addition and multiplication
F3. 0 + a = a and 1a = a  existence of additive and multiplicative identity
F4. a  F b  0  F c, d  F s. t. a + c = 0 and bd = 1  existence of additive and multiplicative inverse
F5. a(b + c) = ab + ac  distributivity of multiplication over addition
We denote a field by F.
 Define a vector space
Example: 2, 3, 4, …, n, where n  Z+ .
Definition: A vector space V over a field “F” consists of a “set” on which two operations “addition and scalar
multiplication” are defined so that x, y  V ! x + y  V and a  F x  V !ax  V such that
VS1. x, y  V x + y = y + x.
VS2. x, y, z  V (x + y) + z = x + (y + z).
VS3. x  V 0  V s. t. x + 0 = x.
(What does 0 look like?)
VS4. x  V y  V s. t. x + y = 0, where y is called the additive inverse.
(What does y look like?)
VS5. x  V 1x = x, where 1 is the multiplicative identity of F.
VS6. a, b  F x  V (ab)x = a(bx).
VS7. a  F x, y  V a(x + y) = ax + ay.
VS8. a, b  F x  V (a + b)x = ax + bx.
  is replaced by F and 2 by V.
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Elements of F are called scalars.
Elements of V are called vectors.
Definition: Let ai  F for i = 1, 2, …, n, where n  Z+.
Then x = (a1, a2, …, an) is an n-tuple of scalars from F.
Two n-tuples x = (a1, a2, …, an) and y = (b1, b2, …, bn) are equal if and only if ai = bi for i = 1, 2, …, n.
Two zero n-tuple is the element 0 = (0, 0, …, 0).
xi = ai for i = 1, 2, …, n.
yi = bi for i = 1, 2, …, n.
x + y = (a1 + b1, a2 + b2, …, an + bn)
(x + y)i = xi + yi = ai + bi for i = 1, 2, …, n.
Let t  F.
tx = (ta1, ta2, …, tan)
(tx)i = txi = tai for i = 1, 2, …, n.
Example: Fn = {x: x = (a1, a2, …, an), ai  F for i = 1, 2, …, n}
Together with the field F and addition and scalar multiplication as defined is a vector space.
i.e. Fn = V
Proof of VS2:
WTS: (x + y) + z = x + (y + z) x, y, z  V.
xi = ai, yi = bi, zi = ci for i = 1, 2, …, n.
(x + y) + z = ((a1 + b1) + c1, (a2 + b2) + c2, …, (an + bn) + cn)
= (a1 + (b1 + c1), a2 + (a2 + c2), …, a3 + (b3 + c3))  by F2
= x + (y + z)
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Lecture (Wednesday, Week 1)
Homework: Section 1.2 #7, 8, 9, 10, 12, 20, 21
Definition: An m  n matrix, denoted A, over the field F is a rectangular array of the form
a11 a12 … a1n
a21 a22 … a2n
A=
where aij  F, 1  i  m, 1  j  n.
…
…
Am1 am2 … amn
Definition: The diagonal elements are those aij such that i = j.
a11 a12 a13
Example: A =
is a 2  3 matrix with aij  F.
a21 a22 a23
a31 a32 a33
Definition: The set of all m  n matrices over the field F is denoted Mmn(F), n  Z+ = {1, 2, …}.
Notation: Z* = {0, 1, 2, 3, …}
Z– = {-1, -2, -3, …}
The components of A are Aij and Aij = aij.
Example: Aij = i + j
The ith row of A is ri = (ai1, ai2, …, ain) , a “row” vector from Fn.
a1j
The jth column of A is cj =
a2j
, a “column” vector from Fn.
…
Amj
Zero matrix is the element 0, where all aij = 0.
Two matrices are equal when all of their components are equal.
i.e. Aij = Bij  aij = bij
Addition and Scalar Multiplication:
Given A, B  Mmn(F), the components of A + B are (A + B)ij = Aij + Bij = aij + bij.
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a11 a12 a13
Example: A + B = a21 a22 a23
b11 b12 b13
+
a31 a32 a33
b21 b22 b23
=
b31 b32 b33
a11 + b11
a12 + b12
a13 + b13
a21 + b21
a22 + b22
a23 + b23
a31 + b31
a32 + b32
a33 + b33
Given A  Mmn(F) and t  F, the components of tA are (tA)ij = tAij = taij.
Example: V = (Mmn(F), +,  ) is a vector space.
VS6  WTS a, b  F x  V (ab)x = a(bx).
Let xij = aij.
[(ab)x]ij = (ab)xij = (ab)aij
[a(bx)]ij = a(bxij) = a(baij)
By F2, (ab)aij = a(baij) ij
Definition: A polynomial (x) with coefficients from F is an “expression” of the form
(x) = anxn + an-1xn-1 + … + a1x + ao, where ak  F, k = 0, 1, 2, …, n.
Degree of the polynomial (x) is the largest value k such that ak  0, denoted deg() = k.
Addition: Let  be a polynomial of degree m.
Let g be a polynomial of degree n.
(x) = amxm + am-1xm-1 + … + a1x + ao
g(x) = bnxn + bn-1xn-1 + … + b1x + bo
Without loss of generality, if n  m, bm, bm-1, …, bn+1 = 0.
g(x) = bmxm + … + bnxn + … + b1x + bo
(x) + g(x) = (am + bm)xm + … + (an + bn)xn + … + (a1 + b1)x + (ao + bo)
Scalar Multiplication: Let t  F and  be a polynomial of degree m.
Then t(x) = tamxm + tam-1xm-1 + … + ta1x + tao.
The zero polynomial is the expression z(x) = 0.
z(x) = mxm + m-1xm-1 + … + 1x + o , where k = 0 for k = 0, 1, …, m.  deg(z) = –1
Two polynomials are equal if and only if all their coefficients are equal.
[(x) + g(x)]k = (ak + bk)xk
[tf(x)]k = takxk
Example: Pn(F) = {(x): (x) = amxm + am-1xm-1 + … + a1x + ao, ak  F for k = 0, 1, …, m}
We do not require am  0.  deg(f)  n
V(Pn(F), +,  ) is a vector space.
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Theorem: (1.1) If x, y, z  V and x + z = y + z, then x = y.
Proof:
z  V v  V such that z + v = 0 (VS4)
x = x + 0 = x + (z + v) = (x + z) + v = (y + z) + v = y + (z + v) = y + 0 = y
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Lecture (Friday, Week 1)
Homework: Section 1.3 #3, 4, 5, 6, 7, 12, 20, 28
Section 1.3 Subspaces of Vector Spaces
Definition: A subset W of a vector space V over a field F is called a subspace of V if and only if W itself is a
vector space over F with the same addition and multiplication as defined on V.
Remark: We find that if W is a vector subspace, then the properties VS1, 2, 5, 6, 7, 8 are automatically inherited.
i.e. (VS1) x, y  V x + y = y + x
Four things are not automatically inherited:
(VSS1) x, y  W x + y  W (closure under addition).
(VSS2) a  F x  W ax  W (closure under multiplication).
(VSS3) x  W 0’  W such that x + 0’ = x.
(VSS4) x  W y’  W such that x + y’ = 0’.
Theorem: (1.3) Let V be a vector space and W  V.
W is a vector subspace of V  the following three conditions hold:
(a) 0  W, where 0 is the zero vector from V.
(b) x + y  W x, y  W.
(c) cx  W c  F x  W.
Note:
This theorem allows us to check only three properties for subspaces.
Proof:
Suppose V is a vector space and W  V.
() Assume W is a vector subspace of V.
VSS1 – 4 are satisfied, hence (b) and (c) automatically hold.
Since W is a vector space, x  W 0’  W such that x + 0’ = x.
Since x  W, x  V, hence 0  V such that x + 0 = x.
x + 0’ = x + 0  0’ = 0  W by Theorem 1.1.
() Assume (a), (b), and (c).
(b)  VSS1, (c)  VSS2, and (a)  VSS3.
Let x  W. Then (–1)x  W by (c).
(Here, since 1  F, –1  F.)
x + (–1)x  W by (a).
x + (–1)x = (1 + (–1))x = 0x = 0, where 0  F.
Thus we have shown VSS4.
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Definition: The transpose of an m  n matrix A is an n  m matrix At whose components satisfy (At)ij = Aji.
Remark: Only square matrices (m = n) can be symmetric.
Example: A =
1
2

2
1
 At = 

3
2
2
3 
 A = At
Example: Snn = {A  Mmn(F): m = n, A = At}
Snn is a vector subspace of Mnn(F).
Snn  Mnn(F) is clear.
Check A, B  Snn A + B  Snn.
(A + B)ij = Aij + Bij = Aji + Bji = (A + B)ji
A + B = (A + B)t  A + B  Snn.
The rest of proof is similar.
Theorem: (1.4) Any intersection of subspaces of a vector space V is a vector space.
Proof:
Let W1, W2 be vector subspaces of vector space V.
Let W = W1  W2.
Then W  W1  V and W  W2  V.
Use Theorem 1.3  Check (a), (b), and (c).
(a) 0  W1, W2  0  W
(b) x, y  W  x, y  W1, W2  x + y  W1, W2  x + y  W
(c) Let c  F and x  W  x  W1, W2  cx  W1, W2  cx  W
Definition: Let A, B  Mmn(F). Trace (A) = tr(A) = A11 + A22 + … + Ann (all diagonal elements).
Exercise 6: TZnn = {A  Mmn(F): tr(A) = 0}
TZnn is a vector subspace of Mmn(F).
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Lecture (Monday, Week 2)
Homework: Section 1.4 #10, 12, 13, 14
Quiz 1: Thursday in discussion (Covering homework 1.2 – 1.4)
Today: Linear Combination (section 1.4)
Definition: Let V be a vector space over a field F and S  V. A vector v  V is a linear combination of vectors
of S if and only if a finite number of vectors u1, …, un  S and a1, …, an  F such that
v = a1u1 + … + anun = i=1n a1u1.
In this case, we say v is a linear combination of u1, …, un (or {u1, …, un}).
Example: Let u1 = (1, 2), u2 = (1, 3), u3 = (2, 4).
Let a1 = 1, a2 = 0, a3 = –1.
i=13 aiui = (1, 2) – (2, 4) = (–1, –2)
Key Problem: Given v V and a subset S  V, V a vector space, is v a linear combination of vectors from S?
Example: v = (2, 6, 8)
u1 = (1, 2, 1), u2 = (–2, –4, –2), u3 = (0, 2, 3), u4 = (2, 0, –3), u5 = (–3, 8, 16).
Is v a linear combination of u1, …, u5?
In other word, is v = a1u1 + a2u2 + a3u3 + a4u4 + a5u5?
2 = a1 – 2a2 + 0a3 + 2a4 – 3a5
6 = 2a1 – 4a2 + 2a3 + 0a4 + 8a5
8 = a1 – 2a2 + 3a3 – 3a4 + 16a5
…  Row-Reduced Echelon Form
a1 – 2a2 + 5a5 = –9
a3 + 3a5 = 7
a4 – 2a5 = 3
Solution set: a1 = 2a2 – a5 – 4
a3 = –3a5 + 7
a4 = 2a5 + 3
(a2, a5) a solution  v is a linear combination of (u1, …, u5).
Solution: Let a2 = a5 = 0  a1 = –4,
a3 = 7, a4 = 3  v = i=15 aiui
v is also a linear combination of u1, u3, u4.
(2, 6, 8) = –4(1, 2, 1) + 7(0, 2, 3) + 3(2, 0, –3)
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Example: 3x3 – 2x2 + 7x + 8 is not a linear combination of 1(x) = x3 – 2x2 – 5x – 3, 2(x) = 3x3 – 5x2 – 4x – 9.
We want to show that there is no a1, a2 such that (x) = a11(x) + a22(x).
Suppose it is true that a1, a2   such that (x) = a11(x) + a22(x).
3x3 – 2x2 + 7x + 8 = (a1 + 3a2)x3 + (–2a1 – 5a2)x2 + (–5a1 – 4a2)x + (–3a1 – 9a2)
3 = a1 + 3a2
–2 = –2a1 – 5a2
7 = –5a1 – 4a2
8 = –3a1 – 9a2
Row-Reduced Echelon Form:
a1 + 3a2 = 3
a2 = 4
11a2 = 22
0 = 17  contradiction!
Definition: Let S  V, V a vector space, S nonempty. Then the span of S denoted span(S) is the set of all linear
combination of the vectors in S.
Define span() = {0}.
Theorem: (1.5) The span of any subset S of a vector space V is a vector subspace of V.
Moreover, any vector subspace T that contains S must also contain span(S).
Proof:
Use Theorem 1.3.
a)
0  span(S)
b) x + y  span(S) x, y  span(S)
c)
cx  span(S) x  span(S) c  F
Trivially, 0  span(S).
Let x, y  span(S) and c  F.
Suppose S = {u1, …, un}.
a1, …, an such that x = i=1n aiui.
b1, …., bn such that y = i=1n biui.
x + y = i=1n (ai + bi)ui, where ai + bi  F  x + y span(S)
cx = i=1n (cai)ui, where cai  F  cx  span(S)
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Discussion (Tuesday, Week 2)
Tutoring Center: RH 414, W 10am – 12pm
Even Functions
 Closure under addition
Let f, g be even functions.
Let x  .
( + g)(–x) = (–x) + g(–x)
= (x) + g(x)
 because , g even
= ( + g)(x)
Hence  + g is a even function.
 Closure under scalar multiplication
Let  be even,   , and x  .
()(–x) = (–x)
= (x)
= ()(x)
Hence  is even.
 because  even
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Section 1.3 #28
If Mt = –M, then we call M  Mnn(F) skew-symmetric.
Let W1 = {M: Mt = –M}  Mnn(F).
WTS W1 is a subspace.
a)
0 = (0)
0t = (0) = –(0)
Hence 0  W1.
b) Let A, B  W1.
(A + B)t = At + Bt
 by problem #3
= –A + (–B)
= –(A + B)
Hence A + B  W1.
c)
Let A  W1 and   F.
(A)t = At
 by problem #3
= (–A)
= –(A)
Hence A  W1.
Let W2 = {M  Mnn(F): Mt = M}.
If F has characteristic > 2, then Mnn(F) = W1  W2.
Let Eij be n  n matrix whose entries are all zero except for having value 1 in the i th row and jth column.
B = {Eij: 1  i, j  n} is the standard basis for Mnn(F).
A  W1, B  W2
Aij = 1/2, Aji = –1/2, and all zero except for the other entries.
Bij = 1/2, Bji = 1/2, and all zero except for the other entries.
Eij = Aij + Bij
Let M  Mnn(F).
Since B is a basis, there are ij  F such that M = i,jn ijEij = i,jn ij(Aij + Bij) = i,jn ijAij + i,jn ijBij,
where i,jn ijAij  W1 and i,jn ijBij  W2.
Hence M  W1  W2.
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Lecture (Wednesday, Week 2)
Homework: Section 1.5 #5, 9, 14, 16
Example: In 3, Let S = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}.
Let (a, b, c)  3.
Then (a, b, c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1).
Definition: A subset S  V, V a vector space, is called linearly dependent if and only if a finite number of
vectors u1, …, un  S and a finite number of scalars a1, …, an  F (not all zero) such that
a1u1 + … + anun = 0, a non-trivial representation of the zero vector.
Remark: Trivially, if ai= 0 i = 1, …, n, then i=1n aiui = 0.
Example: Consider S = {(1, 3, -4, 2), (2, 2, -4, 0), (1, -3, 2, 4), (-1, 0, 1, 0)}.
i=14 aiui = a1(1, 3, -4, 2) + a2(2, 2, -4, 0) + a3(1, -3, 2, 4) + a4(-1, 0, 1, 0) = (0, 0, 0, 0)
a1 + 2a2 + a3 – a4 = 0
3a1 + 2a2 – 3a3 + 0a4 = 0
– 4a1 – 4a2 + 2a3 + a4 = 0
0a1 + 2a2 + 4a3 + 0a4 = 0
 a1 = 4, a2 = -3, a3 = 2, a4 = 0
 S is linearly dependent.
Definition: A subset S  V is called linearly independent if and only if S is not linearly dependent.
finite collection of vectors u1, …, un  S and a1, …, an  F,
i=1n aiui  0 unless (of course) ai = 0 i.
Example: S = {(1, 0, 0, -1), (0, 1, 0, -1), (0, 0, 1, -1), (0, 0, 0, 1)}
i=14 aiui = a1(1, 0, 0, -1) + a2(0, 1, 0, -1) + a3(0, 0, 1, -1) + a4(0, 0, 0, 1) = (0, 0, 0, 0)
a1 = 0
a2 = 0
a3 = 0
– a1 – a2 – a3 + a 4 = 0  a4 = 0
 S is linearly independent.
Example: S = {(1, 0, 0), (0, 1, 0), (-2, -1, 0)}
i=13 aiui = a1(1, 0, 0) + a2(0, 1, 0) + a3(-2, -1, 0) = (0, 0, 0)
a1 – 2a3 = 0
 a1 = 2a3
a2 + a 3 = 0
 a2 = -a3
Clearly, a1 = 2, a2 = -1, a3 = 1 is a solution (actually,  infinitely many solutions).
 S is linearly dependent.
15
Theorem: (1.6) Let V be a vector space and S1  S2  V.
If S1 is linearly dependent, then S2 is also linearly dependent.
Proof:
(Exercise 12)
If S1 is linearly dependent, u1, …, un  S1 and a1, …, an  F such that i=1n aiui = 0 and not all ai = 0.
Since S1  S2, u1, …, un  S2 and a1, …, an  F.
Thus u1, …, un S2 and a1, …, an  F such that i=1n aiui = 0, where not all ai = 0.
 S2 is linearly dependent.
Corollary: S1  S2  V. If S2 is linearly independent, then S1 is also linearly independent.
Proof:
(Exercise 12)
Proof by contra-positive: (A  B)  (B’  A’)
(S1 is linearly dependent)  (S2 is linearly dependent)
Not(S2 is linearly dependent)  Not(S1 is linearly dependent)
(S2 is linearly independent)  (S1 is linearly independent)
Theorem: (1.7) Let S be a linearly independent subset of a vector space V and let v  V such that v  S.
Then S  {v} is linearly dependent if and only if v  span(S).
Proof:
() If v  span(S), then v is a linear combination of vectors from S.
i.e. u1, …, un  S and a1, …, an  F such that v = a1u1 + … + anun.
a1u1 + … + anun + (-1)v = 0.
Set an+1 = -1 and un+1 = v.
i=1n+1 aiui = 0, where not all ai = 0 because an+1  0.
() If S  {v} is linearly dependent, then u1, …, un  S and a1, …, an, an+1  F such that
a1u1 + … + anun + an+1v = 0, where not all ai = 0.
a1u1 + … + anun = (-1)an+1v
v = (–a1/an+1)u1 + … + (–an/an+1)un, where (–ai/an+1)  F i.
v is a linear combination of vectors u1, …, un from S.
 v  span(S)
16
Lecture (Friday, Week 2)
Homework: Read the 1st half of section 1.6.
Section 1.6 Bases and Dimension
Definition: A basis  for a vector space V is a linearly independent subset of V that generates V (span() = V).
If  is a basis for V, we also say that the vectors of  form a basis for V.
Example: span() = {0} and  is linearly independent by definition.
Thus  is a basis for {0}, the zero vector space.
Example: In Fn, let e1 = (1, 0, 0, …, 0), e2 = (0, 1, 0, …, 0), …, en = (0, 0, …, 0, 1).
S = {e1, e2, …, en} is linearly independent  span(S) = Fn
Therefore S is a basis for Fn; called the standard (or canonical) basis.
Example: In Pn(F), the set {1, x, x2, …, xn} is a basis for Pn(F).
(x) = anxn + … + a1x + ao
 = {1, x, x2, …, xn} is called the standard basis for Pn(F).
Example: P(F) is the set of all polynomials of any degree.
The set {1, x, x2, x3, …} is a basis for P(F).
Theorem: (1.8) Let V be a vector space and  = {u1, …, un} be a subset of V.
Then  is a basis  v  V ! linear combination of vectors from .
i.e. ! linear scalars a1, …, an  F such that v = i=1n aiui.
Proof:
() Let  be a basis for V  span() = V
Let v  V. Since span() = V, a1, …, an  F such that v = i=1n aiui.
Suppose another set of scalars b1, …, bn  F such that v = i=1n biui.
v – v = i=1n (ai – bi)ui  ai – bi = 0  ai = bi
() Exercise 19
Let v  V and suppose !ai, …, an  F such that v = i=1n aiui.
We know that span() = V.
Uniqueness implies that if v = i=1n biui, then ai = bi for i = 1, …, n.
In particular, 0  V.
Thus ! linear combination i=1n ciui = 0.
If di = 0 for i = 1, …, n, then we trivially have 0 = i=1n diui.
 ci = di for i = 1, …, n.
  is linearly independent.
17
Theorem: (1.9) If a vector space V is generated by a finite set S (span(S) = V),
then some subset of S is a basis for V.
Hence, V has a finite basis.
Proof:
S =  and S = {0} are trivial.
Algorithm: S contains a nonzero vector u1.
Set 1 = {u1}. Then 1 is linearly independent.
Pick v2  0 such that 2 = {v1, v2} and 2 is linearly independent.
Continue until k = {v1, v2, …, vk}.
Stop when vk+1 = 0 or vk+1  0 but Bk  {vk+1} is linearly dependent.
Claim:
 = k = {u1, …, uk} is a basis for V.
Proof:
 is linearly independent.
It remains to check if span() = V.
We know   S  V
span (S) = V.
If  = S, then span() = span(S) = V.
Now, suppose   S.
Let v  V. Then a1, …, ak, b1, …, bm  F, m > 0 such that v = i=1k aiui + j=1m bjvj,
where vj are the vectors in S\B.
B  {vj} is linearly dependent j = 1, …, m.
 a1, …, ak+1  F such that i=1k aiui + ak+1vj = 0, not all ai = 0.
In particular, ak+1  0.
i=1k aiui = – ak+1vj  i=1k (–ai/ak+1)ui = vj
Let –ai/ak+1 = ai(j). Then vj = i=1k ai(j)ui.
For vector v  V, v = i=1k aiui + j=1m bj(i=1k ai(j)ui), where ui  .
v  V can be written as a linear combination of vectors from .
 span() = V.
18
Lecture (Monday, Week 3)
Homework: Section 1.6 #5, 6, 9, 14, 20, 21, 28
Disregard Lagrange Interpretation Polynomial.
Example: S = {(1, 0, 0), (0, 1, 0), (2, 0, 0), (0, 0, -3)}
span(S) = V = F3
 = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} = {e1, e2, e3}
 is a canonical basis for F3.
’ = {(2, 0, 0), (0, 1, 0), (0, 0, -3)} = {2e1, e2, -3e3}.
A basis is {a1e1, a2e2, a3e3} for any nonzero scalars ai  F.
Theorem: (1.10) Let V be a vector space that is generated by G  V, where #G = n.
And let L  V be linearly independent, where #L = m.
Then m  n and H  G such that #H = n – m and L  H generates V.
Remark: We cannot find a linearly independent set of # larger than n.
Example: S = {(1, 0, 0), (0, 1, 0), (0, 0, 1), (0, -3, 0)}
span(S) = V = F3
S is not linearly independent.
 = {(1, 0, 0), (0, -3, 0), (0, 0, 1)} is a linearly independent subset of V.
Also, ’ = {(1, 0, 0), (0, -3, 0)} is a linearly independent subset of V.
” = {(2, 0, 0), (0, -1, 0) is a linearly independent set of V.
 L is not a subset of G necessarily.
 Replacement Theorem
Corollary 1: Let V be a vector space having a finite basis  (# < ).
If  is another basis for V, then # = #.
Proof:
Let n = #, m = #.
If m > n, we can select a subset S   such that #S = n + 1.
S is linearly independent because  is a basis (hence linearly independent).
Since  generates V and S  V, by the Replacement Theorem, n + 1  n.
Thus m  n.
If we assume m < n, then it leads to n  m.
Therefore m = n.
Definition: A vector space V is called finite-dimensional if and only if it has a basis  such that # < .
The unique number # = n is called the dimension of V, denoted dim(V) = n.
Example: The vector space Pn(F) has dimension n + 1   = {1, x, x2, …, xn}
19
Example: The vector space P(F) is infinite-dimensional   = {1, x, x2, …}
Corollary 2: Let V be a vector space and dim(V) = n.
a)
If G  V, #G < , and span(G) = V, then #G  n. In particular, if #G = n, then G is a basis.
b) If L  V is linearly independent and #L = n, then L is a basis.
c)
Proof:
If L  V s linearly independent and #L < n, then L may be extended to a basis for V.
Let V be a vector space and dim(V) = n.
a)
Suppose #G = m <  and span(G) = V.
By theorem 1.9, H  G such that H is a basis.
By Corollary 1, #H = n.
Thus #G  n.
If #G = n, then H = G.
Thus G is a basis.
Theorem: (1.11) Let W be a subspace of a finite-dimensional vector space V.
Then W is finite-dimensional and dim(W)  dim(V).
If dim(W) = dim(V), then W = V.
Corollary: If W is a subspace of a vector space V, V finite-dimensional,
then any basis for W can be extended to a basis for V.
Example: W = {v = {v1, v2, v3}  F3: v3 = 0}  F3 (like F2)
S = {(1, 0, 0), (0, 1, 0)} is a basis for W.
 = S  {(0, 0, 1)} is a basis for F3.
20
Discussion (Tuesday, Week 3)
Section 1.5 #16
S  V is linearly independent  every finite subset of S is linearly independent.
() Assume S is linearly independent.
Let S’ be a finite subset of S.
Let u1, …, un  S’ and a1, …, an  F such that a1u1 + … + anun = 0.
Since S’  S, u1, …, un  S.
Since S is linearly independent, a1 = … = an = 0.
() Assume every finite subset of S is linearly independent.
Let v1, …, vm  S and a1, …, am  F such that a1v1 + … + amvm = 0.
But S’ = {v1, …, vm}  S is finite and linear independent.
Hence, a1 = … = am = 0.
21
Section 1.6 #21
V is an infinite dimensional vector space  V contains an infinite linearly independent subset.
() Proof by contra-positive
If V is finite dimensional vector space, then V does not contain an infinite linearly independent subset.
Assume dim(V) = n < .
Let S  V be linearly independent. Then #S  n by Replacement Theorem.
Hence S is finite.
() Construct a family of subsets of V in the following way.
Construct S1: Since V is infinite dimensional v1  V such that v1  0
 Let S1 = {v1}.
Inductive step: Suppose we have constructed Sn = {v1, …, vn} linearly independent.
Since span(Sn)  V (because dim(V) > n), vn+1  V such that
{v1, …, vn+1} is linearly independent.
 Let Sn+1 = {v1, …, vn+1}
Now, we have constructed {Sn: n  }.
Set S = n Sn.
Claim 1: S  V.
Proof 1: Let v  S = n Sn.
no such that v  Sn_o  V.
Hence v  V.
Claim 2: #S = 
Proof 2: Assume not; S is finite.  N   such that sm = SN m  N.
Consider SN+1. By the construction, {v1, …, vn+1} is linearly independent.
Hence vn+1  vi for i  n.  SN+1  SN.  Contradiction!
Claim 3: S is linearly independent.
Proof 3: S is linearly independent  every finite subset is linearly independent.
Let S’  S, where #S’ < .
x  S’ n(x) such that x = vn(x).
Let M = maxxS’{n(x)}.
Hence S’  SM is linearly independent by the construction.
S’ is linearly independent.
22
Lecture (Wednesday, Week 3)
Section 2.1 Linear Transformation
Definition: Let V, W be two vector spaces over F. We call T: V  W a linear transformation (linear operator)
if and only if x, y  V and c  F,
a)
T(x + y) = T(x) + T(y)
b) T(cx) = cT(x)
Properties: (Exercise 7)
a)
If T is linear, then T(0V) = 0W.
T(0V) = T(x – x) = T(x) + T(–x) = T(x) + (–1)T(x) = T(x) – T(x) = 0W
b) T is linear if and only if T(cx + y) = cT(x) + T(y) x, y  V c  F.
c)
If T is linear, then T(x – y) = T(x) – T(y) x, y  V.
d) T is linear if and only if x1, …, xn  V a1, …, an  F T(i=1n aixi) = i=1n aiT(xi).
Example: T: 2  2
T((a1, a2)) = (a1, –a2)  Reflection Transformation.
Let x = (a1, a2), y = (b1, b2).
T(cx + y) = T(c(a1, a2) + (b1, b2))
= T(ca1 + b1, ca2 + b2)
= (ca1 + b1, –(ca2 + b2))
= c(a1 – a2) + (b1 – b2)
= cT(x) + T(y)
 This shows that T is a linear transformation.
Example: Let V = C() = {(x)  : (x) is continuous x  }.
Consider T() = ba (t) dt.
Let (x), g(x)  C() and c  F.
T(c + g) = ba [c(t) + g(t)] dt
= ba c(t) dt + ba g(t) dt
= cba (t) dt + ba g(t) dt
= cT() + T(g)
 This shows that T is a linear transformation.
Definition: IV: V  V defined as IV(v) = v v  V is called the identity transformation.
T0: V  W defined as T0(v) = 0W v  V is called the zero transformation.
Both of these are linear transformations.
Definition: Let V, W be vector spaces and let T: V  W be linear. The null space (or kernel) is defined as
N(T) = {x  V: T(x) = 0W}.
The range (or image) is R(T) = {y  W: x  V such that T(x) = y} = {T(x): x  V}.
23
Theorem: (2.1) Let V, W be vector spaces and T: V  W be linear.
Then N(T) is vector subspace of V and R(T) is a vector subspace of W.
Proof:
Let 0V = 0  V, 0W = 0  W.
To show G  V is a vector subspace, need to show (Theorem 1.5)
a)
0G
b) x + y  G x, y  G
c)
cx  G x  G c  F
a)
T(0V) = 0W
Hence 0V  N(T) and 0W  R(T).
b) Let x, y  N(T).
T(x) = T(y) = 0W
T(x + y) = T(x) + T(y) = 0W + 0W = 0W
Hence x + y  N(T).
Let x, y  R(T).
Then x’, y’  V such that T(x’) = x and T(y’) = y.
Set z = x’ + y’. Then z  V.
T(z) = T(x’ + y’) = T(x’) + T(y’) = x + y
z such that T(z) = x + y.
Hence x + y  R(T).
c)
Let c  F and x  N(T).
Then T(x) = 0W
T(cx) = cT(x) = c0W = 0W
Hence cx  N(T).
Let c  F and y  R(T).
x  V such that T(x) = y.
T(cx) = cT(x) = cy.
Set cx = z.
z  V such that T(z) = T(cx) = cy.
Hence cy R(T).
24
Lecture (Friday, Week 3)
Homework: Section 2.1 #7, 8, 13
Midterm: 5th week, Friday
Theorem: (2.2) Let V, W be vector spaces and T: V  W be linear.
If  = {v1, …, vn} is a basis for V, then R(T) = span(T()), where T() = {T(v1), …, T(vn)}.
In other words, the set of image of  is the range.
Example: Consider T: P2()  M22().
T((x)) =
 f (1)  f (2)
0

0 
is linear.
f (0)
 = {1, x, x2} is basis for P2().
R(T) = span(T()) = span({T(1), T(x), T(x2)})
  0
 0

= span  
0
0
= a1 
0   1
,
1  0
0
 1
+ a2 

1
0
  0
 0

= span  
0   1
,
1  0
0   3
,
0 0
0  

0  
0
 3
+ a3 

0
0
0
0
0  

0  
  0
Since these two matrices are linearly independent,  

  0
0   1
,
1  0
0  
  is a basis of R(T).
0  
 dim(R(T)) = 2
Definition: Let V and W be vector spaces, T: V  W be linear.
Then nullity(T) = dim(N(T)) and rank(T) = dim(R(T)).
Remember N(T) is a vector subspace of V and R(T) is a vector subspace of W.
Back to the previous example: dim(R(T)) = rank(T) = 2
What is dim(N(T))?
N(T) = {(x)  P2(): T((x)) = 0}
= {(x)  P2(): (1) – (2) = 0 and (0) = 0}
(0) = ao + a1(0) + a2(0)2  (0) = 0  ao = 0
(1) = ao + a1(1) + a2(1)2 = a1 + a2
(2) = ao + a1(2) + a2(2)2 = 2a1 + 4a2
25
 (1) – (2) = 0  –a1 – 3a2 = 0  a1 = –3a2  a2 = (–1/3)a1
(x) = ao + a1x + a2x2 = a1x – (1/3)a1x2
 = {x – (1/3)x2}  dim(N(T)) = 1
nullity(T) = 1, rank(T) = 2, dim(V) = 3  nullity(T) + rank(T) = dim(V)
Theorem: (2.3) “Dimension Theorem”
Let V and W be vector spaces and T: V  W be linear.
If V is finite-dimensional, then nullity(T) + rank(T) = dim(V).
Proof: Suppose dim(V) = n, dim(N(T)) = k  n.
Remember that N(T) is a vector subspace of V.
Let N = {v1, …, vk} be a basis for N(T).
By the corollary to theorem 1.11 (page 51), N may be extended to  = {v1, …, vk, …, vn} for V.
Claim: S = {T(vk+1), …, T(vn)} is a basis for R(T).
WTS 1) span(S) = R(T).
WTS 2) S is linearly independent.
1) Use theorem 2.2
R(T) = span(T())
= span({T(v1), …, T(vk), …, T(vn)})
= span({0, …, 0, T(vk+1), …, T(vn)}) since v1, …, vk  N(T)
= span({T(vk+1), …, T(vn)})
= span(S)
2) Suppose i=k+1n biT(vi) = 0 for bk+1, …, bn  F.
Since T is linear, T(i=k+1n bivi) = 0.
i=k+1n bivi  N(T).
! C1, …, ck  F such that i=k+1n bivi = i=1k civi.
i=k+1n (–ci)vi + i=k+1n bivi = 0, where v1, …, vk, vk+1, …, vn  , a basis for V.
Since  = {v1, …, vk, vk+1, …, vn} is linearly independent, ai = 0 and bi = 0.
Hence, S = {T(vk+1), …, T(vn)} is linearly independent.
#S = n – k  dim(R(T)) = rank(T) = n – k
nullity(T) = k, rank(T) = n – k, and dim(V) = n.
Hence, nullity(T) + rank(T) = k + (n – k) = n = dim(V).
26
Lecture (Monday, Week 4)
Homework: Section 2.1 #14, 16, 17, 18, 20
Midterm: Section 1.1 – 2.3
Definition: Let T: V  W be a mapping from a set V to a set W. Then T is called one-to-one
 y  R(T) ! x  V such that T(x) = y.
 T(x) = T(y)  x = y
 x  y  T(x)  T(y)
Definition: T is onto  R(T) = W.
Theorem: (2.4) Let V and W be vector spaces and T: V  W be linear.
Then T is one-to-one  N(T) = {0V}
Proof: () Assume T is one-to-one.
Let x  N(T). Then T(x) = 0W.
Now, it is always true that T(0V) = 0W.
T(x) = T(0V)  x = 0v
() Assume N(T) = {0V}.
Let x, y  V such that T(x) = T(y).
0W = T(x) – T(y) = T(x – y) = T(0V)  x – y = 0V
Thus, x = y.
Hence T is one-to-one.
Theorem: (2.5) Let V and W be vector spaces of equal finite dimension. Let T: V  W be linear.
Then the followings are equivalent.
(a) T is one-to-one.
(b) T is onto.
(c) rank(T) = dim(V).
(d)
Proof: From the Dimension Theorem, nullity(T) + rank(T) = dim(V) = dim(W).
(a)  (b) Suppose T is one-to-one.
By theorem 2.4, N(T) = {0V}.
nullity(T) = dim(N(T)) = dim({0V}) = 0
rank(T) = dim(R(T)) = dim(V) = dim(W)
Since R(T) is a vector subspace of W and dim(R(T)) = dim(W), R(T) = W.
Hence, T is onto.
(b)  (c) Suppose T is onto. Then R(T) = W.
rank(T) = dim(W) = dim(V)
27
(c)  (a) Suppose rank(T) = dim(V) = dim(W).
Then nullity(T) = 0  N(T) = {0V}
By theorem 2.4, T is one-to-one.
Example: T(a1, a2) = (a1 + a2, a2)
T: F2  F2 is linear.
Claim: N(T) = {0} = {(0, 0)}
{0}  N(T) is trivial.
WTS N(T)  {0}.
Let x  N(T). Then x = (a1, a2).
T(x) = T(a1, a2) = (a1 + a2, a2) = (0, 0)
a1 + a2 = 0 and a2 = 0
 a1 = a2 = 0  x = (a1, a2) = (0, 0) = 0  F2  x  {0}.
Theorem: (2.6) Let V and W be vector spaces over F, and suppose  = {v1, …, vn} is a basis of V.
For (arbitrary) vectors w1, …, wn  W, ! linear transformation T: V  W s.t. T(vi) = wi, i = 1, …, n.
Idea of Proof: Let x  V. ! a1, …, an  F such that x = i=1n aivi.
Define T: V  W by T(x) = i=1n aiwi.
We need to show 3 things.
(1) T is linear.
(2) T(vi) = wi
If x = vi, then aj = 0, j  i, ai = 1.
T(vi) = T(x) = j=1n ajwj = wj
(3) Uniqueness of T
Corollary: Let V and W be vector spaces, and suppose V has a finite basis  = {v1, …, vn}.
If T: V  W and U: V W are linear and T(vi) = U(vi), for i = 1, …, n, then T = U.
Proof: Let v  V, and suppose T(v) = w  W.
Since  = {v1, …, vn} is a basis for V, ! a1, …, an F such that v = i=1n aivi.
T(v) = T(i=1n aivi) = i=1n aiT(vi) = i=1n aiU(vi) = U(i=1n aivi) = U(v)
Hence, T = U.
28
Discussion (Tuesday, Week 4)
T: V  W over F
N(T) = {v  V: T(v) = 0}
R(T) = {w  W: v  V such that T(v) = w}
T is one-to-one  N(T) = {0}
T is onto  R(T) = W
Section 2.1 Problem #14: Let V and W be vector spaces and T: V  W be linear.
a)
Prove that T is one-to-one if and only if T carries linearly independent subsets of V onto linearly
independent subsets of W.
i.e. T is one-to-one  If A  V is linearly independent,
then T(A) = {w  W: v  A such that T(v) = w}  W is linearly independent.
() Assume T is one-to-one.
Let A  V be linearly independent.
Let w1, …, wn  T(A) and a1, …, an  F such that i=1n aiwi = 0.
WTS ai = 0 i = 1, …, n.
Since i = 1, …, n vi  A such that T(vi) = wi, i=1n aiwi = i=1n aiT(vi).
i=1n aiwi = i=1n aiT(wi) = T(i=1n aivi) = 0
Since T is one-to-one, N(T) = {0}.
T(i=1n aivi) = T(0) = 0  i=1n aivi = 0  ai = 0 i= 1, …, n because A is linearly independent.
() T is one-to-one  N(T) = {0}  (T(v) = 0  v = 0)
Suppose T(v) = 0.
Let  be basis for V.
v1, …, vm   and a1, …, am  F such that v = i=1m aivi.
T(v) = T(i=1m aivi) = i=1m aiT(vi) = 0, where T(vi)  T().
Since  is linearly independent, T() is linearly independent.
Hence, ai = 0 i = 1, …, m.
v = i=1m aivi = i=1m 0vi = 0
b) Suppose that T is one-to-one and that S is a subset of V. Prove that S is linearly independent if and only
if T(S) is linearly independent.
() Always true by part a) ().
() Assume T(S) is linearly independent.
Let s1, …, sn  S and a1, …, an  F such that i=1n aisi = 0.
T(i=1n aisi) = T(0) = 0
Since T is linear, i=1n aiT(si) = 0
Since T(si) is linearly independent, ai = 0 i = 1, …, n.
29
c)
Suppose  = {v1, …, vn} is a basis for V and T is one-to-one and onto. Prove that T() = {T(v1), …,
T(vn)} is a basis for W.
T() is linearly independent by part b) ().
It remains to show that T() spans W.
Let w  W.
Since T is onto, v  V such that T(v) = w.
a1, …, an  F such that v = i=1n aivi.
w = T(v) = T(i=1n aivi) = i=1n aiT(vi), where T(vi)  T() i = 1, …, n.
Hence, T() spans W.
Section 2.1 Problem #18: Give an example of linear transformation T: 2  2 such that N(T) = R(T).
Note that T(N(T)) = 0.
T(N(T)) = T(R(T)) = 0  T2 = 0
Consider a 2  2 upper triangular matrix, where aij = 0 for i = j.
0
0

a  0

0  0
a  0

0  0
0
0
0
 N(T)  R(T) and R(T)  N(T)  N(T) = R(T)
30
Lecture (Wednesday, Week 4)
Homework: Section 2.2 #4, 6, 8, 9, 13
Definition: An ordered basis for V is a basis for V endowed with a specific ordering.
Example: Consider F3.
 = {e1, e2, e3} where e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1).
 = {e2, e3, e1}   as an ordered basis.
 = {ei}i=1n is an ordered basis for Fn.
We call {ei}i=1n the standard ordered basis for Fn.
Example: {xi}i=1n is the standard ordered basis for Pn(F).
Definition: Let  = {u1, …, un} = {ui}i=1n be an ordered basis for V.
x  V let a1, …, an  F be the unique scalars such that x = i=1n aiui.
 a1 
 
The coordinate vector of x relative to  is [x] =  ...  .
 
 an 
Notice [ui] = ei.
The mapping T: V  Fn defined by T(x) = [x] is linear.
Definition: Let V and W be vector spaces with ordered basis  = {vi}i=1n and  = {wj}j=1m, respectively.
Let T: V  W be linear. ! aij  F, 1  i  m such that T(vj) = i=1m aijwi, 1  j  n.
Why? Where?
T(v) = i=1m aiwi  for some j, T(vj) = i=1m aijwi.
The mn matrix A  Mmn(F) defined via Aij = aij is called matrix representation of T in the ordered basis , .
We write A = [T]. If V = W and  = , then we write A = [T].
Example: T: 2  3
T((a1, a2)) = (a1 + 3a2, 0, 2a1 – 4a2)
 = {(1, 0), (0, 1)} is a basis for 2.
 = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is a basis for 3.
Now, T(e1) = T(1, 0) = (1, 0, 2) = 1e1 + 0e2 + 2e3  a11 = 1, a21 = 0, a31 = 2.
T(e2) = T(0, 1) = (3, 0, –4) = 3e1 + 0e2 + (–4)e3  a12 = 3, a22 = 0, a32 = –4.
1

 [T] = 0

2
3 
0 
 4
31
Definition: Let T: V  W and U: V  W be transformation from a vector space V to a vector space W.
Let a  F. Then we define
a)
T + U: V  W by (T + U)(x) = T(x) + U(x) x  V.
b) aT: V  W by (aT)(x) = aT(x) x  V.
Theorem: (2.7) Let V and W be vector spaces over F, and let T: V  W and U: V  W be linear. Then
a)
a  F aT + U is a linear transformation.
b) Using (+) and () defined above, the set of all linear transformations from V to W is a vector space.
To = zero transformation is the zero vector.
Definition: That vector space is called L(V, W). If V = W, we write L(V).
Theorem: (2.8) Let V and W be finite-dimensional vector spaces with ordered bases  and , respectively.
Let T: V  W and U: V  W be linear. Then
a)
[T + U] = [T] + [U]  Mmn(F).
b) [aT] = a[T]  Mmn(F).
Proof: Let  = {vi}i=1n and  = {wj}j=1m.
a)
! aij, bij  F, 1  i  m, 1  j  n such that T(vj) = i=1m aijwi and U(vj) = i=1m bijwi.
Hence, (T + U)(vj) = T(vj) + U(vj) = i=1m aijwi + i=1m bijwi = i=1m (aij + bij)wi
Thus, ([T + U])ij = aij + bij = ([T] + [U])ij
 [T + U] = [T] + [U]
b) (aT)(vj) = aT(vj) = ai=1m aijwi = i=1m aaijwi.
([aT])ij = aaij = (a[T])ij
 [aT] = a[T]
32
Lecture (Friday, Week4)
Section 2.3 Composition of Linear Transformations and Matrix Multiplication
Theorem: (2.9) Let V, W, Z be vector spaces. Let T: V  W and U: W  Z be linear. Then UT is linear.
Remark: UT(v) = U(T(v))  Z, where T(v)  W.
Proof: UT(ax + y) = U(T(ax + y))
= U(aT(x) + T(y)) because T is linear.
= aU(T(x)) + U(T(y)) because U is linear.
= aUT(x) + UT(y)
Theorem: (2.10) Let V be a vector space. Let T, U1, U2  L(v). Then
(a) T(U1 + U2) = TU1 + TU2
(U1 + U2)T = U1(T) + U2(T)
(b) T(U1U2) = (TU1)U2 [= TU1U2]
(c) TI = IT = T, where I(v) = v v  V.
(d) A(U1U2) = (aU1)U2 = U1(aU2) a  F.
Proof: (Exercise 8)
(a) Let v  V.
T(U1 + U2)(v) = T((U1 + U2)(v))
= T(U1(v) + U2(v))
= T(U1(v)) + T(U2(v))
= TU1(v) + TU2(v)
= (TU1 + TU2)(v)
 T(U1 + U2) = TU1 + TU2
33
Let T: V  W and U: W  Z be linear. Let V, W, Z be vector spaces.
Let  = {vi}i=1n,  = {wi}i=1m,  = {zi}i=1p be ordered bases for V, W, Z, respectively.
Let A = [U]  Mpm.
Let B = [T]  Mmn.
Define C  Mpn by Cij = k=1m AikBkj, 1  i  p, 1  j  n.
Definition: C  Mpn is the unique matrix representation of [UT].
UT(vj) = i=1p Cijzi
UT(vj) = U(T(vj))
= U(k=1m Bkjwk)
= k=1m BkjU(wk)
= k=1m Bkj(i=1p Aikzi)
= i=1p (k=1m AikBkj) zi
= i=1p Cijzi
Thus ([UT])ij = Cij.
Definition: A  Mpm and B  Mmn. Then the product of A and B is the p  n matrix C whose components are
Cij = k=1m AikBkj, 1  i  p, 1  j  n.
Example: If A  M32 and B  M22, then C  M32.
34
Lecture (Monday, Week5)
Friday: Midterm Exam I (sections 1.1 – 1.6 & 2.1 – 2.2)
Today (after 5pm): Practice exam on website
Theorem: Suppose A  Mmn, B  Mnp. Then (AB)t = BtAt.
Remark: AB = C  Mmp and BtAt = Ct  Mpm
Proof: Recall (At)ij = Aji and (Bt)ij = Bij.
[(AB)t]ij = (AB)ji = k=1n AjkBki
= k=1n (At)kj(Bt)ik
= k=1n (Bt)ik(At)kj
= (BtAt)ij
Hence, (AB)t = BtAt.
Theorem: (2.4) Let V, W, Z be vector spaces of finite dimension with ordered bases , , , respectively.
Let T: V  W and U: W  Z be linear. Then [UT] = [U][T].
Proof: Done in the previous lecture.
Definition: We define the kronecker delta ij by
ij = 1, for i = j.
ij = 0, for i  j.
The n  n identity matrix In is defined by (In)ij = ij, for 1  i, j  n.
1 0 0


Example: I3 = 0 1 0  multiplicative (matrix product) identity


0 01 
0 0 0 


03 = 0 0 0  the additive identity


0 0 0
35
Theorem: (2.12) Let A  Mmn, B, C  Mnp, and D, E  Mqm. Then
a)
A(B + C) = AB + AC
(D + E)A = DA + EA
b) a(AB) = (aA)B = A(aB) a  F
c)
ImA = A = AIn  Mmn
d) If dim(V) = n, where V is a vector space with ordered basis , then [Iv] = In.
Iv(v) = v  V.
Proof: a) [A(B + C)]ij = k=1n AikMkj
= k=1n Aik(Bkj + Ckj)
= k=1n AikBkj + AikCkj
= k=1n AikBkj + k=1n AikCkj
= (AB)ij + (AC)ij
Hence, A(B + C) = AB + AC.
[(D + E)A]ij = k=1m MikAkj, 1  i  q, 1  j  n.
= k=1m (Dik + Eik)Akj
= k=1m DikAkj + EikAkj
= k=1m DikAkj + k=1n EikAkj
= (DA)ij + (EA)ij
Hence, (D + E)A = DA + EA.
b) – d) See the book & Exercise 5.
Corollary: Let A  Mmn, B1, …, Bk  Mnp, C1, …Ck  Mqm, and a1, …, ak  F.
Then Ai=1k (aiBi) = i=1k aiABi and i=1k (aiCi)A = i=1k aiCiA
Definition: Let A Mmn, k  {0, 1, 2, 3, …}.
Ak = AA … A (k times), for k > 0.
Ak = In, for k = 0
36
Lecture (Wednesday, Week 5)
Theorem: (2.13) Let A  Mmn(F) and B  Mnp(F).
 ( AB ) 

1j 

 and v =
For each j, 1  j  p, let uj = ...
j


 ( AB ) 
mj 

a)
 B1 j 


 ...  . Then


 B nj 


uj = Avj.
b) vj = Bej.
Remark: Avj is a matrix vector multiplication defined by assuming v j is an n  1 matrix.
 n

  A1k B kj 
 ( AB ) 
 B1 j 

1j 
 k 1








...
Proof: a) uj =
= A ...  = Avj

 =  ...



 ( AB ) 
 B mj 
 n



mj 

  Amk B kj 
 k 1

b) See Exercise 6.
Theorem: (2.14) Let V and W be vector spaces of finite-dimension with ordered bases  and , respectively.
Let T: V  W be linear. Then u  V, [T(u)] = [T][u].
Proof: See the book.
Example: Let T: P3()  P2() defined by T((x)) = ’(x).
 = {xi}i=03 and  = {xi}i=02 are basis for P3() and P2().
0 1 0 0 


Let A = 0 0 2 0 .


0 0 0 3 
Claim: [T] = A.
T(vj) = i=1m aijwi
T(1) = 01 + 0x + 0x2
T(x) = 11 + 0x + 0x2
T(x2) = 01 + 2x + 0x2
T(x3) = 01 + 0x + 3x2
Consider p(x) = 2 – 4x + x2 + 3x3.
T(p(x)) = q(x) = p’(x) = – 4 + 2x + 9x2
2 
 4
 and [T(p(x))] = [q(x)] =
[p(x)] = 
1 
 
3 
 4
2 
 
9 
37
 [T][p(x)] = [T(p(x))]
0 1 0 0 
0 0 2 0 


0 0 0 3 
2 
 4
 4
  = 2 
 
1 
9 
 
3 
Definition: Let A  Mmn(F). Denote LA : Fn  Fm by LA(x) = Ax x  Fn.
It is called the left matrix multiplication transformation.
Remark: (m  n)(n  1) = (m  1)  Fm.
Example: A =
1 21 
0 1 2  M23()


LA: 3  2
Ax = b  M21()
 1  
 
1 21 
LA  3  = 

 
0 1 2 
   1 
 
1 
3  = 6 
1 
 
 
1
Theorem: Let A  Mmn(F). Then LA : Fn  Fm is linear.
If B  Mmn(F),  = {ei}i=1n,  = {ei}i=1m, then
a)
[LA] = A
b) LA = LB  A = B
i.e. Ax = Ab  a = B
c)
LA+B = LA + LB
i.e. (A + B)x = Ax + Bx
d) If T: Fn  Fm is linear, then ! C Mmn such that T = LC and C = [T].
i.e. T(x) = Cx
e)
If E  Mmp(F), then LAE = LALE.
f)
If m = n, LI_n = IF^n
i.e. Inx = x = IF^n(x)
Theorem: (2.16) Let A, B, C be matrices such that A(BC) is well-defined (compatible). Then A(BC) = (AB)C
 associativity of matrix products.
Proof: 1) From the definition of AB.
2) Use properties e) and b) in the previous theorem.
38
Lecture (Monday, Week 6)
Homework: Read Appendix B and section 2.4.
Today: Section 2.4
Definition: Let V and W be vector spaces and T: V  W be linear.
A function U: W  V is called an inverse of T if and only if TU = Iw and UT = Iv.
If T has an inverse, it is called invertible.
Remark: If T is invertible, then the inverse of T is unique and is denoted T -1 (Appendix B).
Remark: The following hold for invertible function T and U.
(1) (TU)-1 = U-1T-1
TUU-1T-1 = TIT-1 = TT-1 = I
(2) (T-1)-1 = T
T-1T = T and TT-1 = T
(3) Let T: V  W be linear and let dim(V) = dim(W) = n be finite.
Then T is invertible if and only if rank(T) = dim(V).
(It follows from theorem 2.5.)
Theorem: (2.17) Let V and W be vector spaces and T: V  W be linear and invertible.
Then T-1: W  V is linear.
Proof: Let y1, y2  W and c  F.
Since T is one-to-one and onto, ! x1, x2  V such that T(x1) = y1 and T(x2) = y2.
Thus T-1(y1) = x1 and T-1(y2) = x2.
WTS: T-1(cy1 + y2) = cT-1(y1) + T-1(y2)
T-1(cy1 + y2) = T-1[cT(x1) + T(x2)]
= T-1[T(cx1 + x2)]
= cx1 + x2
= cT-1(y1) + T-1(y2)
Definition: Let A  Mnn. Then A is invertible if and only if B  Mnn such that AB = BA = In.
Remark: If A  Mnn is invertible, and AB = BA = In for some B  Mnn, then B is the unique inverse of A.
We write A-1 = B.
Proof: Suppose C  Mnn such that AC = CA = In.
C = CIn = C(AB) = (CA)B = InB = B.
39
Lemma: Let V and W be vector spaces, and T: V  W be linear and invertible.
Then dim(V) is finite if and only if dim(W) is finite. In this case, dim(V) = dim(W).
Proof: () Suppose dim(V) = n. Let  = {vi}i=1n be an ordered basis for V.
By theorem 2.2 (page 68), span(T()) = R(T) = W.
Hence dim(W)  n by theorem 1.9 (page 44).
() T-1: W  V. Let  = {wi}i=1m be an ordered basis for W.
Then span(T-1()) = R(T-1) = V.
Thus dim(V)  m = dim(W).
Since dim(V)  dim(W) and dim(W)  dim(V), dim(V) = dim(W).
Theorem: (2.18) Let V and W be vector spaces and dim(V) and dim(W) be finite.
Let  and  be ordered bases for V and W, respectively. Let T: V  W be linear.
Then T is invertible if and only if [T] is invertible. Furthermore, [T -1] = ([T])-1.
Proof: () Suppose T is invertible.
dim(V) = dim(W) = n is finite by the previous lemma.
Thus [T]  Mnn.
Let T-1: W  V denote the inverse transformation.
By definition, TT-1 = Iw and T-1T = Iv.
By theorem 2.4, In = [Iv] = [T-1T] = [T-1][T].
Similarly, [T][T-1] = In.
([T])-1[T][T-1] = ([T])-1In
[T-1] = ([T])-1
() Continued on Wednesday.
40
Lecture (Wednesday, Week 6)
Homework: section 2.4 #4, 5, 9, 10, 12, 16, 17, 20
Corollary 1: Let V be a finite dimension vector space with ordered basis  and T: V  V be linear.
Then T is invertible if and only if [T] is invertible.
Furthermore, [T-1] = ([T])-1.
Corollary 2: Let A  Mnn(F). Then A is invertible if and only if LA is invertible.
Furthermore, (LA)-1 = LA-1.
Recall: LA: Fn  Fm, where LA(x) = Ax.
Definition: Let V and W be vector spaces. We say that V is isomorphic to W if and only if
T: V  W that is invertible. T is called an isomorphism from V onto W.
Example: T: F2  P1(F) defined via T((a1, a2) = a1 + a2x is linear and invertible.
Thus F2 is isomorphic to P1(F).
In general, Fn is isomorphic to Pn-1(F) where T((a1, a2, a3, …, an)) = a1 + a2x + a3x2+ … anxn-1.
Theorem: (2.19) Let V and W be finite dimensional vector spaces over F.
Then V is isomorphic to W if and only if dim(V) = dim(W).
Proof: () Assume V is isomorphic to W. This means T: V  W and T is invertible.
By lemma (page 101), dim(v) = dim(W) because T is invertible.
() Suppose dim(V) = dim(W) = n.
Let  = {vi}i=1n and  = {wi}i=1n be ordered bases for V and W, respectively.
By theorem 2.6 (page 72), T: V  W such that T is linear and T(vi) = wi, 1  i  n.
By theorem 2.2 (page 68), R(T) = span(T()) = span() = W.
Hence T is onto.
From theorem 2.5 (page 71), T must be one-to-one because dim(V) = dim(W).
Hence T is invertible.
Thus V is isomorphic to W.
Corollary: Let V be a vector space over F. Then V is isomorphic to Fn if and only if dim(V) = n.
Theorem: (2.20) Let V and W be finite dimensional vector spaces over F where dim(V) = n and dim(W) = m.
Let  and  be ordered bases for V and W, respectively.
Then : L(V, W)  Mmn(F) defined by (T) = [T] for T  L(V, W) is an isomorphism.
In other word, L(V, W) is isomorphic to M mn(F).
Corollary: Let V and W be vector spaces, where dim(V) = n and dim(W) = m.
Then dim(L(V, W)) = m  n.
41
Definition: Let  be an ordered basis for V where dim(V) = n.
The standard representation of V with respect to  is denoted by : V  Fn
and defined by (x) = [x].
Recall: Let  = {vi}i=1n. Then x has the unique representation, x = a1v1 + a2v2 + … anvn.
 a1 
 
a 
[x] =  2   Fn.
 ... 
 
 an 
Theorem: (2.21) For any vector space V, where dim(V) = n and ordered basis ,
 is an isomorphism of V onto Fn.
42
Lecture (Friday, Week 6)
Homework: Section 2.5 #6 (a, b, c), 10, 13
Change of Basis
Example: Consider the relation R: 2x2 – 4xy + 5y2 = 0
x = (2/5)x’ – (1/5)y’
y = (1/5)x’ + (2/5)y’
R’: (x’)2 + 6(y’)2 = 1  ellipse
 2
  1
 , 1/5   }
1 
2 
x
 2  1

[x] =   = (1/5) 
 y
1 2 
’ = {(1/5) 
 x' 
  = Q[x]’
 y'
Q = [I2]’ , I2: 2  2: (’)  ()
[x] = Q[x]’ = [I]’[x]’
Theorem: (2.22) Let  and ’ be two ordered bases for V, a vector space with dim(V) = n < .
Let Q = [Iv]’. Then
a)
Q is invertible.
b) v  V [v] = Q[v]’.
Proof:
a)
Since Iv is invertible, Q is invertible by theorem 2.18 (page 101).
b) By theorem 2.14 (page 91), T: V  W: ()  (), u  V [T(u)] = [T][u].
Let T = Iv, W = V,  = ’, and  = .
Then [u] = [Iv]’[u]’.
Definition: The matrix Q = [Iv]’ is called a change of coordinate matrix.
Q changes ’ coordinates into  coordinates.
Remark: Q-1 changes  coordinates into ’.
Q-1[u] = [u]’
Remark: If  = {x1, …, xn} and ’ = {x1’, …, xn’}, then I: V  V: (’)  (), I(v) = v v  V.
xj’ = i=1n Qijxi, 1  j  n
The jth column of Q is [xj’].
 a1 
 
Recall: If v = i=1n aixi, then [v] =  ...  .
 
 an 
43
Theorem: (2.23) Let T: V  V be linear, V a vector space with dim(V) = n < .
Let  and ’ be ordered bases for V.
Let Q = [Iv]’.
Then [T]’ = Q-1[T]Q.
Proof: Let Iv: V  V. Then T = IvT = TIv.
Q[T]’ = [Iv]’[T]’’ = [IvT]’ = [TIv]’ = [T][Iv]’ = [T]Q
Hence, [T]’ = Q-1[T]Q.
44
Example: T: 2  2 defined via T((a, b)) = (3a – b, a + 3b).
 = {(1, 1), (1, -1)} and ’ = {(2, 4), (3, 1)}.
T((1, 1)) = (2, 4) = a11(1, 1) + a21(1, -1)
 a11 + a21 = 2
 a11 – a21 = 4
 a11 = 3, a21 = -1
T((1, -1)) = (4, -2) = a12(1, 1) + a22(1, -1)
 a12 + a22 = 4
 a12 – a22 = -2
 a12 = 1, a22 = 3
[T] =
3
 1

1
3
Consider Q = [I2]’.
I2((2, 4)) = (2, 4) = a11(1, 1) + a21(1, -1)
 a11 + a21 = 2
 a11 – a21 = 4
 a11 = 3, a21 = -1
I2((3, 1)) = (3, 1) = a12(1, 1) + a22(1, -1)
 a12 + a22 = 3
 a12 – a22 = 1
 a12 = 2, a22 = 1
1  2
2
 Q-1 = (1/5) 


1
1 3 
4 1 
[T]’ = Q-1[T]Q = 
 =B

2
2


Hence, Q = [I2]’ =
3
 1

Check: T((2, 4)) = (2, 14) = a11(2, 4) + a21(3, 1)
 2a11 + 3a21 = 2
 4a11 + a21 = 14
 a21 = -2, a11 = 4
T((3, 1)) = (8, 6) = a12(2, 4) + a22(3, 1)
 2a12 + 3a22 = 8
 4a12 + a22 = 6
 a22 = 2, a12 = 1
45
Corollary: Let A  Mnn(F) and let  be an ordered basis for Fn.
Then [LA] = Q-1AQ.
Recall LA(x) = Ax.
Proof: Let  = {ei}i=1n.
Set Q = [IFn].
Recall A = [LA].
[LA] = Q-1[LA]Q = Q-1AQ
Definition: Let A, B  Mnn(F).
We say that B is similar to A if and only if an invertible Q  Mnn(F) such that B = Q-1AQ.
46
Lecture (Monday, Week 7)
Homework: section 3.1 #2, 3, 4, 5, 6.
Quiz: covers section 2.4, 2.5, and 3.1.
Definition: Let A  Mmn(F). Any one of the following 3 operations on rows [column] of A is called an
elementary row [column] operation:
Type (1): interchanging any two rows [columns] of A.
Type (2): multiplying any row [column] by a nonzero scalar.
Type (3): adding any scalar multiple of a row [column] to another row [column].
1

Example: A =  2
4

2
3
1
0
1
1
4

3
2 
1
1
2
0
3
1
3
1
6
0
3
1
3

4
2 
2

A: (r1  r2, r2  r1)  B = 1
4

2

A: (3c2 c2)  C = 1
4

17

A: (4r3 + r1  r1)  M =  2
4

2
1
0
3

4
2 
7 12 

1 3 
1 2 
Facts: Row or column operations can be reversed (undone).
The reverse operation is the same type as the original.
47
Definition: An n  n elementary matrix is one obtained by performing an elementary row [column] operation on
In. The elementary matrix is of type (1), (2), or (3) according to the type of elementary operation
performed on In.
0

Example: I3: (r1  r2, r2  r1)  E = 1
0

1

I3: (–2r3 + r1  r1)  E =  0
0

1

Fact: I3: (–2c1 + c3  c3)  E =  0
0

1
0
0
0
1
0
0
1
0
0

0
1 
Type (1) elementary matrix
 2

0 
1 
Type (3) elementary matrix
 2

0 
1 
Theorem: (3.1) Let A  Mmn(F) and suppose that B is obtained from A by performing an elementary row
[column] operation.
Then  an elementary matrix E where E  Mmm(F) [E  Mnn(F)] such that B = EA [B = AE].
In fact, E is obtained form Im [In] by performing the same elementary row [column] operation as that
which was performed on A to obtain B.
Conversely, If E is an elementary m m [n  n] matrix, then EA [AE] is the matrix obtained form A
by performing the same elementary row [column] operation as that which produces E from I m [In].
1

Example: A =  2
4

2
3
1
0
1
1
4

3
2 
2

A: (r1  r2, r2  r1)  B = 1
4

0

EA = 1
0

1
0
0
0  2

0  1
1   4
1
1
2
0
3
1
1
1
2
0
3
1
3

4  and I3: (r1  r2, r2  r1)  E =
2 
3
1


4 =  2
4
2 

2
3
1
0
1
1
0

1
0

1
0
0
0

0
1 
4

3
2 
Theorem: (3.2) Elementary matrices of all types are invertible, and the inverse of an elementary matrix is an
elementary matrix of the same type.
48
Lecture (Wednesday, Week 7)
Quiz: covers section 2.4 and 2.5 only.
Today: section 3.2
Homework: read along section 3.2.
A = [LA],  = {ei}i=1n, LA: Fn  Fm, LA(x) = Ax.
Definition: If A  Mmn(F), then rank(A) = rank(LA).
Remark: rank(LA) = dim(R(LA))
Theorem: (3.3) Let T: V  W be linear, V and W finite-dimensional vector spaces with ordered basis  and ,
respectively. Then rank(T) = rank([T]).
Proof: exercise 2.4.20
Theorem: (3.4) Let A  Mmn(F). If P  Mmm(F) and Q  Mnn(F) invertible, then
(a) rank(AQ) = rank(A)
(b) rank(PA) = rank(A)
(c) rank(PAQ) = rank(A)
Proof: (a) rank(AQ) = rank(LAQ) = dim(R(LAQ))
R(LAQ) = R(LALQ) = LALQ(Fn) = LA(LQ(Fn)) = LA(Fn).
Note: since Q is onto, LQ: Fn  Fn is onto, hence LQ(Fn) = Fn.
R(LAQ) = LA(Fn) = R(LA)
rank(AQ) = dim(R(LAQ)) = dim(R(LA)) = rank(LA) = rank(A)
(b) By definition, rank(PA) = rank(LPA).
LPA = LPLA: Fn  Fm
rank(LPA) = rank(LPLA)
R(LPLA) = LPLA(Fn) = Lp(LA(Fn)) = LP(R(LA))
rank(PA) = rank(LPA) = rank(LPLA) = dim(R(LPLA)) = dim(LP(R(LA))) = dim(R(LA))
By exercise 2.4.17 because LP is an isomorphism.
rank(PA) = dim(R(LA)) = rank(LA) = rank(A)
Corollary: Elementary row [column] operations are rank-preserving.
Idea of Proof: Since E is invertible,
B = EA  rank(B) = rank(EA)
B = AE  rank(B) = rank(AE)
49
Theorem: (3.5) The rank of any matrix equals the maximum number of linearly independent columns.
Proof: For any matrix A  Mmn(F), rank(A) = rank(LA) = dim(R(LA)).
Let  = {ei}i=1n be the standard basis for Fn.
By theorem 2.2 (page 68), R(LA) = span(LA()).
span(LA()) = span({LA(e1), …, LA(en)})
= span({Ae1, …, Aen})
= span({a1, …, an}), where aj is the jth column of A.
rank(A) = rank(LA) = dim(R(LA)) = dim(span({a1, …, an}))
1

Example: A = 1
1

1

B = 1
0

1

C = 0
0

1

D = 0
0

1

E = 0
0

1

3
2 
2
0
1
1

0 3  = E1A
 1 1 
2
2
2
1
0
2
1
0
2
1
1

2  = E2E1A
1 
1

2  = E2E1AE3
1 
0

2  = E2E1AE3E4
1 
By the previous corollary, rank is preserved through elementary row [column] operations.
Hence, rank(A) = rank(E) = 2
50
Lecture (Monday, Week 8)
Today: Finish section 3.2
Homework: section 3.2 #2 abc, 5 bcd, 8, 13, 15, 16
Midterm 2: Wednesday, Week 9
Practice Midterm: Friday, this week
Quiz: covers sections 3.1 and 3.2
Theorem: (3.6) Let A  Mmn(F) and rank(A) = r (= rank(LA)).
Then r  m and r  n, and by means of elementary row or column operations,
A may be transformed into D =
Ir


 O2
O  .
O 
1
3
Corollary 1: Let A  Mmn(F) and rank(A) = r.
Then  invertible matrices B Mmm(F) and C  Mnn(F) such that D = BAC Mmn(F).
Corollary 2: Let A  Mmn(F). Then
(a) rank(At) = rank(A)
(b) rank(A) is equal to the number of linearly independent rows.
(c) The rows and columns of any matrix generate subspaces of the same dimensions,
numerically equal to rank(A).
Proof: (a) By corollary 1, B, C invertible such that D = BAC where D =
Dt =
Ir


 O1
Ir


 O2
O 
O 
1
 Mmn(F).
3
O 
O 
2
3
Dt = (BAC)t = CtAtBt
Since B and C are invertible, Bt and Ct are invertible by exercise 2.4.5.
By theorem 3.4, rank(At) = rank(Dt) = r = rank(A).
Corollary 3: Every invertible matrix is a product of elementary matrices.
Proof: Let A  Mnn(F) be invertible.
Note that rank(A) = rank(LA) = dim(R(LA)) = dim(Fn) = n, where LA: Fn  Fn.
By corollary 1, B, C Mnn(F) which are invertible such that D = BAC = In.
B = EpEp-1…E1 and C = G1G2…Gq where Ei and Gj are elementary matrices for 1  i  p and 1  j  q.
D = EpEp-1…E1AG1G2…Gq = In
A = E1-1…Ep-1InGq-1…Gq-1 = E1-1…Ep-1Gq-1…Gq-1
51
Theorem: (3.7) Let T: V  W and U: W  Z be linear transformations on finite-dimensional vector spaces.
Let A and B be matrices such that AB is defined. Then
(a) rank(UT)  rank(U)
(b) rank(UT)  rank(T)
(c) rank(AB)  rank(A)
(d) rank(AB)  rank(B)
Proof: (a) Note that R(T) is a vector subspace of W.
R(UT) = U(T(V)) = U(R(T))  U(W) = R(U).
rank(UT) = dim(R(UT))  dim(R(U)) = rank(U).
(c) By proof (a), rank(AB) = rank(LAB) = rank(LALB)  rank(LA) = rank(A)
(d) By proof (c) and corollary 2 (page 158), rank(AB) = rank((AB) t) = rank(BtAt)  rank(Bt) = rank(B).
Matrix Inversions
Definition: Let A and B be m  n and m  p matrices, respectively.
By the augmented matrix (A  B), we mean the m  (n + p) matrix (AB).
Let A  Mnn(F) be an invertible matrix.
Consider C = (A  In) Mn2n(F).
By exercise 15, A-1C = (A-1A  A-1In)
= (In  A-1)
A-1 = Ep…E1, where Ei is elementary matrix for 1  i  p.
Ep…E1(A  In) = A-1C = (In  A-1)
52
Lecture (Wednesday, Week 8)
Today: Section 3.3 Linear Systems of Equations
a11x1 + a12x2 + … + a1nxn = b1
a21x1 + a22x2 + … + a2nxn = b2
…
am1x1 + am2x2 + … + amnxn = bm
aij, xj, bi  F for 1  i  m and 1  j  n.
 a11

a
Ax = b, where A =  21
...


 a m1
a
a
12
a
22
m2


...
2n 
, x =

... a mn 
...
a
a
1n
 x1 
 
x 
 2  , and b =
 ... 
 
 xn 
 b1 
 
b 
 2
 ... 
 
 bm 
A  Mmn(F), x  Fn, and b  Fm.
LA: Fn  Fm, where LA(x) = Ax = b.
b is known and x is unknown.
 s1 
 
s 
Definition: A solution of (S) is any n-tuples S =  2   Fn such that AS = b.
 ... 
 
 sm 
The set of all solutions of (S) is called the solution set of (S).
 = {S  Fn: AS = b} is the solution set and   Fn.
The system (S) is called consistent if and only if   ; otherwise (S) is called inconsistent.
Example: (1b) 2x1 + 3x2 + x3 = 1
x1 – x2 + 2x3 = 6
Ax = b 
2

1
 x1 
3 1  
1 
  x 2  =  
1 2  
6
 
 x3 
  6
 
S1 =  2  and S2 =
7 
 
8 
 
  4
 3
 
53
Definition: A system Ax = b of m linear equations in n unknowns is said to be homogeneous
if and only if b = 0; otherwise the system is called inhomogeneous (non-homogeneous).
Theorem: (3.8) Let Ax = 0 where A  Mmn(F) and x  Fn.
Let KH = {S  Fn: AS = 0} = ker(A). Then KH = N(LA).
Hence, KH is a subspace of Fn of dimension n – rank(LA) = n – rank(A) = nullity(LA).
[dim(KH) = n – rank(A)]
Proof: Recall from the Dimension Theorem that dim(V) = rank(T) + nullity(T), T: V  W.
Recall LA: Fn  Fm, where LA(x) = Ax.
n = dim(Fn) = rank(LA) + nullity(LA)
= rank(A) + nullity (LA)
nullity(LA) = n – rank(A)
N(LA) = {x  Fn: LA(x) = 0  Fm} = KH
dim(N(LA)) = nullity(LA) = dim(KH)
Definition: KH = kernel of A = ker(A)
Example: Consider A =
1

1
2 1
 and x  F3.
 1 1
What are the solution to Ax = 0  F2.
rank(A) = # of linearly independent rows or column = 2
dim(KH) = n – rank(A) = 3 – 2 = 1
Equation 1: x1 + 2x2 + x3 = 0
Equation 2: x1 – x2 – x3 = 0
 3x2 – 2x3 = 0
Let x3 = t  F.
x2 = – (2/3)t
x1 + 2x2 + x3 = 0
x1 – 2(2/3)t + t = 0
x1 – (1/3)t = 0
x1 = (1/3)t
 (1 / 3)t 
 1/ 3 
 1/ 3 






St =  (2 / 3)t  = KH = {x  F3: x = t   2 / 3  , where t  F} = span({   2 / 3  })
 t 
 1 
 1 






 1/ 3 


Hence,  = {   2 / 3  } a the basis for KH.
 1 


54
Lecture (Friday, Week 8)
Theorem: (3.9) Let K be the solution set of Ax = b. Let KH be the solution set of Ax = 0.
Then for any solution s to Ax = b, K = {s} + KH = {s + sH: sH  KH}.
Proof: Fix s, a solution to Ax = b.
Let w  K. Then Aw = b. Hence, A(w – s) = Aw – As = b – b = 0.
Thus w – s  KH.
sH  KH such that w – s  SH.
w = s + sH  w {s} + KH. Hence, K  {s} + KH.
Conversely, suppose w  {s} + KH.
Then w = s + sH for some sH  KH.
However, Aw = A(s + sH) = As + AsH = b + 0 = b.
w  K. Hence, {s} + KH  K.
Remark: Note that we always have A0n = 0m for 0n  Fn and 0m  Fm.
Hence, 0n  KH.
Suppose that for a given b  Fm s  Fn such that As = b.
Further, suppose KH = {0n}.
Then s is the only solution of Ax = b because K = {s} + KH = {s} + {0n} = {s}.
Theorem: (3.10) Let Ax = b be a system of n linear equations in n unknowns. (i.e. LA: Fn  Fn, A  Mnn(F))
Then A is invertible if and only if the system has exactly one solution.
Proof: () Suppose that A is invertible.
Then A-1 exists such that AA-1 = A-1A = In.
Let s = A-1b.
Then As = A(A-1b) = (AA-1) = Inb = b.
Let s1 and s2 be any two solutions of Ax = b.
Then As1 = b and As2 = b.
LA(s1 – s2) = A(s1 – s2) = 0
Since LA is one-to-one and onto, s1 – s2 = 0  s1 = s2
() Suppose there is only one solution to Ax = b.
Let s be the solution.
Then K = {s} = {s} + KH.
Thus KH = {0}.
N(LA) = KH = {0}  LA is invertible  A is invertible.
55
Theorem: (3.11) Let Ax = b be a system of linear equations.
Then the system is consistent if and only if rank(A) = rank(Ab).
Proof: () Suppose the system is consistent.
Then s  Fn such that As = b.
This implies b  R(LA)  Fm.
R(LA) = span({a1, …, an}), where ai is a column of A.
Thus b  span({a1, …, an}).
span({a1, …, an}) = span({a1, …, an, b})
dim(span({a1, …, an})) = dim(span({a1, …, an, b}))
Hence, rank(A) = rank(Ab)
() Since the proof of () is an equivalent argument, () holds automatically.