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Lecture 8
8
...
We now want to answer
the reverse question: Are all linear mappings matrix mappings in disguise? If T : Rn → Rm
is a linear mapping, then to show that T is in fact a matrix mapping we must show that
there is some matrix A ∈ Mm×n such that T(x) = Ax
...
, en in Rn :
0
0
0
1
0
0
1
0
e1 = 0 , e2 = 0 , e3 = 1 , · · · , en = 0
...
...
...
...
...
, en } because:
x1
1
0
0
x2
0
1
0
x =
...
+ x2
...
= x1 e1 + x2 e2 + · · · + xn en
...
...
xn
0
0
1
With this notation we prove the following
...
14: Every linear mapping is a matrix mapping
...
Let T : Rn → Rm be a linear mapping
...
, vn = T(en )
...
Now, for arbitrary x ∈ Rn we can write
x = x1 e1 + x2 e2 + · · · + xn en
...
Define the matrix A ∈ Mm×n by A = v1 v2 · · · vn
...
Therefore, T is a matrix mapping with the matrix A ∈ Mm×n
...
In words, the columns of A are the images of the
standard unit vectors e1 , e2 ,
...
The punchline is that if T is a linear mapping,
then to derive properties of T we need only know the standard matrix A corresponding to
T
...
15
...
Use the standard unit vectors e1 =
1
0
matrix A ∈ R2×2 corresponding to T
...
We have
A = T(e1 ) T(e2 ) =
Tθ (e1 )
b
e1
#
"
cos(θ) − sin(θ)
sin(θ)
cos(θ)
Example 8
...
Let T : R3 → R3 be a dilation of factor k = 2
...
Solution
...
Then
1
2
0
0
0
0
T(e1 ) = 2 0 = 0 , T(e2 ) = 2 1 = 2 , T(e3 ) = 2 0 = 0
0
0
0
0
1
2
Therefore,
2
0
0
A = T(e1 ) T(e2 ) T(e3 ) = 0 2 0
0 0 2
is the standard matrix of T
...
1
Sums of Matrices
We begin with the definition of matrix addition
...
1: Given matrices
a11 a12 · · · a1n
a21 a22 · · · a2n
A =
...
,
...
...
...
am1 am2 · · · amn
b11
b21
B =
...
b12
b22
...
bm1 bm2
both of the same dimension m × n, the sum A + B
a11 + b11 a12 + b12
a21 + b21 a22 + b22
A+B=
...
...
am1 + bm1 am2 + bm2
· · · b1n
· · · b2n
...
...
· · · bmn
is defined as
···
···
...
a1n + b1n
a2n + b2n
...
· · · amn + bmn
...
Definition 9
...
...
...
...
...
...
...
...
3
...
We compute:
find 3A − 2B
...
Theorem 9
...
Then
(a) A + B = B + A
(b) (A + B) + C = A + (B + C)
(c) A + 0 = A
9
...
If x ∈ Rp then TB (x) ∈ Rn
and thus we can apply TA to TB (x)
...
Hence, each
x ∈ Rp can be mapped to a point in Rm , and because TB and TA are linear mappings the
resulting mapping is also linear
...
1)
...
Because (TA ◦ TB ) : Rp → Rm is a linear mapping it has an associated standard matrix,
which we denote for now by C
...
, ep under
the linear mapping
...
Applying this to x = ei for all i = 1, 2,
...
76
Lecture 9
Rm
Rp
TB
Rn
TA
b
x
b
TB (x)
b
TA (TB(x))
(TA ◦ TB )(x)
Figure 9
...
Now Be1 is
Be1 = b1 b2 · · · bp e1 = b1
...
, p
...
This computation motivates the following definition
...
5: For A ∈ Rm×n and B ∈ Rn×p , with B = b1 b2 · · · bp , we define the
product AB by the formula
AB = Ab1 Ab2 · · · Abp
...
The following diagram is useful for remembering this:
(m × n) · (n × p) → m × p
From our definition of AB, the standard matrix of the composite mapping TA ◦ TB is
C = AB
...
Example 9
...
For A and B below compute AB and BA
...
First AB = [Ab1 Ab2 Ab3 Ab4 ]:
AB =
1 2 −2
1 1 −3
=
2
7
2 0 4
7 9 10
=
=
=
2 0
7 9
2 0 4 4
7 9 10 2
−4 2
4 −4
−1 −5 −3 3
−4 −4 −3 −1
On the other hand, BA is not defined! B has 4 columns and A has 2 rows
...
7
...
−4
4
3
−1 −1
0
0 −2
A = 3 −3 −1 , B = −3
−2 −1
1
−2
1 −2
Solution
...
e
...
An important matrix that arises frequently is the
n:
1 0 0 ···
0 1 0 · · ·
In =
...
· · ·
0 0 0 ···
identity matrix In ∈ Rn×n of size
0
0
...
1
You should verify that for any A ∈ Rn×n it holds that AIn = In A = A
...
Theorem 9
...
Then
(1) A(BC) = (AB)C
(2) A(B + C) = AB + AC
(3) (B + C)A = BA + CA
(4) α(AB) = (αA)B = A(αB)
(5) In A = AIn = A
If A ∈ Rn×n is a square matrix, the kth power of A is
Ak = |AAA
{z· · · A}
k times
79
Matrix Algebra
Example 9
...
Compute A3 if
A=
−2 3
1 0
...
Compute A2 :
2
A =
−2 3
1 0
−2 3
1 0
=
7 −6
−2
3
And then A3 :
A =A A=
7 −6
−2
3
=
−20 21
7 −6
3
2
−2 3
1 0
We could also do:
3
2
A = AA =
9
...
Matrix Transpose
We begin with the definition of the transpose of a matrix
...
10: Given a matrix A ∈ Rm×n , the transpose of A is the matrix AT whose
ith column is the ith row of A
...
For example, if
0 −1
8 −7 −4
−4
6 −10 −9
6
A=
9
5 −2 −3
5
−8
8
4
7
7
then
AT =
0 −4
9 −8
−1
6
5
8
8 −10 −2
4
...
11
...
, B = −1 −2
0
0 −1
80
Lecture 9
Solution
...
Theorem 9
...
The following hold:
(1) (AT )T = A
(2) (A + B)T = AT + BT
(3) (αA)T = αAT
(4) (AB)T = BT AT
A consequence of property (4) is that
(A1 A2
...
Example 9
...
Let T : R2 → R2 be the linear mapping that first contracts vectors by a
factor of k = 3 and then rotates by an angle θ
...
Let e1 = (1, 0) and e2 = (0, 1) denote the standard
unit vectors in R2
...
Recall that the standard matrix
of a rotation by θ is
cos(θ) − sin(θ)
sin(θ) cos(θ)
Contracting e1 by a factor of k = 3 results in ( 13 , 0) and then rotation by θ results in
1
cos(θ)
3
= T(e1 )
...
1
cos(θ)
3
Therefore,
A = T(e1 ) T(e2 ) =
"1
3
cos(θ) − 13 sin(θ)
1
3
sin(θ)
1
3
cos(θ)
#
On the other hand, the standard matrix corresponding to a contraction by a factor k =
"1
Therefore,
3
0
0
1
3
# "1 #
"
0
cos(θ) − sin(θ)
3
sin(θ)
|
0 1
cos(θ)
{z
} | {z 3 }
rotation
=
1
3
is
#
"1
3
cos(θ) − 13 sin(θ)
1
3
sin(θ)
1
3
cos(θ)
#
=A
contraction
After this lecture you should know the following:
• know how to add and multiply matrices
• that matrix multiplication corresponds to composition of linear mappings
• the algebraic properties of matrix multiplication (Theorem 9
...
12)
82
Lecture 10
Lecture 10
Invertible Matrices
10
...
Formally speaking, the inverse of a non-zero number a ∈ R is the unique
number c ∈ R such that ac = ca = 1
...
This motivates the following definition
...
1: A matrix A ∈ Rn×n is called invertible if there exists a matrix C ∈
Rn×n such that AC = In and CA = In
...
Then
C2 = C2 (AC1 ) = (C2 A)C1 = In C1 = C1
...
This motivates the following definition
...
2: If A is invertible then we denote the inverse of A by A−1
...
Example 10
...
Given A and C below, show that C is the inverse of A
...
Compute AC:
Compute CA:
1 −3
0
−14 −3 −6
1 0 0
2 −2 −5 −1 −2 = 0 1 0
AC = −1
−2
6
1
2
0
1
0 0 1
−14 −3 −6
1 −3
0
1 0 0
CA = −5 −1 −2 −1
2 −2 = 0 1 0
2
0
1
−2
6
1
0 0 1
Therefore, by definition C = A−1
...
4: Let A ∈ Rn×n and suppose that A is invertible
...
Proof: Let b ∈ Rn be arbitrary
...
Therefore, with x = A−1 b we have that
Ax = A(A−1 b) = AA−1 b = In b = b
˜ is another solution of the equation, that is, A˜
and thus x = A−1 b is a solution
...
Thus, x = x
˜
...
5
...
3
...
A = −1
−2
6
1
−1
Solution
...
3 that
−14 −3 −6
A−1 = −5 −1 −2
...
Theorem 10
...
Then:
(1) The matrix A−1 is invertible and its inverse is A:
(A−1 )−1 = A
...
(3) The matrix AT is invertible and its inverse is (A−1 )T :
(AT )−1 = (A−1)T
...
To prove (3) we compute
AT (A−1)T = (A−1 A)T = ITn = In
...
2
Computing the Inverse of a Matrix
If A ∈ Mn×n is invertible, how do we find A−1 ? Let A−1 = c1 c2 · · · cn and we will
find expressions for ci
...
On the other hand,
we also have AA−1 = In = e1 e2 · · · en
...
, cn such
that
Ac1 Ac2 · · · Acn = e1 e2 · · · en
...
Here the image vector “b”
is ei
...
85
Invertible Matrices
We will need to do this for each
c2 ,
...
In summary, to determine if A−1 exists and to simultaneously compute it, we compute the
RREF of the augmented matrix
A In ,
that is, A augmented with the n × n identity matrix
...
If the RREF of A is not In then A is not invertible
...
7
...
−1 −2
Solution
...
The inverse is
−2 −3
−1
A =
1
1
Verify:
AA
−1
1
3
=
−1 −2
1 0
−2 −3
...
Example 10
...
Find the inverse of A = 1 1
−2 0 −7
86
Lecture 10
Solution
...
The inverse is
7 0 3
A−1 = −7 1 −3
−2 0 −1
Verify:
AA−1
1 0 3
7 0 3
1 0 0
= 1 1 0 −7 1 −3 = 0 1 0
−2 0 −7
−2 0 −1
0 0 1
1 0
1
Example 10
...
Find the inverse of A = 1 1 −2 if it exists
...
Form the augmented matrix A I3 and row reduce:
1 0 1 1 0 0
1 0 1
1 0 0
−R1 +R2 , 2R1 +R2
1 1 −2 0 1 0 −
−−−−−−−−−→ 0 1 −3 −1 1 0
−2 0 −2 0 0 1
0 0 0
2 0 1
We need not go further since the rref(A) is not I3 (rank(A) = 2 )
...
10
...
Let TA−1 : Rn → Rn be the matrix mapping with standard matrix A−1
...
87
Invertible Matrices
Therefore, (TA−1 ◦ TA )(x) = In x = x
...
Similarly, the standard matrix of (TA ◦ TA−1 ) is also In
...
Moreover, since Ax = b always has a solution, TA is
onto
...
The following theorem summarizes equivalent conditions for matrix invertibility
...
10: Let A ∈ Rn×n
...
(b) A is row equivalent to In , that is, rref(A) = In
...
(d) The linear transformation TA (x) = Ax is one-to-one
...
(f) The matrix equation Ax = b is always solvable
...
(h) The columns of A are linearly independent
...
Proof: This is a summary of all the statements we have proved about matrices and matrix
mappings specialized to the case of square matrices A ∈ Rn×n
...
Example 10
...
Without doing any arithmetic, write down the inverse of the dilation
matrix
"
#
3 0
A=
...
12
...
A=
sin(θ) cos(θ)
After this lecture you should know the following:
• how to compute the inverse of a matrix
• properties of matrix inversion and matrix multiplication
• relate invertibility of a matrix with properties of the associated linear mapping (1-1,
onto)
• the characterizations of invertible matrices Theorem 10
...
1
Determinants of 2 × 2 and 3 × 3 Matrices
Consider a general 2 × 2 linear system
a11 x1 + a12 x2 = b1
a21 x1 + a22 x2 = b2
...
Notice the denominator is the same in both expressions
...
This motivates the following definition
...
1: Given a 2 × 2 matrix
a
a
A = 11 12
a21 a22
we define the determinant of A as
a11 a12
det A = det
= a11 a22 − a12 a21
...
=
det
a21 a22
a21 a22
89
Determinants
Example 11
...
Compute the determinant of A
...
For (i):
3 −1
= (3)(2) − (8)(−1) = 14
det(A) =
8 2
For (ii):
3
1
= (3)(−2) − (−6)(1) = 0
det(A) =
−6 −2
For (iii):
−110 0
= (−110)(0) − (568)(0) = 0
det(A) =
568 0
As in the 2 × 2 case, the solution of a 3 × 3 linear system Ax = b can be shown to be
x1 =
Numerator2
Numerator3
Numerator1
, x2 =
, x3 =
D
D
D
where
D = a11 (a22 a33 − a23 a32 ) − a12 (a21 a33 − a23 a31 ) + a13 (a21 a32 − a22 a31 )
...
| {z }
| {z }
| {z }
a22 a23
a21 a23
a21 a22
a32 a33
a31 a33
a31 a32
Let
A11
a22 a23
,
=
a32 a33
Then we can write
The matrix A11
A12
a21 a23
,
=
a31 a33
and A13
a21 a22
...
a22 a23
is obtained from A by deleting the 1st row and the 1st column:
=
a32 a33
a11 a12 a13
a
a
22
23
A = a21 a22 a23 −→ A11 =
...
A = a21 a22 a23 −→ A12 =
a31 a33
a31 a32 a33
a21 a22
is obtained from A by deleting the 1st row and the 3rd
Finally, the matrix A13 =
a31 a32
column:
a11 a12 a13
a
a
21
22
A = a21 a22 a23 −→
...
This motivates
the following definition
...
3: Let A be a 3 × 3 matrix
...
Define the cofactor of ajk to be the number
Cjk = (−1)j+k det Ajk
...
This definition of the determinant is called the expansion of the determinant along the
first row
...
For example, the cofactor of a12
is
C12 = (−1)1+2 det A12 = − det A12
and the cofactor of a13 is
C13 = (−1)1+3 det A13 = det A13
...
For example,
the cofactor of a23 is
C23 = (−1)2+3 det A23 = − det A23
...
+
This works not just for 3 × 3 matrices but for any square n × n matrix
...
4
...
From the definition of the determinant
det A = a11 C11 + a12 C12 + a13 C13
= (4) det A11 − (−2) det A12 + (3) det A13
2 3
2 5
3 5
+2
= 4
1 6 + 3 1 0
0 6
= 4(3 · 6 − 5 · 0) + 2(2 · 6 − 1 · 5) + 3(2 · 0 − 1 · 3)
= 72 + 14 − 9
= 77
We can compute the determinant of a matrix A by expanding along any row or column
...
a22 a23
And along the 2nd column:
a11 a13
a11 a13
a21 a23
...
If a particular row or column contains zeros, say entry ajk , then the computation of
the determinant is simplified if you expand along either row j or column k because ajk Cjk = 0
and we need not compute Cjk
...
5
...
In Example 11
...
92
Lecture 11
Notice that a32 = 0
...
2
Determinants of n × n Matrices
Using the 3 × 3 case as a guide, we define the determinant of a general n × n matrix as
follows
...
6: Let A be a n × n matrix
...
The determinant of A is defined to be
det A = a11 C11 + a12 C12 + · · · + a1n C1n
...
Theorem 11
...
Then det A may be obtained by a cofactor
expansion along any row or any column of A:
det A = aj1 Cj1 + aj2 Cj2 + · · · + ajn Cjn
...
Corollary 11
...
Proof
...
93
Determinants
Corollary 11
...
Sketch of the proof
...
Example 11
...
Compute the determinant of
1
3
0 −2
1
2 −2 −1
A=
0
0
2
1
−1 −3
1
0
Solution
...
11
...
Expanding along the second row:
det A = − det A21 + 2 det A22 − (−2) det A23 − 1 det A24
1 0 −2
3 0 −2
1
1 + 2 0 2
= − 0 2
−1 1
−3 1
0
0
1
1
3
0
3
−2
0 2
0
1 − 0
+ 2 0
−1 −3
0 −1 −3 1
= −1(−3 − 12) + 2(−1 − 4) + 2(0) − (0)
=5
11
...
Definition 11
...
In other words, all the entries of A below the diagonal entries aii are
zero
...
For example, a 4 × 4 upper triangular matrix takes the form
a11 a12 a13 a14
0 a22 a23 a24
A=
0
0 a33 a34
0
0
0 a44
Expanding along the first column,
a22 a23
det A = a11 0 a33
0
0
we compute
a24
a33 a34
= a11 a22 a33 a44
...
Theorem 11
...
After this lecture you should know the following:
• how to compute the determinant of any sized matrix
• that the determinant of A is equal to the determinant of AT
• the determinant of a triangular matrix is the product of its diagonal entries
95