Transcript Linear Algebra Chapter 3 Determinants and Eigenvectors 大葉大學 資訊工程系 黃鈴玲 3.1 Introduction to Determinants Definition The determinant (行列式) of a 2  2 matrix A is.

Linear Algebra
Chapter 3
Determinants and Eigenvectors
大葉大學 資訊工程系
黃鈴玲
3.1 Introduction to Determinants
Definition
The determinant (行列式) of a 2  2 matrix A is denoted |A| and
is given by
a11
a12
a21
a22
 a11a22  a12 a21
Observe that the determinant of a 2  2 matrix is given by the
different of the products of the two diagonals of the matrix.
The notation det(A) is also used for the determinant of A.
Example 1
A   2 4
 3 1
det( A)  2 4  (2 1)  (4  (3))  2  12  14
3 1
Ch03_2
Definition
Let A be a square matrix.
The minor (子式) of the element aij is denoted Mij and is the
determinant of the matrix that remains after deleting row i and
column j of A.
The cofactor (餘因子) of aij is denoted Cij and is given by
Cij = (–1)i+j Mij
Note that Cij = Mij or Mij .
Ch03_3
Example 2
Determine the minors and cofactors of the elements a11 and a32
of the following matrix A.
0 3
1
A   4  1 2
0  2 1
Solution
1
0 3
Minor of a11 : M 11  4  1 2   1 2  (11)  (2  (2))  3
0 2 1 2 1
Cofactor of a11 : C11  (1)11 M 11  (1) 2 (3)  3
1
0 3
Minor of a32 : M 32  4  1 2  1 3  (1 2)  (3  4)  10
0 2 1 4 2
Cofactor of a32 : C32  (1)32 M 32  (1)5 (10)  10
隨堂作業：3(c)
Ch03_4
Definition
The determinant of a square matrix is the sum of the products
of the elements of the first row and their cofactors.
If A is 3  3, A  a11C11  a12C12  a13C13
If A is 4  4, A  a11C11  a12C12  a13C13  a14C14

If A is n  n, A  a11C11  a12C12  a13C13    a1nC1n
These equations are called cofactor expansions (餘因子展開式)
of |A|.
Ch03_5
Example 3
Evaluate the determinant of the following matrix A.
 1 2  1
A  3 0
1
1
4 2
Solution
A  a11C11  a12C12  a13C13
 1(1) 2 0 1  2(1)3 3 1  (1)(1) 4 3 0
2 1
4 1
4 2
 [(0 1)  (1 2)]  2[(3 1)  (1 4)]  [(3  2)  (0  4)]
 2  2  6
 6
Ch03_6
Theorem 3.1
The determinant of a square matrix is the sum of the products of
the elements of any row or column and their cofactors.
A  ai1Ci1  ai 2Ci 2    ain Cin
ith row expansion:
jth column expansion: A  a1 j C1 j  a2 j C2 j    anj Cnj
Example 4
Find the determinant of the following matrix using the second row.
 1 2  1
A  3 0
1
1
4 2
Solution
A  a21C21  a22C22  a23C23
 3 2  1  0 1  1  1 1 2
2
1
4
1 4 2
 3[(2 1)  (1 2)]  0[(11)  (1 4)]  1[(1 2)  (2  4)]
 12  0  6  6
隨堂作業：9(d) Ch03_7
Example 5
Evaluate the determinant of the following 4  4 matrix.
1
2
0  1
7  2
1
0
0
4
0
2
3
5
0  3
Solution
A  a13C13  a23C23  a33C33  a43C43
 0(C13 )  0(C23 )  3(C33 )  0(C43 )
2
1
4
 3 0 1
2
0
1 3
2  6(3  2)  6
 3(2)  1
1 3
隨堂作業：11(b)
Ch03_8
Example 6
Solve the following equation for the variable x.
x x 1  7
1 x  2
Solution
Expand the determinant to get the equation
x( x  2)  ( x  1)(1)  7
Proceed to simplify this equation and solve for x.
x2  2x  x 1  7
x2  x  6  0
( x  2)( x  3)  0
x  2 or 3
There are two solutions to this equation, x = – 2 or 3.
隨堂作業：14
Ch03_9
Computing Determinants of 2  2
and 3  3 Matrices
a11

A
a21
a12 
a22   A  a11a22  a12 a21
 a11
A  a21
a
 31
a12
a22
a32
a13 
 a11
a23   a21
a
a33 
 31
a12
a22
a32
a13  a11
a23  a21
a33  a31
a12
a22
a32
 A  a11a22 a33  a12 a23 a31  a13 a21a32
(diagonalproductsfromleft toright)
 a13 a22 a31  a11a23 a32  a12 a21a33
(diagonalproductsfrom right toleft)
Note：此法不可用在4  4及更大的矩陣!
Ch03_10
Homework
Exercise 3.1:
3, 9, 11, 14
Ch03_11
3.2 Properties of Determinants
Theorem 3.2
Let A be an n  n matrix and k be a nonzero scalar.
(a) If A  B or A  B , then |B| = k|A|.
kRj
kCj
(b) If A  B or A  B, then |B| = –|A|.
Ri Rj
Ci  Cj
(c) If A  B or A  B, then |B| = |A|.
Ri kRj
Ci  kCj
Proof (a)
|A| = aj1Cj1 + aj2Cj2 + … + ajnCjn
|B| = kaj1Cj1 + kaj2Cj2 + … + kajnCjn
|B| = k|A|.
Ch03_12
Example 1
3
4 2
Evaluate the determinant  1  6
3.
2
9 3
Solution
3
4 2
1  6
2
3
9 3
3 0 2

C2  2 C3
1 0
3  (3)
2 3 3
3
2
1
3
 21
隨堂作業：4(a)(b)(c)
Ch03_13
Example 2
4 3
 1
2 5, and |A| = 12 is known.
If A   0
 2  4 10
Evaluate the determinants of the following matrices.
4 3
 1 12 3
 1
 1 4 3
(a ) B1   0
6 5 (b) B2   2  4 10 (c) B3  0 2 5
 2  12 10
 0
0 4 16
2 5
Solution
(a) A  B1
Thus |B1| = 3|A| = 36.
3C 2
(b) A
 B2 Thus |B2| = – |A| = –12.
R 2 R 3
(c) A R 32 R1 B3 Thus |B3| = |A| = 12.
隨堂作業：10(b,d)
Ch03_14
Definition
A square matrix A is said to be singular (奇異) if |A|=0.
A is nonsingular if |A|0.
Theorem 3.3
Let A be a square matrix. A is singular if
(a) all the elements of a row (column) are zero.
(b) two rows (columns) are equal.
(c) two rows (columns) are proportional (成比例的). (i.e., Ri=cRj)
Proof
(a) Let all elements of the kth row of A be zero.
A  ak1Ck1  ak 2Ck 2    aknCkn  0Ck1  0Ck 2    0Ckn  0
(c) If Ri=cRj, then
 |A|=|B|=0
A  B , row i of B is [0 0 … 0].
Ri cRj
Ch03_15
Example 3
Show that the following matrices are singular.
 2 0  7
2  1 3
(a ) A   3 0
1 (b) B   1 2 4
 4 0
2 4 8
9
Solution
(a) All the elements in column 2 of A are zero. Thus |A| = 0.
(b) Row 2 and row 3 are proportional. Thus |B| = 0.
Ch03_16
Theorem 3.4
Let A and B be n  n matrices and c be a nonzero scalar.
(a) |cA| = cn|A|.
(b) |AB| = |A||B|.
(c) |At| = |A|.
1
1
(d) A  A (assuming A–1 exists)
Proof
(a)
(d)
A

cR1, cR 2 , ...,cRn
cA  cA  c n A
A  A1  A  A1  I  1
 A1 
1
A
Ch03_17
Example 4
If A is a 2  2 matrix with |A| = 4, use Theorem 3.4 to compute
the following determinants.
(a) |3A|
(b) |A2|
(c) |5AtA–1|, assuming A–1 exists
Solution
(a) |3A| = (32)|A| = 9  4 = 36.
(b) |A2| = |AA| =|A| |A|= 4  4 = 16.
1
(c) |5AtA–1| = (52)|AtA–1| = 25|At||A–1|  25 A  25.
A
Example 5
Prove that |A–1AtA| = |A|
隨堂作業：9(a,b,e)
Solution
1
1
1
1
A A A  (A A )A  A A A  A
t
t
t
1
A A
AA A
A
t
Ch03_18
Example 6
Prove that if A and B are square matrices of the same size, with A
being singular, then AB is also singular. Is the converse true?
Solution
()
()
|A| = 0  |AB| = |A||B| = 0
Thus the matrix AB is singular.
|AB| = 0  |A||B| = 0  |A| = 0 or |B| = 0
Thus AB being singular implies that either A or B is singular.
The inverse is not true.
Ch03_19
Determinant of an Upper Triangular
Matrix
Definition
A square matrix is called an upper triangular matrix (上三角矩
陣) if all the elements below the main diagonal are zero.
It is called a lower triangular matrix (下三角矩陣) if all the
elements above the main diagonal are zero.
3
0

0
1
0

0

0
4 0 7
8 2

2
3
5

1 5,
0 0 9
0 9

0 0 1
upper  t riangular
8
1

7

4
0 0 0
7 0 0 

4
0
0
 2 1 0 ,



0 2 0
 3 9 8

5 8 1
lower  t riangular
Ch03_20
Example
2  1 9
Let A  0 3 0
0 0  5
0 0 0
4
6, find A .
3
1
3
0 6
| A | 2 0  5 3  2  3  5 3  2  3  (5) 1  30
0 1
0
0 1
Note: The determinant of a triangular matrix is the product of its
diagonal elements.
快速求行列式的方法：
利用elementary row operations 將矩陣三角化(對角線上的
數字可以不等於1)，再將對角線上的數字相乘即可
Ch03_21
Numerical Evaluation of a Determinant
Example 8
2
4 1
Evaluation the determinant  2  5 4 .
4
9 10
Solution
2
2
4
4
1

2
 5 4 R2  R1 0
9 10 R 3  ( 2) R1 0

4
1
1 5
1 8
1  2  (1) 13  26
R3  R 2 0  1 5
0 0 13
2
4
Ch03_22
Example 9
1 2
4
2 5
Evaluation the determinant.  1
2  2 11
Solution
1 2
4

1 2 4
1
2  5 R2  R1 0
0 1
2  2 11 R 3  (2)R1 0
2 3

1 2
( 1) 0
R2  R3
0
4
2
3
0 1
 (1) 1 2  (1)  2
Ch03_23
Example 10
1 1
1
1
Evaluation the determinant.
2 2
6 6
0
2
3
5
2
3
4
1
Solution
1 1
1
1
2 2
6 6
0
2
3
5
2

3
R2  R1
4 R3  (2)R1
1 R4  (6)R1
1 1
0 0
0 0
0 0
0
2
2
5 0
3
0
5  11
diagonal element is zero and
all elements below this
diagonal element are zero.
隨堂作業：13(a,b)
Ch03_24
Homework
Exercise 3.2:
4, 9, 10, 13, 14
Ch03_25
3.3 Determinants, Matrix Inverse,
and Systems of Linear Equations
Definition
Let A be an n  n matrix and Cij be the cofactor of aij.
The matrix whose (i, j)th element is Cij is called the matrix of
cofactor of A.
The transpose of this matrix is called the adjoint of A and is
denoted adj(A).
 C11 C12  C1n 
C21 C22  C2 n 
 




C
C

C
n2
nn 
 n1
matrixof cofactors
 C11 C12  C1n 
C

C

C
22
2n 
 21
 




C
C

C
n2
nn 
 n1
adjoint matrix
t
Ch03_26
Example 1
Give the matrix of cofactors and the adjoint matrix of the
following matrix A.
0
3
 2
A   1
4  2
5
 1  3
Solution The cofactors of A are as follows.
C11  4  2  14 C12    1  2  3 C13   1 4  1
3
5
1
5
1 3
0 6
C21   0 3  9 C22  2 3  7
C23   2
3 5
1 5
1 3
2
3
0
3
2 0 8
C31 
 12 C32  
C

1
33
4 2
1 4
1  2
The matrix of cofactors
The adjoint of A is
of A is  14 3  1
 14  9  12
  9 7 6
adj( A)   3
7
1


6
8
 1
 12 1 8


Ch03_27
Theorem 3.5
Let A be a square matrix with |A|  0. A is invertible with
1
1
A  adj( A)
A
Proof
Consider the matrix product Aadj(A). The (i, j)th element of this
product is
(i, j ) th element  (row i of A)  (column j of adj( A))
 C j1 
C j 2 
 ai1 ai 2  ain  
  
C jn 
 ai1C j1  ai 2C j 2    ain C jn
Ch03_28
Proof of Theorem 3.6
If i = j, ai1C j1  ai 2C j 2   ainC jn  A.
If i  j, let A

Rj is replacedby Ri
B. Matrices A and B have the same cofactors
Cj1, Cj2, …, Cjn.
So ai1C j1  ai 2C j 2   ainC jn  B  0.
 A if i  j
Therefore (i. j ) th element  
0 if i  j
1

Since |A|  0, A adj( A)   I n
A



1

Similarly,  adj( A)  A  I n .
A

row i = row j in B
 A adj(A) = |A|In
1
Thus A  adj( A)
A
1
Ch03_29
Theorem 3.6
A square matrix A is invertible if and only if |A|  0.
Proof
() Assume that A is invertible.
 AA–1 = In.
 |AA–1| = |In|.
 |A||A–1| = 1
 |A|  0.
() Theorem 3.6 tells us that if |A|  0, then A is invertible.
A–1 exists if and only if |A|  0.
Ch03_30
Example 2
Use a determinant to find out which of the following matrices are
invertible.
 2 4  3
 1 2  1
1

1
4
2




A
B
C   4 12  7 D   1 1 2
3 2
2 1
1
 1 0
 2 8 0
Solution
|A| = 5  0.
|B| = 0.
|C| = 0.
|D| = 2  0.
A is invertible.
B is singular. The inverse does not exist.
C is singular. The inverse does not exist.
D is invertible.
Ch03_31
Example 3
Use the formula for the inverse of a matrix to compute the inverse
0
3
 2
of the matrix
A   1
4  2
5
 1  3
Solution
|A| = 25, so the inverse of A exists.We found adj(A) in Example 1
 14  9  12
adj( A)   3
7
1
6
8
 1
1
1 14  9  12
A  adj( A) 
3
7
1
A
25  1

6
8


1
隨堂作業：7(a)
Ch03_32
Determinants and Systems of Linear Equations
Theorem 3.7
Let AX = B be a system of n linear equations in n variables.
(1) If |A|  0, there is a unique solution.
(2) If |A| = 0, there may be many or no solutions.
Proof
(1) If |A|  0
 A–1 exists (Thm 3.6)
 there is then a unique solution given by X = A–1B (Thm 2.8).
(2) If |A| = 0
 since A  C implies that if |A|0 then |C|0 (Thm 3.2).
 the reduced echelon form of A is not In.
 The solution to the system AX = B is not unique.
 many or no solutions.
Ch03_33
Example 4
Determine whether or not the following system of equations has
an unique solution.
3 x1  3 x2  2 x3  2
4 x1  x2  3x3  5
7 x1  4 x2  x3  9
Solution
Since
3 3 2
4 1
3 0
7 4
1
Thus the system does not have an unique solution.
隨堂作業：14(b)
Ch03_34
Theorem 3.8 Cramer’s Rule
Let AX = B be a system of n linear equations in n variables such
that |A|  0. The system has a unique solution given by
A1
A2
An
x1 
, x2 
, ... , xn 
A
A
A
Where Ai is the matrix obtained by replacing column i of A with B.
Proof
|A|  0  the solution to AX = B is unique and is given by
X  A1 B
1
 adj( A) B
A
Ch03_35
Proof of Cramer’s Rule
xi, the ith element of X, is given by
1
xi  [row i of adj( A)]  B
A
 b1 
1
 
 C1i C2i Cni b2 

A
b 
 n
1
 (b1C1i  b2C2i    bnCni )
A
Ai
Thus xi 
A
the cofactor expansion of |Ai|
in terms of the ith column
Ch03_36
Example 5
Solving the following system of equations using Cramer’s rule.
x1  3 x2  x3  2
2 x1  5 x2  x3  5
x1  2 x2  3x3  6
Solution
The matrix of coefficients A and column matrix of constants B are
 1 3 1
  2
A  2 5 1 and B    5
 1 2 3
 6
It is found that |A| = –3  0. Thus Cramer’s rule be applied. We
get
 2 3 1
 1  2 1
 1 3  2
A1    5 5 1 A2  2  5 1 A3  2 5  5
6 3
6
 6 2 3
 1
 1 2
Ch03_37
Giving A1  3, A2  6, A3  9
Cramer’s rule now gives
A1  3
A2
A3  9
6
x1 

 1, x2 

 2, x3 

3
A 3
A 3
A 3
The unique solution is x1  1, x2  2, x3  3.
隨堂作業：12(c)
Ch03_38
Homogeneous Systems of Linear Equations
Example 6
Determine values of  for which the following system of
equations has nontrivial solutions. Find the solutions for each
value of .
(  2) x  (  4) x  0
1
Solution
2
2 x1  (  1) x2  0
homogeneous system
 x1 = 0, x2 = 0 is the trivial solution.
 nontrivial solutions exist  many solutions
 2 4
0
2  1
 (  2)(  1)  2(  4)  0  2    6  0  (  2)(  3)  0
  = – 3 or  = 2.
Ch03_39
 = – 3 results in the system
 x1  x2  0
2 x1  2 x2  0
This system has many solutions, x1 = r, x2 = r.
 = 2 results in the system
4 x1  6 x2  0
2 x1  3x2  0
This system has many solutions, x1 = – 3r/2, x2 = r.
隨堂作業：15
Ch03_40
Homework
Exercise 3.3:
7, 14, 15
Ch03_41
3.4 Eigenvalues and Eigenvectors
Definition
Let A be an n  n matrix. A scalar  is called an eigenvalue (特徵
值,固有值) of A if there exists a nonzero vector x in Rn such that
Ax = x.
The vector x is called an eigenvector corresponding to .
>0
<0
Figure 3.1
Ch03_42
Computation of Eigenvalues and Eigenvectors
Let A be an n  n matrix with eigenvalue  and corresponding
eigenvector x.
 Ax = x
 Ax – x = 0
 (A – In)x = 0
 a system of linear equations, and x=0 is a solution.
 we need nonzero solutions  many solutions
 |A – In| = 0
 Solve |A – In| = 0 for  leads to all the eigenvalues of A.
On expending the determinant |A – In|, we get a polynomial in .
This polynomial is called the characteristic polynomial of A.
The equation |A – In| = 0 is called the characteristic equation of
A.
Ch03_43
Example 1
Find the eigenvalues and eigenvectors of the matrix
  4  6
A
5 
3
Solution
Find the characteristic polynomial of A:
A  I 2   4  6   1 0   4    6 
5 
5   
 3
0 1  3
A  I 2  (4   )(5   )  18  2    2
2    2  0

(  2)(  1)  0    2 or 1
The eigenvalues of A are 2 and –1.
The corresponding eigenvectors are found by using these values
of  in the equation(A – I2)x = 0.
Ch03_44
(1)  = 2
(A – 2I2)x = 0
 6  6  x1 
0


3

3   x2 

 6 x1  6 x2  0

 3 x1  3 x2  0
 x1  r
 
, rR
 x2   r
Thus the eigenvectors of A corresponding to  = 2 are
 1
x  r  , r  R, r  0
 1
Ch03_45
(2)  = –1
(A + 1I2)x = 0
 3  6  x1 
0


3

6   x2 

  3 x1  6 x2  0

 3 x1  6 x2  0
x  2s
  1
, sR
 x2  s
Thus the eigenvectors of A corresponding to  = –1 are
  2
x  s  , s  R, s  0
 1
隨堂作業：9
先不求eigenspaces
Ch03_46
Eigenspaces
Theorem 3.9
Let A be an n  n matrix and  an eigenvalue of A. The set V of
all eigenvectors corresponding to , together with the zero vector,
is a subspace of Rn. This subspace is called the eigenspace of .
Proof
Let x1, x2 V and let c be a scalar. Then Ax1 = x1 and Ax2 = x2.
Hence,
Ax1  Ax 2  x1  x 2
A(x1  x 2 )   (x1  x 2 )
Thus x1+x2 V. The set is closed under addition. Further,
cAx1  cx1
A(cx1 )   (cx1 )
Therefore cx1 V. The set is closed under scalar multiplication.
Thus V is a subspace of Rn.
Ch03_47
Example 2
Find the eigenvalues and corresponding eigenspaces of the matrix
 5 4 2
A  4 5 2
2 2 2


Solution
5
4
2
1  1 
0
A  I 3  4 5   2  4
5
2
2
2
2
2
2
2

1 
0
4
9
2
4
0
2
2
 (1   )[(9   )(2   )  8]  (1   )[2  11  10]
 (1   )(  10)(  1)  (  10)(  1) 2  0
=10 or 1
Ch03_48
(1)  = 10
( A  10I 3 )x  0
 5 4 2   x1 
 4  5 2   x2   0
 2 2  8  x3 

 x1  2r

 x2  2 r , r  R
x  r
 3
Thus the eigenspace of  = 10 is
 2
{r 2 | r  R}
1 
2
The set{ 2 } is a basis, and the dimension is 1.
1 
Ch03_49
(2)  = 1
( A  1I 3 )x  0
4 4 2  x1 
 4 4 2   x2   0
2 2 1   x3 

 x1   s  t

s,t  R
 x2  s ,
 x  2t
 3
 1  1
Thus the eigenspace of  = 1 is {s  1   t  0  | s, t  R}
 0   2 
 1  1
The set { 1 ,  0 } is a basis, and the dimension is 2.
 0   2 
隨堂作業：10
If an eigenvalue occurs as a k times repeated root of the characteristic
equation, we say that it is of multiplicity (重根數) k.
Thus =10 has multiplicity 1, while =1 has multiplicity 2.
Ch03_50
Example 3
Let A be an n  n matrix with eigenvalues 1, …, n, and
corresponding eigenvectors x1, …, xn. Prove that if c  0, then the
eigenvalues of cA are c1, …, cn with corresponding
eigenvectors x1, …, xn.
Solution
Let i be one of eigenvalues of A with corresponding eigenvectors xi.
 Axi = ixi.
 cAxi = cixi
Thus ci is an eigenvalues of cA with corresponding eigenvector xi.
Further, since cA is an n  n matrix, the characteristic polynomial of
A is of degree n. The characteristic equation has n roots, implying
that cA has n eigenvalues. The eigenvalues of cA are therefore c1, …,
cn with corresponding eigenvectors x1, …, xn.
Ch03_51
Homework
Exercise 3.4:
9, 10, 13, 24, 26, 32
Ex24: Prove that if A is a diagonal matrix, then its eigenvalues are
the diagonal elements.
Ex26: Prove that if A and At have the same eigenvalues.
Ex32: Prove that the constant term of the characteristic polynomial
of a matrix A is |A|.
(3.5節跳過)
Ch03_52