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By:
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2. Chapter 7
Eigenvalues and Eigenvectors
7.1 Eigenvalues and Eigenvectors
7.2 Diagonalization
7.3 Symmetric Matrices and Orthogonal Diagonalization
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Elementary Linear Algebra 投影片設計編製者
R. Larsen et al. (6 Edition) 淡江大學 電機系 翁慶昌 教
授

3. 3/53
7.1 Eigenvalues and Eigenvectors
 Eigenvalue problem:
If A is an n´n matrix, do there exist nonzero vectors x in Rn
such that Ax is a scalar multiple of x ？
 Eigenvalue and eigenvector:
A：an n´n matrix
 ：a scalar
x ： a nonzero vector in Rn
Eigenvalue
Ax =lx
Eigenvector
 Geometrical Interpretation
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Elementary Linear Algebra: Section 7.1, pp.421-422

4. 4/53
 Ex 1: (Verifying eigenvalues and eigenvectors)
ù
2 0
1
= é0
A úû
=
0 1
úû
é
êë
-
ù
êë
1 x
úûù
0
êë= é1
Ax = é
2 0 ù
é
1
úû
úû
2
-
0
= 2 é 1 = x 2
1 0 1
0
0 1 ù
êë
ù
êëé = úû
ù
êë
úû
êë
Eigenvalue
Eigenvector
úû
úû
Ax = é
2 0 ù
é
0
é
úû
0
é 0 -
= - 1
= - 1 = - x ( 1)
2 0 1
1
1 2 ù
êë
ù
êë
ù
êë
úû
êë
Eigenvalue
Eigenvector
2 x
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Elementary Linear Algebra: Section 7.1, p.423

5. 5/53
 Thm 7.1: (The eigenspace of A corresponding to l)
If A is an n´n matrix with an eigenvalue l, then the set of all
eigenvectors of l together with the zero vector is a subspace
of Rn. This subspace is called the eigenspace of l .
Pf:
x1 and x2 are eigenvectors corresponding to l
( . . , ) 1 1 2 2 i e Ax = lx Ax =lx
A x + x = Ax + Ax = l x +l x = l x +
x
(1) ( ) ( )
1 2 1 2 1 2 1 2
i e x +
x λ
( . . is an eigenvector corresponding to )
1 2
A cx = c Ax = c x = cx
(2) ( ) ( ) ( ) ( )
1 1 1 1
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( . . is an eigenvector corresponding to )
1
l
l l
i e cx
Elementary Linear Algebra: Section 7.1, p.424

6. 6/53
 Ex 3: (An example of eigenspaces in the plane)
Find the eigenvalues and corresponding eigenspaces of
ù
úû
é-
=
êë
1 0
0 1
A
ù
úû
é-
= úû
êë
ù
If v = (x, y)
êë é
úû ù
é-
=
êë
x
y
x
y
A
1 0
0 1
v
For a vector on the x-axis Eigenvalue 1 1 l = -
ù
úû
1 0 x x x
é
- = úû
êë
ù
é-
= úû
êë
ù
é
êë
ù
úû
êë é-
0
1
0 0
0 1
Sol:
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Elementary Linear Algebra: Section 7.1, p.425

7. 7/53
For a vector on the y-axis
ù
úû
0
1
é
= úû
êë
ù
0 0
é
= úû
êë
ù
é
êë
ù
úû
é-
êë
y y y
1 0
0 1
Eigenvalue 1 2 l =
Geometrically, multiplying a vector (x,
y) in R2 by the matrix A corresponds to a
reflection in the y-axis.
1 1 l = -
The eigenspace corresponding to is the x-axis.
The eigenspace corresponding to is the y-axis.
1 2 l =
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Elementary Linear Algebra: Section 7.1, p.425

8.  Thm 7.2: (Finding eigenvalues and eigenvectors of a matrix AÎMn´n )
Let A is an n´n matrix.
(1) An eigenvalue of A is a scalar l such that d e t ( l I - A ) = 0 .
(2) The eigenvectors of A corresponding to l are the nonzero
8/53
solutions of .
 Note:
Ax = lx Þ (lI - A)x = 0
(homogeneous system)
 Characteristic polynomial of AÎMn´n:
det( I - A) = ( I - A) = n + c n - + + c +
c
n
1 1 0
1
- l l l l  l
 Characteristic equation of A:
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det(lI - A) = 0
(lI - A)x = 0
If ( l I - A ) x = 0 has nonzero solutions iff d e t ( l I - A ) = 0 .
Elementary Linear Algebra: Section 7.1, p.426

9. 9/53
 Ex 4: (Finding eigenvalues and eigenvectors)
ù
úû
é
2 12
=
1 5
êë
-
-
A
Sol: Characteristic equation:
2 12
l
-
- l
+
1 5
3 2 ( 1)( 2) 0
- =
det( I )
= l 2 + l + = l + l
+ =
l A
Þl = -1, - 2
Eigenvalues : 1, 2 1 2 l = - l = -
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Elementary Linear Algebra: Section 7.1, p.426

10. 10/53
(2) 2 2 l = -
t
4 4
é
= úû
t
ù
3 3
ù
é
= úû
, 0
1
3 12
é
-
é -
1 4
~
ù
ù
4 12
é
-
é -
1 3
~
0 0
3 12
ù
4 12
1 3
0
ù
é
= úû
é
= úû
0
0
1 3
é
-
( I )
x
1
é
é
= úû
x
1
2
x
1
x
1
2
2
¹ úû
êë
ù
ù
é
= úû
êë
ù
é
êë
ù
é
ù
Þ úû
êë
ù
úû
é
-
êë
-
úû
êë
ù
êë
úû
êë
-
Þ - =
t t
t
x
x
A x

l
(1) 1 1 l = -
, 0
1
0 0
1 4
0
1 4
( I )
2
2
1
¹ úû
êë
ù
êë
ù
êë
Þ úû
êë
úû
êë
-
úû
êë
ù
êë é
úû
êë
-
Þ - =
t t
t
x
x
A x

l
Check : Ax x i l
=
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Elementary Linear Algebra: Section 7.1, p.426

11. 11/53
 Ex 5: (Finding eigenvalues and eigenvectors)
Find the eigenvalues and corresponding eigenvectors for
the matrix A. What is the dimension of the eigenspace of
each eigenvalue?
ù
ú ú
û
2 1 0
é
=
0 0 2
ê ê
ë
0 2 0
A
Sol: Characteristic equation:
l
- -
I - = = l
- 3 =
( 2) 0
2 1 0
l
-
0 2 0
l
-
0 0 2
l A
Eigenvalue: l = 2
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Elementary Linear Algebra: Section 7.1, p.428

12. 12/53
The eigenspace of A corresponding to l = 2 :
ù
ú ú ú
û
é
=
ê ê ê
ë
ù
ú ú ú û
é
ê ê ê
ë
ù
ú ú ú
û
é -
ê ê ê
ë
- =
0
0
0
0 1 0
0 0 0
0 0 0
( I )
x
1
2
x
3
x
l A x
0
ù
é
+
0
, , 0
1
1
0
0
s
0
0 1 0
~
0 0 0
0 0 0
é -
0 1 0
0 0 0
0 0 0
x
1
2
3
¹
ú ú ú
û
ê ê ê
ë
ù
ú ú ú
û
é
=
ê ê ê
ë
ù
ú ú ú
û
é
=
ê ê ê
ë
ù
ú ú ú
û
é
Þ
ê ê ê
ë
ù
ú ú ú
û
é
ê ê ê ë
ù
ú ú ú
û
ê ê ê
ë
s t s t
t
x
x

ü
é
s t s t R l
, : the eigenspace of A corresponding to 2
0
0
é
+
1
1
0
0
=
ïþ
ïý
ì
ïî
ïí
Î
ù
ú ú ú
û
ê ê ê
ë
ù
ú ú ú û
ê ê ê
ë
Thus, the dimension of its eigenspace is 2.
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Elementary Linear Algebra: Section 7.1, p.428

13. 13/53
 Notes:
(1) If an eigenvalue l1 occurs as a multiple root (k times) for
the characteristic polynominal, then l1 has multiplicity k.
(2) The multiplicity of an eigenvalue is greater than or equal
to the dimension of its eigenspace.
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Elementary Linear Algebra: Section 7.1, p.428

14. 14/53
 Ex 6：Find the eigenvalues of the matrix A and find a basis
for each of the corresponding eigenspaces.
ù
ú ú ú ú
û
1 0 0 0
é
= -
1 0 0 3
ê ê ê ê
1 0 2 0
ë
0 1 5 10
A
Sol: Characteristic equation:
1 0 0 0
- -
-
0 1 5 10
- -
1 0 2 0
- -
1 0 0 3
( 1) ( 2)( 3) 0
I
- =
= l - 2 l - l
- =
l
l
l
l
l A
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Eigenvalues : 1, 2, 3 1 2 3 l = l = l =
Elementary Linear Algebra: Section 7.1, p.429

15. 15/53
(1) 1 1 l =
ù
ú ú ú ú
û
é
=
ê ê ê ê
ë
ù
ú ú ú ú
û
é
ê ê ê ê
ë
ù
ú ú ú ú
û
é
ê ê ê ê
ë
0 0 0 0
-
0 0 5 10
- -
1 0 1 0
- -
Þ - =
0
0
0
0
1 0 0 2
( I )
x
1
2
x
x
3
4
1
x
l A x
2
é-
+
0
ù
, , 0
2
1
0
é
=
1
0
0
2
é-
=
s
t
2
1 0 0 2
~
0 0 1 2
0 0 0 0
0 0 0 0
0 0 0 0
-
0 0 5 10
- -
1 0 1 0
1 0 0 2
x
1
2
x
x
3
4
¹
ú ú ú ú
û
ê ê ê ê
ë
ù
ú ú ú ú
û
ê ê ê ê
ë
ù
ú ú ú ú
û
ê ê ê ê
ë
ù
ú ú ú ú
û
é
Þ
ê ê ê ê
ë
ù
ú ú ú ú
û
é
ê ê ê ê ë
-
ù
ú ú ú ú
û
é
ê ê ê ê
ë
- -
s t s t
t
t
x
2
é-
0
Þ is a basis for the eigenspace
2
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1
0
é
1
ù
,
0
0
ü
ï ï
ý
ï ï
þ
ì
ï ï
í
ï ï
î
ù
ú ú ú ú
û
ê ê ê ê
ë
ú ú ú ú
û
ê ê ê ê
ë
of A corresponding to l =1
Elementary Linear Algebra: Section 7.1, p.429

16. 16/53
(2) 2 2 l =
ù
ú ú ú ú
û
é
=
ê ê ê ê
ë
ù
ú ú ú ú
û
é
ê ê ê ê
ë
ù
ú ú ú ú
û
é
ê ê ê ê
ë
1 0 0 0
0 1 5 10
-
-
1 0 0 0
- -
Þ - =
0
0
0
0
1 0 0 1
( I )
x
1
2
x
x
3
4
2
x
l A x
0
5
ù
, 0
0
é
=
1
0
é
=
t
5
0
1 0 0 0
0 1 5 0
~
0 0 0 1
0 0 0 0
1 0 0 0
-
0 1 5 10
1 0 0 0
ù
1 0 0 1
x
1
2
x
x
3
4
¹
ú ú ú ú
û
ê ê ê ê
ë
ù
ú ú ú ú
û
ê ê ê ê
ë
ù
ú ú ú ú
û
é
Þ
ê ê ê ê
ë
ù
ú ú ú ú
û
é
ê ê ê ê
ë
-
ú ú ú ú
û
é
ê ê ê ê
ë
-
- -
t t
t
x
0
5
é
Þ is a basis for the eigenspace
1
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ü
ï ï
ý
ï ï
þ
ì
ï ï
í
ï ï
î
ù
ú ú ú ú
û
ê ê ê ê
ë
of A corresponding to l = 2
Elementary Linear Algebra: Section 7.1, p.429

17. 17/53
(3) 3 3 l =
ù
ú ú ú ú
û
é
=
ê ê ê ê
ë
ù
ú ú ú ú
û
é
ê ê ê ê
ë
ù
ú ú ú ú
û
é
ê ê ê ê
ë
2 0 0 0
0 2 5 10
-
-
-
Þ - =
0
0
0
0
1 0 1 0
1 0 0 0
( I )
x
1
2
x
x
3
4
3
x
l A x
0
ù
5
é
-
=
, 0
0
1
0
5
é
-
=
0
1 0 0 0
0 1 0 5
~
0 0 1 0
0 0 0 0
2 0 0 0
0 2 5 10
1 0 1 0
1 0 0 0
x
1
2
x
x
3
4
¹
ú ú ú ú
û
ê ê ê ê
ë
ù
ú ú ú ú
û
ê ê ê ê
ë
ù
ú ú ú ú
û
é
Þ
ê ê ê ê
ë
ù
ú ú ú ú
û
é
ê ê ê ê
ë
ù
ú ú ú ú
û
é
ê ê ê ê
ë
-
-
-
t t
t
t
x
0
ù
5
é
-
Þ is a basis for the eigenspace
0
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com
ü
ï ï
ý
ï ï
þ
ì
ï ï
í
ï ï
î
ú ú ú ú
û
ê ê ê ê
ë
of A corresponding to l = 3
Elementary Linear Algebra: Section 7.1, p.429

18. 18/53
 Thm 7.3: (Eigenvalues of triangular matrices)
If A is an n´n triangular matrix, then its eigenvalues are
the entries on its main diagonal.
 Ex 7: (Finding eigenvalues for diagonal and triangular matrices)
ù
ú ú ú
û
é
ê ê ê
ë
2 0 0
1 1 0
-
= -
5 3 3
(a)A
ù
ú ú ú ú
û
é
ê ê ê ê
ë
1 0 0 0 0
0 2 0 0 0
0 0 0 0 0
-
-
=
0 0 0 4 0
0 0 0 0 3
(b)A
Sol:
-
( ) I - = = l - l - l
+
( 2)( 1)( 3)
2 0 0
l
-
1 1 0
- - l
+
5 3 3
l
a l A
2, 1, 3 1 2 3 l = l = l = -
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( ) 1, 2, 0, 4, 3 1 2 3 4 5 b l = - l = l = l = - l =
Elementary Linear Algebra: Section 7.1, p.430

19. 19/53
 Eigenvalues and eigenvectors of linear transformations:
A number is called an eigenvalue of a linear transformation
® x x = x
T V V T
: if there is a nonzero vector such that ( ) .
The vector is called an eigenvector of corresponding to ,
and the setof all eigenvectors of (with the zero vector) is
l
called the eigenspace of .
l
l
l
l
T
x
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Elementary Linear Algebra: Section 7.1, p.431

20. 20/53
 Ex 8: (Finding eigenvalues and eigenspaces)
Find the eigenvalues and corresponding eigenspaces
ù
.
1 3 0
3 1 0
0 0 2
ú ú ú û
é
ê ê ê
ë
-
A =
Sol:
- = l l
( 2) ( 4)
- -
1 3 0
3 1 0
0 0 2
= + 2 -
+
- -
l
l
l
lI A
eigenvalues : 4, 2 1 2 l = l = -
The eigenspaces for these two eigenvalues are as follows.
= l
=
{(1, 1, 0)} Basis for 4
B
1 1
= - l
= -
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{(1, 1, 0),(0, 0, 1)} Basis for 2
B
2 2
Elementary Linear Algebra: Section 7.1, p.431

21. ®
(1) Let be the linear transformation whose standard matrix
is A in Ex. 8, and let be the basis of made up of three linear
21/53
 Notes:
independent eigenvectors found in Ex. 8. Then , the matrix of
relative to the basis ,is diagonal.
' {(1, 1, 0),(1, 1, 0),(0, 0, 1)}
3
3 3
= -
B
B'
A' T
B' R
T:R R
ù
ú ú
û
é
ê ê
4 0 0
= -
0 0 2
ë
0 2 0
-
A'
Eigenvectors of A Eigenvalues of A
(2) The main diagonal entries of the matrix A' are the eigenvalues of A.
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Elementary Linear Algebra: Section 7.1, p.431

22. 22/53
Keywords in Section 7.1:
 eigenvalue problem: 特徵值問題
 eigenvalue: 特徵值
 eigenvector: 特徵向量
 characteristic polynomial: 特徵多項式
 characteristic equation: 特徵方程式
 eigenspace: 特徵空間
 multiplicity: 重數
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23. 23/53
7.2 Diagonalization
 Diagonalization problem:
For a square matrix A, does there exist an invertible matrix P
such that P-1AP is diagonal?
 Diagonalizable matrix:
A square matrix A is called diagonalizable if there exists an
invertible matrix P such that P-1AP is a diagonal matrix.
(P diagonalizes A)  Notes:
B = P-1AP
(1) If there exists an invertible matrix P such that ,
then two square matrices A and B are called similar.
(2) The eigenvalue problem is related closely to the
diagonalization problem.
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Elementary Linear Algebra: Section 7.2, p.435

24. 24/53
 Thm 7.4: (Similar matrices have the same eigenvalues)
If A and B are similar n´n matrices, then they have the
same eigenvalues.
Pf:
A and B are similarÞ B = P-1AP
1 1 1 1
B P AP P P P AP P A P
l l l l
- = - = - = -
I I I ( I )
- - -
1 1 1
P A P P P A P P A
= - = - = -
A
= -
- - - -
I
I I I
l
l l l
A and B have the same characteristic polynomial.
Thus A and B have the same eigenvalues.
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Elementary Linear Algebra: Section 7.2, p.436

25. 25/53
 Ex 1: (A diagonalizable matrix)
ù
ú ú
û
1 3 0
é
ê ê
=
0 0 2
ë
3 1 0
-
A
Sol: Characteristic equation:
l
- -
I - = = l - l
+ 2 =
( 4)( 2) 0
1 3 0
- l
-
3 1 0
l
+
0 0 2
l A
Eigenvalues : 4, 2, 2 1 2 3 l = l = - l = -
(See p.403 Ex.5)
1
é
1
0
(1) 4 Eigenvector : 1
ù
ú ú ú
û
ê ê ê
ë
l = Þ p =
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Elementary Linear Algebra: Section 7.2, p.436

26. 26/53
(See p.403 Ex.5)
0
0
é
=
1
1
ù
é
l = - Þ p = - p
(2) 2 Eigenvector : 2 3
1
,
0
ù
ú ú ú
û
ê ê ê
ë
ú ú ú û
ê ê ê
ë
ù
ú ú ú
û
é
ê ê ê
ë
4 0 0
0 2 0
-
[ ] 1
1 2 3 P p p p P AP
Þ = -
ù
ú ú ú
û
1 1 0
é
ê ê ê ë
= = - -
0 0 2
1 1 0
0 0 1
ù
ù
ú ú ú
û
é
é
ê ê ê
ë
2 0 0
0 4 0
2 0 0
-
-
-
 Notes:
P p p p P AP
Þ =
ù
ù
ú ú ú
û
é
1 1 0
= = -
é
ê ê ê
1 1 0
1 0 1
= = -
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ë
ú ú ú
û
ê ê ê
ë
-
-
Þ =
ú ú ú
û
ê ê ê
ë
-
0 2 0
0 0 4
1 0 1
0 1 0
(2) [ ]
0 0 2
0 0 1
(1) [ ]
1
2 3 1
1
2 1 3
P p p p P AP
Elementary Linear Algebra: Section 7.2, p.436

27. 27/53
 Thm 7.5: (Condition for diagonalization)
An n´n matrix A is diagonalizable if and only if
it has n linearly independent eigenvectors.
Pf:
(Þ)A is diagonalizable
1
= -
P D P AP
there exists an invertible s.t. is diagonal
P = p p  p D = diag
l l  l
n n Let [ ] and ( , , , )
1 2 1 2
0 
0
0 
0
   
0 0
PD p p p
é
[ ]
[ ]
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2 2
2
1
1 2
n n
n
n
p p p
l l l
l
l
l



=
ù
ú ú ú ú
û
ê ê ê ê
ë
=
Elementary Linear Algebra: Section 7.2, p.437

28. 28/53
[ ] [ ] 1 2 n 1 2 n AP = A p p  p = Ap Ap  Ap
AP PD
=
l

= =
Ap p , i 1, 2, ,
n
i e p P A
( . . the column vector of are eigenvectors of )
i
i i i 
is invertible , , , are linearly independent. 1 2 n P Þ p p  p
A has n linearly independent eigenvectors.
n A n p p p
1 2 Ü
( ) has linearly independent eigenvectors , ,
l l l
n

with corresponding eigenvalues , ,
1 2
Ap p i n i i i i.e. = l , =1, 2,,
Let [ ] 1 2 n P = p p  p
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Elementary Linear Algebra: Section 7.2, p.438

29. 29/53
AP A p p p
[ 
]
1 2
Ap Ap Ap
[ ]
n

1 2
p p p
ù
l l l
[ ]
é
0 
0
0 
0
1 1 2 2
p p p PD
   
n
n
n n
n
=
ú ú ú ú
û
ê ê ê ê
ë
=
=
=
=
l
l
l



0 0
[ ]
2
1
1 2
p p p P n
, , , are linearly independent is invertible
1
 
1 1
P AP D
=
is diagonalizable
A
Þ
Þ
-
Note: If n linearly independent vectors do not exist,
v i d e oth.eend anh on´len .mcaotrmix A is not diagonalizable.
Elementary Linear Algebra: Section 7.2, p.438

30. I 1 l A I A p
A does not have two (n=2) linearly independent eigenvectors,
so A is not diagonalizable.
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 Ex 4: (A matrix that is not diagonalizable)
Show that the following matrix is not diagonalizable.
ù
úû
1 2
= é
0 1
êë
A
Sol: Characteristic equation:
l A l
Eigenvalue : 1 1 l =
- = - - l l
I 1 2 = - 2 = -
0 1 ( 1) 0
ù
úû
1
é
= Þ úû
êë
ù
é
0 1
~
êë
ù
úû
é -
êë
- = - =
0
Eigenvector :
0 0
0 2
0 0
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Elementary Linear Algebra: Section 7.2, p.439

31. 

ù
é
- = = l
1 
31/53
 Steps for diagonalizing an n´n square matrix:
Step 1: Find n linearly independent eigenvectors
for A with corresponding eigenvalues
Step 2: Let [ ] 1 2 n P = p p  p
n p , p , , p 1 2 
Step 3:
0 0
1
0 0
2
P AP D Ap p i n i i i
   
n
, where , 1, 2, ,
0 0

= =
ú ú ú ú
û
ê ê ê ê
ë
l
l
l
Elementary Linear Algebra: Section 7.2, p.439
n l ,l , ,l 1 2 
Note:
The order of the eigenvalues used to form P will determine
vthide eorode.er idn hwohilceh. tchoe meigenvalues appear on the main diagonal of D.

32. 32/53
 Ex 5: (Diagonalizing a matrix)
- -
1 1 1
1 3 1
ù
ú ú ú
- -
3 1 1
-
û
P P 1AP
A
é
ê ê ê
ë
=
Find a matrix such that is diagonal.
Sol: Characteristic equation:
l
-
I - = = l - l + l
- =
( 2)( 2)( 3) 0
1 1 1
- l
- -
1 3 1
- l
+
3 1 1
l A
Eigenvalues : 2, 2, 3 1 2 3 l = l = - l =
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Elementary Linear Algebra: Section 7.2, p.440

33. 1
ù
ù
1
Þ = -
1
33/53
2 1 l =
ù
ú ú
û
é
1 0 1
~
ê ê
ë
ù
ú ú
1 1 1
1 1 1
û
é
ê ê
Þ - = - - -
ë
-
0 1 0
0 0 0
3 1 3
I 1 l A
ú ú ú
û
é-
ê ê ê
ë
Þ =
1
ù
ú ú ú
0 1
û
é-
=
ê ê ê
ë
ù
ú ú ú
û
é-
=
ê ê ê ë
ù
ú ú ú
û
é
Þ
ê ê ê
ë
0
1
Eigenvector :
0
1
x
1
2
3
t p
t
t
x
x
2 2 l = -
ù
ú ú
û
é -
1 0
~
ê ê
I 4
ë
ù
ú ú
3 1 1
1 5 1
û
é
ê ê
ë
- - -
- -
-
Þ - =
0 1
1
0 0 0
3 1 1
1
4
2 l A
ú ú ú
û
é
ê ê ê
ë
ù
ú ú ú
1
1
4 2
û
é
ê ê ê
= -
ë
ù
ú ú ú
û
é
= -
ê ê ê
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ë
ù
ú ú ú
û
é
Þ
ê ê ê
ë
4
Eigenvector :
4
1
1
4
1
4
x
1
2
3
t p
t
t
t
x
x
Elementary Linear Algebra: Section 7.2, p.440

34. 34/53
3 3 l =
ù
ú ú
1 0 1
~
0 1 1
û
é
ê ê
ë
-
ù
ú ú
2 1 1
1 0 1
û
é
ê ê
Þ - = - -
ë
-
0 0 0
3 1 4
I 3 l A
1
ù
ú ú ú
û
é-
ê ê ê
ë
Þ =
1
ù
ú ú ú
û
é-
=
ê ê ê
ë
ù
ú ú ú
û
é-
=
ê ê ê ë
ù
ú ú ú
û
é
Þ
ê ê ê
ë
1
1
Eigenvector :
1
1
3
x
1
2
3
t p
t
t
t
x
x
é
- -
ù
ú ú ú
û
P p p p
é
ê ê ê
2 0 0
Þ = -
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ë
1 1 1
ù
ú ú ú
û
ê ê ê
ë
-
= =
-
0 2 0
0 0 3
0 1 1
1 4 1
Let [ ]
1
1 2 3
P AP
Elementary Linear Algebra: Section 7.2, p.440

35. 35/53
 Notes: k is a positive integer
ù
ú ú ú ú
û
é
ê ê ê ê
(1) 2
ë
D
Þ =
ù
ú ú ú ú û
é
=
ê ê ê ê
ë
0 
0
0 
0
   
k
n
k
k
k
d
d
0 
0
0 
0
   
n d
d
d
d
D


0 0
0 0
1
2
1
k -
1
k
D P AP
D P AP
( )
- - -
1 1 1
Þ =
P AP P AP P AP
= ×××
( )( ) ( )
- - - -
1 1 1 1
P A PP A PP PP AP
= ×××
-
1
P AA AP
= ×××
-
1
P A P
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1
1
( ) ( ) ( )
(2)
-
-
k k
=
=
=
A PD P
k
Elementary Linear Algebra: Section 7.2, p.445

36. 36/53
 Thm 7.6: (Sufficient conditions for diagonalization)
If an n´n matrix A has n distinct eigenvalues, then the
corresponding eigenvectors are linearly independent and
A is diagonalizable.
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Elementary Linear Algebra: Section 7.2, p.442

37. 37/53
 Ex 7: (Determining whether a matrix is diagonalizable)
ù
ú ú ú
û
é
ê ê ê
ë
1 2 1
-
-
=
0 0 1
0 0 3
A
Sol: Because A is a triangular matrix,
its eigenvalues are the main diagonal entries.
1, 0, 3 1 2 3 l = l = l = -
These three values are distinct, so A is diagonalizable. (Thm.7.6)
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Elementary Linear Algebra: Section 7.2, p.443

38.  Ex 8: (Finding a diagonalizing matrix for a linear transformation)
38/53
3 3
T:R ®
R
Let be the linear transformation given by
T x ,x ,x = x - x - x , x + x + x , - x + x -
x
( ) ( 3 3 )
1 2 3 1 2 3 1 2 3 1 2 3
3
B R T
Find a basis for such that the matrix for
B
relative to is diagonal.
Sol:
T
The standard matrix for is given by
[ ]
1 1 1
ù
ú ú ú
û
é
ê ê ê
ë
- -
1 3 1
- -
1 2 3 A T e T e T e
= =
3 1 1
( ) ( ) ( )
From Ex. 5, there are three distinct eigenvalues
2, 2, 3 1 2 3 l = l = - l =
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so A is diagonalizable. (Thm. 7.6)
Elementary Linear Algebra: Section 7.2, p.443

39. 39/53
Thus, the three linearly independent eigenvectors found in Ex.
5
( 1, 0, 1), (1, 1, 4), ( 1, 1, 1) 1 2 3 p = - p = - p = -
can be used to form the basis B. That is
{ , , } {( 1, 0, 1),(1, 1, 4),( 1, 1, 1)} 1 2 3 B = p p p = - - -
The matrix for T relative to this basis is
[ ]
[ ]
[ ]
D T p T p T p
[ ( )] [ ( )] [ ( )]
B B B
1 2 3
Ap Ap Ap
[ ] [ ] [ ]
B B B
1 2 3
p p p
l l l
[ ] [ ] [ ]
B B B
1 1 2 2 3 3
ù
ú ú ú
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û
=
=
2 0 0
é
= -
ê ê ê
ë
=
0 2 0
0 0 3
Elementary Linear Algebra: Section 7.2, p.443

40. 40/53
Keywords in Section 7.2:
 diagonalization problem: 對角化問題
 diagonalization: 對角化
 diagonalizable matrix: 可對角化矩陣
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41. 7.3 Symmetric Matrices and Orthogonal Diagonalization
41/53
 Symmetric matrix:
A square matrix A is symmetric if it is equal to its
transpose: A = AT
 Ex 1: (Symmetric matrices and nonsymetric
matrices)
ù
ú ú
û
0 1 2
é
-
ê ê
ë
-
=
1 3 0
2 0 5
A
B 4 3
úûù
= é 3 1
êë
ù
ú ú
û
3 2 1
é
= -
1 0 5
ê ê
1 4 0
ë
C
(symmetric)
(symmetric)
(nonsymmetric)
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Elementary Linear Algebra: Section 7.3, p.447

42. 42/53
 Thm 7.7: (Eigenvalues of symmetric matrices)
If A is an n´n symmetric matrix, then the following properties
are true.
(1) A is diagonalizable.
(2) All eigenvalues of A are real.
(3) If l is an eigenvalue of A with multiplicity k, then l has k
linearly independent eigenvectors. That is, the eigenspace
of l has dimension k.
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Elementary Linear Algebra: Section 7.3, p.447

43. 43/53
 Ex 2:
Prove that a symmetric matrix is diagonalizable.
ù
úû
A a c
= é c b
êë
Pf: Characteristic equation:
I A a c l l l
l l
- = - - a b ab c c b
- - = 2 - ( + ) + - 2 = 0
As a quadratic in l, this polynomial has a discriminant of
2 2 2 2 2
a + b - ab - c = a + ab + b - ab +
c
( ) 4( ) 2 4 4
2 2 2
a ab b c
= - + +
2 4
³ 0 2 2
a b c
= - +
( ) 4
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Elementary Linear Algebra: Section 7.3, p.448

44. 44/53
(1) (a - b)2 + 4c2 = 0
Þ a = b, c = 0
is a matrix of diagonal.
0
0
ù
úû
é
=
êë
a
a
A
(2) (a - b)2 + 4c2 > 0
The characteristic polynomial of A has two distinct real
roots, which implies that A has two distinct real eigenvalues.
Thus, A is diagonalizable.
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Elementary Linear Algebra: Section 7.3, p.448

45. 0 4
3
0 1 0
ù
45/53
 Orthogonal matrix:
A square matrix P is called orthogonal if it is invertible and
P-1 = PT
 Ex 4: (Orthogonal matrices)
0 1
é -
= = úû
é
-
P = P- PT
(a) 1 úû
é -
0 4
3
P = 0 1 0
P- PT
0 3
4
ù
Elementary Linear Algebra: Section 7.3, p.449
.
0 1
1 0
is orthogonal because
1 0
ù
êë
ù
êë
.
5
0 3
4
5
5
5
is orthogonal because
5
5
5
5
(b) 1
ú ú ú ú ú
û
é
ê ê ê ê ê
ë
-
= =
ú ú ú ú ú
û
ê ê ê ê ê
ë
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46. 46/53
 Thm 7.8: (Properties of orthogonal matrices)
An n´n matrix P is orthogonal if and only if
its column vectors form an orthogonal set.
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Elementary Linear Algebra: Section 7.3, p.450

47. 47/53
 Ex 5: (An orthogonal matrix)
ù
ú ú ú
û
é
=
ê ê ê
ë
1
-
1
2
- -
5
3 5
4
3 5
2
3 5
5
5
3 2
3 2
3
P 0
Sol: If P is a orthogonal matrix, then P-1 = PT Þ PPT = I
I
1 0 0
0 1 0
0 0 1
é
2
= 1
-
0
0
2
4
5
3 5
1
3 2
3 5
5
3 2
3 5
5
3
5
3 5
1
4
3 5
1
2
2
PPT -
3 5
5
5
3 2
3 2
3
=
ù
ú ú
û
é
=
ê ê
ë
ù
ú ú ú
û
é
ê ê ê
ë
ù
ú ú ú
û
ê ê ê
ë
- -
- -
ù
ú ú ú
û
é
=
ê ê ê
1 Let p , p , p 0
ë
ù
ú ú ú
û
é
=
ê ê ê
ë
ù
ú ú ú
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û
é
=
ê ê ê
ë
1
3
-
3 2
1
2
5
- -
3 2
5
3 5
3
5
4
3 5
2
2
3 5
Elementary Linear Algebra: Section 7.3, p.451

48. 48/53
p p p p p p
× = × = × =
p p p
1 2 1 3 2 3
1
0
produces
= = =
1 2 3
{ , , } is an orthonormal set. 1 2 3 p p p
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Elementary Linear Algebra: Section 7.3, p.451

49. 49/53
 Thm 7.9: (Properties of symmetric matrices)
Let A be an n´n symmetric matrix. If l1 and l2 are distinct
eigenvalues of A, then their corresponding eigenvectors x1
and x2 are orthogonal.
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Elementary Linear Algebra: Section 7.3, p.452

50. 0 1
ù
é
-
 Ex 6: (Eigenvectors of a symmetric matrix)
A =
úû
Show that any two eigenvecto rs of 1 0
êë
corresponding to distinct eigenvalues are orthogonal.
 Sol: Characteristic function
l
- -
l A
Eigenvalues : 2, 4 1 2 Þ l = l =
I - = = l 2 - l + = l - l
- =
6 8 ( 2)( 4) 0
3 1
- l
-
1 3
1
ù
úû
ù
é-
Þ = é
ù
- -
é
- -
l = Þl - A = s s
(1) 2 I 1 1 1 ¹ úû
, 0
1
x
1 1
~
0 0
1 1
úû
1 1
êë
êë
êë
1
ù
úû
ù
é
Þ = é -
ù
é
-
-
l = Þl - A = t t
x x 0 x and x are orthogonal. 1 2 1 2 Þ = - = úû
(2) 4 I 2 2 2 ¹ úû
, 0
1
x
1 1
~
0 0
1 1
1 1
êë
êë
úû
êë
ù
t
é
× úû
s
× = st st
êë
ù
é-
êë
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t
Elementary Linear Algebra: Section 7.3, p.452 50/53

51. 51/53
 Thm 7.10: (Fundamental theorem of symmetric matrices)
Let A be an n´n matrix. Then A is orthogonally diagonalizable
and has real eigenvalue if and only if A is symmetric.
 Orthogonal diagonalization of a symmetric matrix:
Let A be an n´n symmetric matrix.
(1) Find all eigenvalues of A and determine the multiplicity of each.
(2) For each eigenvalue of multiplicity 1, choose a unit eigenvector.
(3) For each eigenvalue of multiplicity k³2, find a set of k linearly
independent eigenvectors. If this set is not orthonormal, apply Gram-
Schmidt orthonormalization process.
(4) The composite of steps 2 and 3 produces an orthonormal set of n
eigenvectors. Use these eigenvectors to form the columns of P. The
matrix will P-1AP = PT AP = D be diagonal.
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Elementary Linear Algebra: Section 7.3, p.453

52. 52/53
 Ex 7: (Determining whether a matrix is orthogonally
diagonalizable)
1 1 1
é
=
1 1 1
ù
ú ú
1 0 1
û
ê ê
ë
1 A
ù
ú ú
û
5 2 1
é
-
ê ê
ë
=
2 1 8
1 8 0
2 A
ù
úû
3 2 0
êë= é 2 0 1
3 A
ù
úû
é
= 0 - 2
êë
0 0
4 A
Orthogonally
diagonalizable
Symmetric
matrix
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Elementary Linear Algebra: Section 7.3, p.454

53. 53/53
 Ex 9: (Orthogonal diagonalization)
Find an orthogonal matrix that diagonalizes A.
ù
ú ú ú û
é
ê ê ê
ë
2 2 2
-
-
2 1 4
- -
=
2 4 1
A
P
Sol:
(1) lI - A = (l -3)2 (l + 6) = 0
6, 3 (has a multiplicity of 2) 1 2 l = - l =
l v u v
= - = - Þ = = -
3 (2) 2
6, (1, 2, 2) 1
( 1
, 2
, ) 3
3
1
1 1 1
v
(3) 3, (2, 1, 0), ( 2, 0, 1) 2 2 3 l = v = v = -
Linear Independent
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Elementary Linear Algebra: Section 7.3, p.455

54. 54/53
Gram-Schmidt Process:
w v w v v w
= = = - × w
(2, 1, 0), 3 2
( 2
, 4
, 1) 2
5
5
w w
2 2
2 2 3 3
= -
×
u w
( , , 0), 3
( 2
, 4
, 5
) 3 5
3 5
3 5
3
1
5 3
2
5
u w
2
2
2
= = = = -
w
w
ù
ú ú ú
û
é
ê ê ê
1
= = -
ë
-
2
4
5
3 5
2
3 2
3 5
2
1
5
3
3 5
5
3
1 2 3
0
(4) P [ p p p ]
ù
ú ú ú
û
é-
ê ê ê
ë
Þ - = =
6 0 0
0 3 0
0 0 3
P 1AP PT AP
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Elementary Linear Algebra: Section 7.3, p.456

55. 55/53
Keywords in Section 7.3:
 symmetric matrix: 對稱矩陣
 orthogonal matrix: 正交矩陣
 orthonormal set: 單範正交集
 orthogonal diagonalization: 正交對角化
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