5.3 Diagonalization of an Operator in U 189
We rst verify that M is a normal matrix:
2.1 213 4+1 0 50
Ef'i _ _
[i2] [f2] [0 1+4] [05
2t' 2i 4+1 0 50
i _ _ _ T_ T
[i2][iz][O14][05]:>.ci.Qf.cf.
Then, we
5.3 Diagonalization of an Operator in U 183
Furthermore, we shall show that the necessary and sufcient condition for a
matrix .c in R3 to have an orthonormal eigenbasis, in which it is represented by
192 5 The Ei gen Problem or Diagonal Form of Representing Matrices
We have to use now the GramSchmidt procedure for orthonorrnalization to get
:11, ON basis vectors in the eigenspace EM
When we do thi
190 5 The Ei gen Problem or Diagonal Form of Representing Matrices
Since 522 : El (it is selfadjoint). it is obviously a normal matrix (ail :
l).
The characteristic equation is
det[.e M3] =
1 7L 0 i
5.1 Eigenvalues, Eigenveetors, and Eigenspaces 1?
Two Examples
1. Solve the Eigen problem for the following 3 X 3 real matrix
2 1 0
112? = 1 0 1
1 3 1
The characteristic equation of w is
2 7L 1 0
deMg
5.5 The Most Important Subsets of Normal Operators in U 207
DISTRIBLJTIVITY
ran P1 ["1 (Uneran P2) ran P1 Uneran P1 ran P2 =
ran P1 Gran P2. A
5.5.5 The Spectral Form of a Normal OperatorA
Going back
203 5 The Ei gen Problem or Diagonal Form of Representing Matrices
eigenvalue spectrum of A) cfw_7&1 , l2 , . . , M, k <_< n, and the corresponding set of
eigen projection operators cfw_PM , P7, , , P
6.4. FACTORING TRINOMIALS H 393
Report the results on your homework as follows.
y y=4x2717m715
IIIIIIO
>35
10 10

o
:3.,

01
Hence, the solutions of 4:1:2 17:5 15 : 0 are a: : 0.75 and
6.4. FACTORING TRINOMIALS H 391
Next, use the 2zzero utility from the CALC menu to nd the mintercepts
of the graph.
Report the results on your homework as follows.
1;
31:23:2791775
.P
>$
10
6.4. FACTORING TRINOMIALS H 389
Thus, the solutions of 7m 3 = 6:t:2 are a: = 1/3 and a: = 3/2.
35. Because there is a power of .38 larger than one, the equation is nonlinear.
Make one side zero.
26:5
390 CHAPTER 6. FACTORIN G
We have a product that equals zero. Use the zero product property to complete
the solution.
356120 or 2x+9:0
3.17:1 2m=9
331 5679
"3 _ 2
Thus, the solutions of 6:172 = 25:12
392 CHAPTER 6. FACTORIN G
60 and the integer pair 3 and 20 has product 60 and sum 17, the coef
cient of :5. Split the middle term up using this pair.
eP17m15=0
m3+3mam15=0
Factor by grouping.
x(4a:+3
7.1. NEGATIVE EXPONENTS 419
41. All the operators involved are multiplication, so the commutative and asso
ciative properties of multiplication allow us to change the order and grouping.
Well show th
430 CHAPTER 7. RATIONAL FUNCTIONS
To convert to a single power of 10, repeat the base and add the exponents.
= 1.1 x 102+1
: 1.1 x 101
Thus, 0.011 x 101 : 1.1 x 101.
33. The notation 1.134E~1 is the c
ITS 5 The Ei gen Problem or Diagonal Form of Representing Matrices
These coefcients form the basis vector in 15.1, i.e.,
E_1=cfw_s % sER=ker(f+13)
1
(see Chap. 2, 17B, about ker A).
Choosing s : 4, w
182 5 The Ei gen Problem or Diagonal Form of Representing Matrices
ab
Example To show when a 2 X 2 matrix .0? = [ ] is diagonalizable, we shall
(3 d
calculate its characteristic equation
411? b
c dl
180 5 The Ei gen Problem or Diagonal Form of Representing Matrices
Therefore, the vector space R3 is decomposed into the direct sum of E2 and E1,
i.e., R3 : E2 +E1, since 3 is linearly independent fro
184 5 The Ei gen Problem or Diagonal Form of Representing Matrices
Denition A normal operatorA in U is characterized by the property that it com
mutes with its adjointAF (Sect. 4.2):
AAT :ATA
Theorem
186 5 The Ei gen Problem or Diagonal Form of Representing Matrices
Un Z/ZV 4
A
'0*0'0"0* 0'00'4'
90000 00 00
90000 0
090%
L_J.u
0 Q0?
000000
>000000
L A. L. A A. A
90.0.02000
0
0
4
0390'0
02
070
0:0
5.5 The Most Important Subsets of Normal Operators in U 195
(2) =>~ (3)
R if
(4)
From (2) follows (3):
(3) _
(AitAfFLEATAJ?)(i,y)1V.ity6Un
From (3) we have (4):
_ _ _ (3) _ _ _ _
(MAX) = Hzixll2 = Ha
188 5 The Ei gen Problem or Diagonal Form of Representing Matrices
This bij ection g is, furthermore, an isomorphism:
1. g(A+C) = f(A+C)f'1=fAf'1+fo'1=g(A)+g(C);
2 3%) = f(aA)f'1=a(fAf)= aw);
3. MAC)
5.4 The Actual Method for Diagonalization of a Normal Operator 191
1 i 0 mi
3*: 0x/E 0
as 1 0 1
The matrix 57 will diagonalize 5213 by the unitary similarity
000
ylwy:ywy:a: 020
002
This can be eas
196 5 The Ei gen Problem or Diagonal Form of Representing Matrices
21* U 0
0 252*. 0
9129*: , , , _ ,
0 0 .z,cfw_
and 91 has on its diagonal the inverse eigenvalues
zl'l 0 0
9_]_ 0 Zg_1. 0
0 0 2.1]
l,
5.4 The Actual Method for Diagonalisation of a Normal Operator 193
This demonstrates that the normal n x a matrix as? is diagonalizable by the unitary
similarity.
The matrix 57 is in fact the unitary
5.2 Diagonalization of Square Matrices 181
If we form the matrix, whose columns are linearly independent eigenvectors
012
1,2,319= l 0 l,
0 ll
this matrix will be invertible (rank 2 3), and the three
5.3 Diagonalization of an Operator in U 185
3. We shall prove that (A My) commutes with its adjoint:
(A MuNA Nerf = (A  Mimi  Wu) =
:AAT _ MT are +muu 2m :UA MTHLWU =
: (Ar 4mm My) 2 (A MU)T(A Mu)
5.3 Diagonalization of an Operator in UN 18?
Remark as a
Theorem (on isomorphism of algebras) The representation of operators acting
in VH(F) by matrices from FM [caused by the choice of a basis in VH
5.1 Eigenvalues, Eigenveetors, and Eigenspaees 115
The equation ded M) = 0 is in fact an equation of the degree n in the
unknown 1. It is called the characteristic equation of A. Its explicit form is
194 5 The Ei gen Problem or Diagonal Form of Representing Matrices
Thus, we are back to the statement that the normal n X n matrix M can be brought to
a diagonal form 33 by a unitary similarity transf
428 CHAPTER 7. RATIONAL FUNCTIONS
23. Converting a number into scientifc notation requires that we convert the
given number into the form a x 10', where k is an integer and 1 S a < 10.
The requireme
7.1. NEGATIVE EXPONENTS 417
29. In this case, we are raising a power to a power, so well use the following
law of exponents:
(01711)! = almn
That is, well repeat the base, then multiply the exponents.
432 CHAPTER 7. RATIONAL FUNCTIONS
3.2E'5/2.5E'?
I
1.23:2
Hence, (3.2 x 105)/(2.5 x 107) z 1.28 x 102. Dont forget to return your
calculator to its original mode by selecting NORMAL and FLOAT in the
7.1. NEGATIVE EXPONENTS 421
51. The law of exponents (111) = (1% says that when you raise a product to
a power, you must raise each factor to that power. So we begin by raising each
factor to the four
7.2. SCIENTIFIC NOTATION 431
Hence, (2.5 x 101)(1.6 x 107) = 4 x 10s.
41. Well use the approximations 1.4 m 1 and 1.8 m 2, which enable us to
write:
(1.4 x 107)(1.8 x 10) m (1 x107)(2 x 104)
m 2 x 1