程序代写代做代考 C Key issues:

Key issues:
Lecture III
• Eigenvalues, eigenvectors and the characteristic polynomial of a square matrix
• Similarity
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Definition:
Eigenvalue and Eigenvector
Given a square n  n matrix A, the set of eigenvalues    are such that the characteristic equation
Ac  c , or equivalently I  Ac  0 has a nonzero solution c  0.
Such a vector c  n is called an eigenvector associated with eigenvalue .
As a result, for any eigenvalue,
detIA0, A.
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How to compute an eigenvalue?
They are the roots of the characteristic polynomial of A: The eigenvalues  of a matrix Ann areJ1
the roots of its characteristic polynomial:  detI  A
n 1n1 n1n.
So, in total there are n eigenvalues (possibly repeated
and complex-valued).
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Slide 117 J1
Jiang, 9/21/2020

Comment on Characteristic Polynomial
Indeed, det I  A  0 implies, by means of permutation: detIAn1n1n1n 0.
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Comment 1
For any given eigenvalue λ, the eigenvectors are nonzero solutions to the homogeneous
equation:
(λI – A)c=0.
Clearly, any vector of the form μ x c for a
nonzero scalar μ is still an eigenvector.
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Comment 2
For any given eigenvalue λ, the eigenvectors are nonzero solutions to the homogeneous
equation:
(λI-A)c=0. cNull IA,c0. 
So, the maximum number of linearly independent eigenvectors is equal to:
n – rank(λI – A).
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eigenvectors are:
An Example
For the identity matrix A   1 0  , the eigenvalues 0 1

are 1   2  1, for which two linearly independent
 
c1  1 , c2  0 .  
01
 
Other choices of independent eigenvectors are:
 
c1  1 , c2  3 .  
21
 

A Useful Observation
A square matrix A of dimension n is singular if and only if it has a zero eigenvalue.
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Proof
Necessity :
Assume that A is singular. Then, there is a nonzero solution x  0 to the equation (*) Ax  0, or equivalently, Ax  0x.
So,   0 is an eigenvalue.
Sufficiency :
If   0 is an eigenvalue, then there is an eigenvector x  0, which is a solution to (*).
Thus, A must be a singular matrix.
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M1 1.
1 1 
Exercise
Find the eigenvalues and the associated eigenvectors of the following matrix

More about Eigenvalues
 A real square matrix can have complex eigenvalues. For example, the eigenvalues of
0 1 1 0

are1 j, 2  j.
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More about Eigenvalues
 A real square matrix can have multiple eigenvalues. For example, the eigenvalues of
A1 a 0 1

are1 2 1foranya.
We say that 1  1 is an eigenvalue of multiplicity 2. In this case, the characteristic polynomial is
2 det IA  1 .

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More about Eigenvalues
 More generally, the characteristic polynomial of a matrix Ann takes the form
mmm detIA1 2r
12r where 1, , r are different, with m  n,
 aneigenvalueof(algebraic)multiplicitym. ii
Whenrnandm m 1, thematrixAissaidto 1n
n have distinct eigenvalues i i1 .

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i i

More about Eigenvectors
 Back to the example of A  1 a. It has an eigenvalue 0 1

 1of(algebraic!)multiplicity2,foranya.
1
Case 1: For a  0, the associated eigenvectors are distinct:  
c1,c0. 12
01
 
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More about Eigenvectors
BacktoA1 a.Ithasaneigenvalue 1 0 1 1

of (algebraic) multiplicity 2, for any a  .
Case 2: For a  0, there is only one distinct 
eigenvector: c1  1 0

Its “geometric” multiplicity is 1 i.e., all other eigenvectors take the form r  c1 , for some
scalar r  0.
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Exercise
Do you know the eigenvalues of the following matrix
1 1 1 M1 1 1?

1 1 1 
Can you try to find the eigenvectors for each eigenvalue?
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A General Result
Assume Ann has n distinct eigenvalues 1, …, n. Then,
(1) A must have n linearly independent eigenvectors 1n
c , , c .
(2) In addition, each eigenvector c j associated with 
j
is unique apart from a nonzero scalar multiplier.
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 i1
not all zero,
Proof of Statement 1
Let c1  0, , cn  0 be eigenvectors satisfying Aci ci, fori1,2,,n.
i
We prove the statement by contradiction.
n
Assume that ci are linearly dependent. Let k  n
 i1
be the least positive integer such that k of the c’s are dependent. Without loss of generality, assume that
k
ci are dependent, that is, 
i c1c2ck 0.
12k
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Proof of Statement 1 (cont’d)
 notallzero, c1  c2  ck 0. i12k
Thus,k2andalli 0(otherwise,contradiction with k being the least).
Now, multiplytheeq.by AkIleadsto:
1 1k ck1 k1k c 0 1 k1
k1
which, in turn, implies that c i1 are dependent.
 i
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A contradiction.

Proof of Statement 2
We must show the “uniqueness” of ci : Acc, c0  cci, with0.
i
As it was proved in statement (1), c i1 are
linearly independent and thus form a basis. So, cc1 ci  cn.
1in
Multiplying the above eq. by A  i I  gives:
0   c1 0   cn 11i nni
Therefore: k  0, k  i.
In other words, c   ci , as wished.
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134
i
n i

Corollary
Under the above conditions, define matrix Pc1 c2 cn

which is nonsingular & implies P1 AP  diag  .
i
In this case, we say that A is similar to diag  , i
while P is a similarity matrix.
Denote A  diag  . A is called “diagonalizable”
i
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Remark 1
As we will see in Lecture IV, diag   is i
a canonical form for matrices, which have distinct eigenvalues,
and for symmetric matrices,
which may have repeated eigenvalues.
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Remark 2
Any two similar matrices A and B must have the same eigenvalues.
Indeed,ABP, suchthatBP1AP. It then follows that
detBIdetP1API
 det P 1  A   I  P  det P 1 det  A   I  det P
detAI,
because detP1 det P  det P1P 1.
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Questions
Are you ready for some tricky questions?
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Question 1
If the set of all eigenvalues of A  33 , or thespectrum A  1,2,3 ,
  what is  A A3 ?
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A General Result
k
For any polynomial p t    a t i , and any n  n
matrixA, denotep(A)
If , x is a pair of eigenvalue and eigenvector
of A, then p(), x is a pair of eigenvalue and eigenvector of p(A).
(Its proof is left as an exercise.)
“Matrix Polynomial”
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k
i0
aAi, withA0 I. i
i i0

Question 2
For any idempotent matrix A, that is, A2  A, what are the possible eigenvalues?
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Idempotent Matrix: 2×2 case
If A  a c  is idempotent, then b d
 aa2 bc,
babbd, implyingb0, ord1a  c  c a  c d , i m p l y i n g c  0 , o r d  1  a
d bcd2. 
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Examples: Idempotent Matrix (1)Aa 0 isidempotent,ifa,d 0,1.
0 d 
(2)A11cos sin . 2 sin 1cos

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Answer
An idempotent matrix can only have 0 or 1 as its eigenvalues.
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Question 3
AnilpotentmatrixAissuchthatAq 0forapositive integer q. Such a smallest q is called the
index of nilpotency.
What are the eigenvalues of a nilpotent matrix A?
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Indeed,ifAxx, x0, then, usingAq 0, weobtainqx0
Answer
All eigenvalues of a nilpotent matrix are 0.
which, in turn, implies   0.
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Examples: Nilpotent Matrix 0 *  *
  (1)M * 0 *
00  0 
5 3 2 (2)M15 9 6

10 6 4 
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Equivalence Relation: Nilpotent Matrix
The following statements are equivalent:
(1) M nn is nilpotent.
(2) The minimal polynomial of M is sq , for some q  n.
(3) The characteristic polynomial of M is sn . (4) The only eigenvalue of M is 0.
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0 2 0 0 A  0 0 2 1
0 0 0 2 
Question 4
Compute the algebraic and geometric multiplicities of the eigenvalue   2 for the matrix
2100

Useful Identities about Matrix Eigenvalues
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For any nn matrix A, we have n
trace(A)   ,
detAn .
i i1
i i1
See p.42 of (Horn-Johnson, 1st ed., 1985), or p. 50 of (Horn-Johnson, 2nd ed., 2013).

Note:
Similarity
Two matrices A and B are said to be similar, if B  P1 AP, for some invertible matrix P.
Notation: A  B
The set of all similar matrices to a given square matrix A :
S P AP:Pisinvertible  1 
Similar matrices are just different basis representations of
a single linear mapping.
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Similarity: Physical meaning
Let T :V  V be a linear transformation, and
B1 v1,,vn, B2 w1,,wnbetwobasesforV. Denote
xB1 col1,…,n,withx1v1…nvn. Then, by linearity,
TxTv … Tv 11 nn
For any basis B2 of V , Tv  colt ,…,t ,
 jB2 1j nj
 T x   n   T v    t  c o l   , . . . ,   . j j ij 1 n
B2 j1  B2 nxn
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Similarity: Physical meaning
n
Tx   Tv   t col  ,…,

.
depends on T
j j ij 1 n B2 j1  B2 nxn
It is important to note that the matrix tij  nxn
and the choice of the bases B1 and B2 , but not x. Define the B1 -B2 basis representation of T as:
Tt Tv,…,Tv B1ijnxn1 n
2B B2 B2 So, Tx B2 = B2 T B1 x B1 ,xV.
 
For the special case when B1  B2 ,
B1 T B1 is called the B1 representation of T . 
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Similarity: Identities
For the identity linear transformation Ix  x, x V ,
it can be shown that
B2IB1B1IB2 In,B1IB2B2IB1 In, and

B2 T B2 =B2 I B1 B1 T B1 B1 I B2 . In other words,
  
1
BP AP,wherePB1 IB2,AB1 TB1,BB2 TB2.
B -B change of basis matrix 21
For a proof, see (Horn & Johnson, 2nd ed, 2013, page 40).
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is diagonal.
Exercise
Find an invertible matrix P such that 1 1 1
P10 2 2P 
0 0 3 

Homework #3
1. Give all the solutions of the system 1 2 3 10 13
4 5 6x11 14. 
7 8 9 12 15 
2. Prove that the following eq. has no solution:
 1 3x1.
26 3 
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Fall 2020

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Homework #3
3. Find a least-squares fit
b  x  x a1  x a2 012
for the data: 110
 0 2 0
b , a1  , a2  . 0 3 0
 041

Homework #3
4. Find independent eigenvectors for A1 2.
3 1 

Can you express x  1 as a linear combination
2 
of these eigenvectors of A?
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