Transcript 5.3 ppt File

5 Eigenvalues and Eigenvectors
5.3
DIAGONALIZATION
© 2012 Pearson Education, Inc.
DIAGONALIZATION

 7 2
Example 1: Let A  
. Find a formula for

 4 1
Ak,
given that A  PDP , where
1
 1 1
5 0
P
and D  



1

2
0
3





Solution: The standard formula for the inverse of a
2  2 matrix yields
 2 1
P 


1

1


1
© 2012 Pearson Education, Inc.
Slide 5.3- 2
DIAGONALIZATION
 Then, by associativity of matrix multiplication,
A2  ( PDP 1 )( PDP 1 )  PD ( P 1 P) DP 1  PDDP 1
I
2
1
1
5
0   2 1



2 1
 PD P  



2

1

2

1

1
0
3




 Again,
A3  ( PDP 1 ) A2  ( PD P 1 ) P D 2 P 1  PDD 2 P 1  PD 3 P 1
I
© 2012 Pearson Education, Inc.
Slide 5.3- 3
DIAGONALIZATION
 In general, for k  1,
 1 1 5
A  PD P  


 1 2   0
k
k
k
1
 25  3
 k
k
2  3  2  5
k
k
0   2 1

k
3   1 1
5 3 
k
k
23  5 
k
k
 A square matrix A is said to be diagonalizable if A is
1
similar to a diagonal matrix, that is, if A  PDP
for some invertible matrix P and some diagonal,
matrix D.
© 2012 Pearson Education, Inc.
Slide 5.3- 4
THE DIAGONALIZATION THEOREM
 Theorem 5: An n  n matrix A is diagonalizable if
and only if A has n linearly independent eigenvectors.
1
In fact, A  PDP , with D a diagonal matrix, if
and only if the columns of P and n linearly
independent eigenvectors of A. In this case, the
diagonal entries of D are eigenvalues of A that
correspond, respectively, to the eigenvectors in P.
In other words, A is diagonalizable if and only if
n
there are enough eigenvectors to form a basis of
.
n
We call such a basis an eigenvector basis of
.
© 2012 Pearson Education, Inc.
Slide 5.3- 5
THE DIAGONALIZATION THEOREM
 Proof: First, observe that if P is any n  n matrix with
columns v1, …, vn, and if D is any diagonal matrix with
diagonal entries λ1, …, λn, then
AP  A v1
v2
vn    Av1
Av2
0

0
   λ1v1


λn 
λ 2 v2
Avn 
----(1)
while
 λ1
0
PD  P 

0

0
λ2
0
© 2012 Pearson Education, Inc.
λ n vn 
----(2)
Slide 5.3- 6
THE DIAGONALIZATION THEOREM
1
A

PDP
 Now suppose A is diagonalizable and
. Then
right-multiplying this relation by P, we have
AP  PD.
 In this case, equations (1) and (2) imply that
 Av
1
Av2
Avn    λ1v1 λ 2 v2
λ n vn 
----(3)
 Equating columns, we find that
Av1  λ1v1 , Av2  λ 2 v2 , , Av n  λ n v n
----(4)
 Since P is invertible, its columns v1, …, vn must be
linearly independent.
© 2012 Pearson Education, Inc.
Slide 5.3- 7
THE DIAGONALIZATION THEOREM
 Also, since these columns are nonzero, the equations
in (4) show that λ1, …, λn are eigenvalues and v1, …,
vn are corresponding eigenvectors.
 This argument proves the “only if ” parts of the first
and second statements, along with the third statement,
of the theorem.
 Finally, given any n eigenvectors v1, …, vn, use them
to construct the columns of P and use corresponding
eigenvalues λ1, …, λn to construct D.
© 2012 Pearson Education, Inc.
Slide 5.3- 8
THE DIAGONALIZATION THEOREM
 By equation (1)–(3), AP  PD .
 This is true without any condition on the
eigenvectors.
 If, in fact, the eigenvectors are linearly independent,
then P is invertible (by the Invertible Matrix
Theorem), and AP  PD implies that A  PDP 1.
© 2012 Pearson Education, Inc.
Slide 5.3- 9
DIAGONALIZING MATRICES
 Example 2: Diagonalize the following matrix, if
possible.
1 3 3


A   3 5 3 


1
 3 3
That is, find an invertible matrix P and a diagonal
1
matrix D such that A  PDP .
 Solution: There are four steps to implement the
description in Theorem 5.
 Step 1. Find the eigenvalues of A.
 Here, the characteristic equation turns out to involve a
cubic polynomial that can be factored:
© 2012 Pearson Education, Inc.
Slide 5.3- 10
DIAGONALIZING MATRICES
0  det( A  λI )   λ  3λ  4
3
2
 (λ  1)(λ  2) 2
 The eigenvalues are λ  1 and λ  2 .
 Step 2. Find three linearly independent eigenvectors
of A.
 Three vectors are needed because A is a 3  3 matrix.
 This is a critical step.
 If it fails, then Theorem 5 says that A cannot be
diagonalized.
© 2012 Pearson Education, Inc.
Slide 5.3- 11
DIAGONALIZING MATRICES
 1
 
 Basis for λ  1: v1  1
 
 1
 1
 1




λ


2
:
v

1
v

0
 Basis for
and 3
2
 
 
 0 
 1
 You can check that {v1, v2, v3} is a linearly
independent set.
© 2012 Pearson Education, Inc.
Slide 5.3- 12
DIAGONALIZING MATRICES
 Step 3. Construct P from the vectors in step 2.
 The order of the vectors is unimportant.
 Using the order chosen in step 2, form
P   v1
v2
 1 1 1


v3   1 1 0


 1 0 1
 Step 4. Construct D from the corresponding eigenvalues.
 In this step, it is essential that the order of the eigenvalues
matches the order chosen for the columns of P.
© 2012 Pearson Education, Inc.
Slide 5.3- 13
DIAGONALIZING MATRICES
 Use the eigenvalue λ  2 twice, once for each of the
eigenvectors corresponding to λ  2 :
 1 0 0
D   0 2 0 


 0 0 2 
1
 To avoid computing P , simply verify that AD  PD.
 Compute
 1 3 3  1 1 1  1 2 2 






AP  3 5 3 1 1 0  1 2 0


 

1  1 0 1  1 0 2 
 3 3
© 2012 Pearson Education, Inc.
Slide 5.3- 14
DIAGONALIZING MATRICES
 1 1 1  1 0 0   1 2 2 
PD   1 1 0  0 2 0    1 2 0 


 

 1 0 1 0 0 2   1 0 2 
 Theorem 6: An n  n matrix with n distinct
eigenvalues is diagonalizable.
 Proof: Let v1, …, vn be eigenvectors corresponding
to the n distinct eigenvalues of a matrix A.
 Then {v1, …, vn} is linearly independent, by
Theorem 2 in Section 5.1.
 Hence A is diagonalizable, by Theorem 5.
© 2012 Pearson Education, Inc.
Slide 5.3- 15
MATRICES WHOSE EIGENVALUES ARE NOT
DISTINCT
 It is not necessary for an n  n matrix to have n
distinct eigenvalues in order to be diagonalizable.
 Theorem 6 provides a sufficient condition for a matrix
to be diagonalizable.
 If an n  n matrix A has n distinct eigenvalues, with
corresponding eigenvectors v1, …, vn, and if
P   v1
v2 , then P is automatically invertible
because its columns are linearly independent, by
Theorem 2.
© 2012 Pearson Education, Inc.
Slide 5.3- 16
MATRICES WHOSE EIGENVALUES ARE NOT
DISTINCT

When A is diagonalizable but has fewer than n
distinct eigenvalues, it is still possible to build P in
a way that makes P automatically invertible, as the
next theorem shows.

Theorem 7: Let A be an n  n matrix whose distinct
eigenvalues are λ1, …, λp.
a. For 1  k  p , the dimension of the eigenspace
for λk is less than or equal to the multiplicity
of the eigenvalue λk.
© 2012 Pearson Education, Inc.
Slide 5.3- 17
MATRICES WHOSE EIGENVALUES ARE NOT
DISTINCT
b. The matrix A is diagonalizable if and only if
the sum of the dimensions of the eigenspaces
equals n, and this happens if and only if (i) the
characteristic polynomial factors completely
into linear factors and (ii) the dimension of the
eigenspace for each λk equals the multiplicity
of λk.
c. If A is diagonalizable and Bk is a basis for the
eigenspace corresponding to Bk for each k,
then the total collection of vectors in the sets
n
B1, …, Bp forms an eigenvector basis for
.
© 2012 Pearson Education, Inc.
Slide 5.3- 18