Elementary Linear Algebra
Anton & Rorres, 9th Edition
Lecture Set – 07
Chapter 7:
Eigenvalues, Eigenvectors
Chapter Content



Eigenvalues and Eigenvectors
Diagonalization
Orthogonal Digonalization
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7-1 Eigenvalue and Eigenvector

If A is an nn matrix
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a nonzero vector x in Rn is called an eigenvector of A if Ax
is a scalar multiple of x;
that is, Ax = x for some scalar .
The scalar  is called an eigenvalue of A, and x is said to be
an eigenvector of A corresponding to .
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7-1 Eigenvalue and Eigenvector

Remark

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To find the eigenvalues of an nn matrix A we rewrite Ax =
x as Ax = Ix or equivalently, (I – A)x = 0.
For  to be an eigenvalue, there must be a nonzero solution
of this equation. However, by Theorem 6.4.5, the above
equation has a nonzero solution if and only if
det (I – A) = 0.
This is called the characteristic equation of A; the scalar
satisfying this equation are the eigenvalues of A. When
expanded, the determinant det (I – A) is a polynomial p in
 called the characteristic polynomial of A.
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7-1 Example 2

Find the eigenvalues of
0 1 0
A  0 0 1 
 4 17 8 

Solution:
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The characteristic polynomial of A is
0 
  1
det( I  A)  det  0 
1    3  8 2  17  4
 4 17   8
The eigenvalues of A must therefore satisfy the cubic equation
3 – 82 + 17 – 4 =0
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7-1 Example 3

Find the eigenvalues of the upper triangular matrix
a11 a12 a13
0 a a
22
23
A
 0 0 a33

0 0 0
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a14 
a24 
a34 

a44 
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Theorem 7.1.1

If A is an nn triangular matrix (upper triangular, low
triangular, or diagonal)


then the eigenvalues of A are entries on the main diagonal
of A.
Example 4

The eigenvalues of the lower triangular matrix
1

0
0
2



2
A   1
0 
3


1
 5 8  

4 
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Theorem 7.1.2 (Equivalent Statements)

If A is an nn matrix and  is a real number, then the
following are equivalent.
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 is an eigenvalue of A.
The system of equations (I – A)x = 0 has nontrivial solutions.
There is a nonzero vector x in Rn such that Ax = x.
 is a solution of the characteristic equation det(I – A) = 0.
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7-1 Finding Bases for Eigenspaces



The eigenvectors of A corresponding to an eigenvalue 
are the nonzero x that satisfy Ax = x.
Equivalently, the eigenvectors corresponding to  are the
nonzero vectors in the solution space of (I – A)x = 0.
We call this solution space the eigenspace of A
corresponding to .
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7-1 Example 5

Find bases for the eigenspaces of

Solution:
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0 0 2 
A  1 2 1 
1 0 3 
The characteristic equation of matrix A is 3 – 52 + 8 – 4 = 0, or in
factored form, ( – 1)( – 2)2 = 0; thus, the eigenvalues of A are  = 1
and  = 2, so there are two eigenspaces of A.
0
2   x1  0

 1   2 1   x   0
(3)
(I – A)x = 0  
 2  
 1
0
  3  x3  0
If  = 2, then (3) becomes  2
0 2   x1  0
 1 0 1  x   0

 2  
 1 0 1  x3  0
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7-1 Example 5
 2 0 2   x1  0
 1 0 1  x   0

 2  
 1 0 1  x3  0

Solving the system yield

x1 = -s, x2 = t, x3 = s
Thus, the eigenvectors of A corresponding to  = 2 are the nonzero
vectors of the form
  s    s  0
 1 0
x   t    0    t   s  0   t 1
 s   s  0
 1  0

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The vectors [-1 0 1]T and [0 1 0]T are linearly independent and form a
basis for the eigenspace corresponding to  = 2.
Similarly, the eigenvectors of A corresponding to  = 1 are the nonzero
vectors of the form x = s [-2 1 1]T
Thus, [-2 1 1]T is a basis for the eigenspace corresponding to  = 1.
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Theorem 7.1.3

If k is a positive integer,  is an eigenvalue of a
matrix A, and x is corresponding eigenvector


then k is an eigenvalue of Ak and x is a corresponding
eigenvector.
Example 6 (use Theorem 7.1.3)
0 0 2 
A  1 2 1 
1 0 3 
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Theorem 7.1.4

A square matrix A is invertible if and only if  = 0 is
not an eigenvalue of A.


(use Theorem 7.1.2)
Example 7
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The matrix A in the previous example is invertible since it has
eigenvalues  = 1 and  = 2, neither of which is zero.
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Theorem 7.1.5 (Equivalent Statements)

If A is an mn matrix, and if TA : Rn  Rn is multiplication
by A, then the following are equivalent:

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
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A is invertible.
Ax = 0 has only the trivial solution.
The reduced row-echelon form of A is In.
A is expressible as a product of elementary matrices.
Ax = b is consistent for every n1 matrix b.
Ax = b has exactly one solution for every n1 matrix b.
det(A)≠0.
The range of TA is Rn.
TA is one-to-one.
The column vectors of A are linearly independent.
The row vectors of A are linearly independent.
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Theorem 7.1.5 (Equivalent Statements)

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
The column vectors of A span Rn.
The row vectors of A span Rn.
The column vectors of A form a basis for Rn.
The row vectors of A form a basis for Rn.
A has rank n.
A has nullity 0.
The orthogonal complement of the nullspace of A is Rn.

The orthogonal complement of the row space of A is {0}.
ATA is invertible.

 = 0 is not eigenvalue of A.

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Chapter Content



Eigenvalues and Eigenvectors
Diagonalization
Orthogonal Digonalization
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Elementary Linear Algebra
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7-2 Diagonalization

A square matrix A is called diagonalizable

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
if there is an invertible matrix P such that P-1AP is a
diagonal matrix (i.e., P-1AP = D);
the matrix P is said to diagonalize A.
Theorem 7.2.1
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If A is an nn matrix, then the following are equivalent.
 A is diagonalizable.
 A has n linearly independent eigenvectors.
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7-2 Procedure for Diagonalizing a Matrix

The preceding theorem guarantees that an nn matrix A with n linearly
independent eigenvectors is diagonalizable, and the proof provides the
following method for diagonalizing A.
 Step 1. Find n linear independent eigenvectors of A, say, p1, p2, …, pn.


Step 2. From the matrix P having p1, p2, …, pn as its column vectors.
Step 3. The matrix P-1AP will then be diagonal with 1, 2, …, n as
its successive diagonal entries, where i is the eigenvalue
corresponding to pi, for i = 1, 2, …, n.
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7-2 Example 1


Find a matrix P that diagonalizes
Solution:

From the previous example, we have the following bases for the
eigenspaces:


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0 0 2 
A  1 2 1 
1 0 3 
 = 2:
Thus,
Also,
 1
0
p1   0 , p 2  1
 1 
0
 = 1:
  2
p 3   1 
 1 
  1 0  2
P   0 1 1 
 1 0 1 
 1 0 2  0 0  2  1 0  2 2 0 0
P 1 AP   1 1 1  1 2 1   0 1 1   0 2 0  D
 1 0  1 1 0 3   1 0 1  0 0 1
Elementary Linear Algebra
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7-2 Example 2
(A Non-Diagonalizable Matrix)

Find a matrix P that diagonalizes

Solution:

The characteristic polynomial of A is
 1
det( I  A)  1
3
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0
 2
5
0
0  (  1)(  2) 2
 2
The bases for the eigenspaces are


 1 0 0
A   1 2 0
 3 5 2
 = 1:
 1/ 8 
p1   1 / 8
 1 
 = 2:
0 
p 2  0
1
Since there are only two basis vectors in total, A is not diagonalizable.
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7-2 Theorems

Theorem 7.2.2


If v1, v2, …, vk, are eigenvectors of A corresponding to
distinct eigenvalues 1, 2, …, k,
 then {v1, v2, …, vk} is a linearly independent set.
Theorem 7.2.3
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If an nn matrix A has n distinct eigenvalues
 then A is diagonalizable.
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7-2 Example 3

Since the matrix

has three distinct eigenvalues,   4,   2  3,   2  3
Therefore, A is diagonalizable.
Further,
4
0
0

0 1 0
A  0 0 1 
 4 17 8 


P 1 AP   0
0

2 3
0


0 
2  3 
for some invertible matrix P, and the matrix P can be found using
the procedure for diagonalizing a matrix.
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7-2 Example 4 (A Diagonalizable Matrix)

Since the eigenvalues of a triangular matrix are the entries
on its main diagonal (Theorem 7.1.1).

Thus, a triangular matrix with distinct entries on the main
diagonal is diagonalizable.

For example,
 1
0
A
0

0





0 0 2 
2 4
3 1
0 5
0
7
8
is a diagonalizable matrix.
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7-2 Example 5
(Repeated Eigenvalues and Diagonalizability)

Whether the following matrices are diagonalizable?
1 0 0
I 3  0 1 0
0 0 1
1 1 0
J 3  0 1 1
0 0 1
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7-2 Geometric and Algebraic Multiplicity

If 0 is an eigenvalue of an nn matrix A


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then the dimension of the eigenspace corresponding to 0 is
called the geometric multiplicity of 0, and
the number of times that  – 0 appears as a factor in the
characteristic polynomial of A is called the algebraic
multiplicity of A.
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Theorem 7.2.4
(Geometric and Algebraic Multiplicity)

If A is a square matrix, then :

For every eigenvalue of A the geometric multiplicity is less
than or equal to the algebraic multiplicity.

A is diagonalizable if and only if the geometric multiplicity
is equal to the algebraic multiplicity for every eigenvalue.
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7-2 Computing Powers of a Matrix




If A is an nn matrix and P is an invertible matrix, then (P-1AP)k = P1AkP for any positive integer k.
If A is diagonalizable, and P-1AP = D is a diagonal matrix, then
P-1AkP = (P-1AP)k = Dk
Thus,
Ak = PDkP-1
The matrix Dk is easy to compute; for example, if
 d1k 0 ... 0 
 d1 0 ... 0 


 0 d ... 0 
k
0
d
...
0
2
2

 , and Dk  
D
 :
:
:
:
:
: 



k
0
0
...
d

 0 0 ... d n 
n

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7-2 Example 6 (Power of a Matrix)

Find A13
0 0 2 
A  1 2 1 
1 0 3 
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Chapter Content



Eigenvalues and Eigenvectors
Diagonalization
Orthogonal Digonalization
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7-3 The Orthogonal Diagonalization
Matrix Form

Given an nn matrix A, if there exist an orthogonal matrix
P such that the matrix
P-1AP = PTAP
then A is said to be orthogonally diagonalizable and P is
said to orthogonally diagonalize A.
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Theorem 7.3.1 & 7.3.2

If A is an nn matrix, then the following are equivalent.




A is orthogonally diagonalizable.
A has an orthonormal set of n eigenvectors.
A is symmetric.
If A is a symmetric matrix, then:

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The eigenvalues of A are real numbers.
Eigenvectors from different eigenspaces are orthogonal.
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7-3 Diagonalization of Symmetric Matrices

The following procedure is used for orthogonally
diagonalizing a symmetric matrix.



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Step 1. Find a basis for each eigenspace of A.
Step 2. Apply the Gram-Schmidt process to each of these bases
to obtain an orthonormal basis for each eigenspace.
Step 3. Form the matrix P whose columns are the basis vectors
constructed in Step2; this matrix orthogonally diagonalizes A.
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7-3 Example 1

Find an orthogonal matrix P that diagonalizes

Solution:



 4 2 2
A   2 4 2
 2 2 4
The characteristic equation of A is
2 
  4 2
det( I  A)  det  2   4 2   (  2) 2 (  8)  0
 1
 1
 2
2   4
 1  and u   0 
u

2
The basis of the eigenspace corresponding to  = 2 is 1  
 
 0 
 1 
Applying the Gram-Schmidt process to {u1, u2} yields
the following orthonormal eigenvectors:
 1/ 2 
 1/ 6 




v1   1/ 2  and v 2   1/ 6 
 0 


2
/
6




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7-3 Example 1


1
The basis of the eigenspace corresponding to  = 8 is u3  1
1
Applying the Gram-Schmidt process to {u } yields:
1/ 3 


v 3  1/ 3 


1/
3



3
Thus,
P  v1
v2
 1 / 2

v3    1/ 2
 0

1/ 6 1/ 3

1/ 6 1/ 3
2 / 6 1 / 3 
orthogonally diagonalizes A.
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