23290356 eigenvalues and eigenvectors and their applications
TRANSCRIPT
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Eigenvalues and
Eigenvectors and theirApplications
ByDr. P.K.Sharma
Sr. Lecturer in Mathematics
D.A.V. College Jalandhar.Email Id: [email protected]
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The purpose of my lecture is tomake you to understand the
following : What are eigenvectors and eigenvalues ? What is the origin of eigenvectors and eigenvalues ?
Do every matrix have eigenvectors and eigenvalues ?
Are eigenvectors corresponding to a given eigenvalue unique?
How many L.I. eigenvectors corresponding to a giveneigenvalue exists ?
What are the eigenvalues corresponding to special types ofmatrices like symmetric , skew symmetric , orthoganal andunitary matrices etc.
Some important Theorems relating to eigenvalues Why eigenvectors and eigenvalues are important ?
What are the application of eigenvectors and eigenvalues ?
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Linear algebra studies linear transformations,which are represented by matrices acting on
vectors. Eigenvalues, eigenvectors and
eigenspaces are properties of a matrix. In general, a matrix acts on a vector by changingboth its magnitude and its direction. However, amatrix may act on certain vectors by changing onlytheir magnitude, and leaving their directionunchanged (or possibly reversing it). These vectorsare the eigenvectors of the matrix. A matrix actson an eigenvector by multiplying its magnitude bya factor, which is positive if its direction isunchanged and negative if its direction is reversed.This factor is the eigenvalue associated with thateigenvector.
http://en.wikipedia.org/wiki/Linear_transformationshttp://en.wikipedia.org/wiki/Matrix_(mathematics)http://en.wikipedia.org/wiki/Vectorhttp://en.wikipedia.org/wiki/Vectorhttp://en.wikipedia.org/wiki/Matrix_(mathematics)http://en.wikipedia.org/wiki/Linear_transformations -
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Definition If A is an n n matrix, then a nonzero vector x in Rn is
called an eigenvector of A if Ax is a scalar multiple ofx ;that is ,
A x = x , for some scalar .
The scalar is called an eigenvalue of A , and x is
called the eigenvector of A corresponding to theeigenvalue .
In short : (1) An Eigenvector is a vector that maintains its
direction after undergoing a linear transformation.
(2) An Eigenvalue is the scalar value that the eigenvector was
multiplied by during the linear transformation
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Example 1Eigenvector of a 22 Matrix
The vector is an eigenvector ofthe matrix
Corresponding to the eigenvalue =3, since
1
2 =
x
3 0
8 1A
=
3 0 1 33
8 1 2 6
A
= = =
x x
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xX is not an Eigenvector of a 22
Matrix The vector is not eigenvector of the matrix
, there donot exist scalar such that
Hence the vector x is not eigenvector of the matrix A
Note: Not all matrices have eigenvalues . Only squarematrices have eigenvalues and eigenvectors.
3 0
8 1
A
=
2
3 =
x
3 0 2 6=
8 1 3 13
A
= =
x x
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Eigenvalues and eigenvectors have their
origins in physics, in particular in problems wheremotion is involved, although their uses extendfrom solutions to stress and strain problems todifferential equations and quantum mechanics.
Recall from last class that we used matrices todeform a body - the concept of STRAIN.Eigenvectors are vectors that point in
directions where there is no rotation.Eigenvalues are the change in length of theeigenvector from the original length.
The basic equation in eigenvalue problems=
Origin of Eigenvalues and Eigenvec
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How to find eigenvalues of asquare matrix of order n
To find the eigenvalues of an n n matrix A . werewrite Ax = x as Ax = I x
or equivalently, (I - A)x = 0 (1)
Equation (1) has a nonzero solution if and only if
det (I - A) = 0 (2)Equation (2) is called the characteristic equation of A; the scalar
satisfying this equation are the eigenvalues of A. When expanded,
det (I - A) is a polynomial p in called the characteristic
polynomial of A.The set of all eigenvalues of A is called the Spectrum of A .
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Example 3Eigenvalues of a 33 Matrix
Find the eigenvalues of
Solution.The characteristic polynomial of A is
The eigenvalues of A must therefore satisfy thecharacteristic equation
0 1 0
0 0 1
4 17 8
A
=
3 2
1 0
det( ) det 0 1 8 17 4
4 17 8
I A
= = +
3 28 17 4 0 (2) + =
solving we find = 4 , 2 + 3 , 2 - 3On
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xamp eEigenvalues of an Upper Triangular
Matrix Find the eigenvalues of the upper triangular matrix
Solution.
Recalling that the determinant of a triangular matrix is theproduct of the entries on the main diagonal , we obtain
Thus, the characteristic equation is (-a11 )(-a22 ) (-a33 ) (-a44 )=0and the eigenvalues are =a11 , =a22 , =a33 , =a44which are precisely the diagonal entries of A.
11 12 13 14
22 23 24
33 34
44
0
0 0
0 0 0
a a a a
a a aA
a a
a
=
11 12 13 14
22 23 24
11 22 33 4
33 34
44
0det( ) det ( )( )( )(
0 0
0 0 0
a a a a
a a a I A a a a
a a
a
= =
Ei l f i l t f
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Eigenvalues of special types ofmatrices
Types of Matrices
Symmetric
Skew Symmetric
Orthogonal
Hermitian
Skew Hermatian
Unitary
Nature of Eigenvalues Reals
Purely Imaginary orZero
Unit modulus Reals
Purely imaginary orzero
Unit modulus
( = A )TA
( = -A )TA
( = -A )A( = A )A
( A = I )A
( A = I )TA
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If A is an nn triangular matrix (uppertriangular, lower triangular, or diagonal),
then the eigenvalues of A are entries onthe main diagonal of A.
Theorem 2
Ifkis a positive integer, is an eigenvalue of a matrix A,and x is corresponding eigenvector, then k is aneigenvalue of Akand x is a corresponding eigenvector.
Theorem 3
A square matrix A is invertible if and only if =0 is not aneigenvalue of A.
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Theorem 4Equivalent Statements
If A is an n n matrix and is areal number, then the following areequivalent.
a) is an eigenvalue of A.b) The system of equations (I-A)x = 0
has nontrivial solutions.
c) There is a nonzero vector x in Rn
such that Ax=x.
d) is a solution of the characteristic
equation det(I-A)=0.
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corresponding to giveneigenvector (or Bases for
Eigenspaces) The eigenvectors of A corresponding to an eigenvalue arethe nonzero vectors x that satisfy Ax = x.
Equivalently, the eigenvectors corresponding to are thenonzero vectors in the solution space of (I-A)x =0. We callthis solution space the eigenspace of A corresponding to .
Are eigenvector x corresponding to eigenvalue unique ?No ; every scalar multiple of eigenvector x is also
eigvevector corresponding to eigenvalue , forA(kx) = k(Ax) = k(x) = (kx).
The set of L.I. eigenvectors forms the bases of theeigenspace.
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Finding Eigenvectors of a square
matrix
(or Bases for Eigenspaces) Find eigenvectors and hence the bases for the
eigenspaces of
(1)Solution.
The characteristic equation of matrix A is
3-52+8-4=0 or (-1)(-2)2=0 ; ------(2)
Thus, the eigenvalues of A are =1 and =2, so weneed to find the eigenvectors corresponding to thesetwo distinct eigenvalues.
0 0 2
1 2 1
1 0 3
A
=
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By definition,
Is an eigenvector of A corresponding to if and only ifxis a nontrivial solution of (I-A)x=0, that is, of
1
2
3
x
x
x
=
x
1
2
3
0 2 0
1 2 1 0 (3)1 0 3 0
x
xx
=
1
2
3
2 0 2 0
1 0 1 0
1 0 1 0
x
x
x
=
If =2, then (3) becomes
Solving this system yield x1= - s , x
2= t ,
x3= s
Thus, the eigenvectors of A corresponding to =2 are
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0 1 0
0 0 1
0 1 0
s s
t t s t
s s
= = + = +
x
1 0
0 and 1
1 0
are linearly independent eigenvectors.
and so these vectors form a basis for the eigenspace corresponding to
Similarly, the eigenvectors corresponding to 1 are the nonzero-vector so it form a basis for the eigenspace
corresponding to 1.
Note: The number of L.I. eigenvector corresponding to eigenvalue equal to n rank(I A) , where n is the order of square matix A.
-2
1
1
Since
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Theorem (5/1)Equivalent Statements
If A is an n n matrix, and if TA: Rn Rn is multiplication by A,
then the following are equivalent.
a) A is invertible.
b) Ax = 0 has only the trivial solution.
c) The reduced row-echelon form of A is In.
d) A is expressible as a product of elementary matrix.
e) Ax = B is consistent for every n 1 matrix B.
f) Ax = B has exactly one solution for every n 1 matrix B.
g) det(A)0.
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Theorem ( 5/2 )Equivalent Statements
h) The range of TA is Rn.
i) TA is one-to-one.
j) The column vectors of A are linearly independent.
k) The row vectors of A are linearly independent.
l) The column vectors of A span Rn.
m) The row vectors of A span Rn.
n) The column vectors of A form a basis for Rn.
o) The row vectors of A form a basis for Rn.
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Theorem (5/3)Equivalent Statements
p) A has rank n.
q) A has nullity 0.
r) The orthogonal complement of thenullspace of A is Rn.
s) The orthogonal complement of the rowspace of A is {0}.
t) ATA is invertible.u) = 0 is not eigenvalue of A.
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Diagonalization
Definition : A square matrix A is
called diagonalizable if there is
an invertible matrix P such that
P-1
AP is a diagonal matrix; thematrix P is said to diagonalize A.
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Ifv1, v2, vk, are eigenvectors of A
corresponding to distinct eigenvalues 1, 2,
, k , then {v1, v2, vk} is a linearlyindependent set.
Theorem 7
If an n n matrix A has n distincteigenvalues, then A is diagonalizable.
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Theorem 8
If A is an n n matrix, then the
following are equivalent.
a) A is diagonalizable.
b) A has n linearly independent
eigenvectors.
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Procedure for Diagonalizing aMatrix
The preceding theorem guarantees that an n n matrix A with n L.I.
eigenvectors is diagonalizable, and the proof provides the following
method for diagonalizing A.
Step 1. Find n L.I. 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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Example 6Finding a Matrix P That Diagonalizes a
Matrix A Find a matrix P that diagonalizes
Solution.From Example 5 of the preceding section we found the
characteristic equation of A to be (-1)(-2)2=0
and we found the following bases for the eigenspaces:
0 0 2
1 2 1
1 0 3
A
=
1 2 3
1 0 2
2 : 0 , 1 =1: 1
1 0 1
= = = =
p p p
E l 6 (C )
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Example 6 (Cont.)Finding a Matrix P That Diagonalizes a
Matrix AThere are three basis vectors in total, so the matrix A is
diagonalizable and
diagonalizes A. As a check, the reader should verify that
1 0 2
0 1 1
1 0 1
P
=
1
1 0 2 0 0 2 1 0 2 2 0 0
1 1 1 1 2 1 0 1 1 0 2 0
1 0 1 1 0 3 1 0 1 0 0 1
P AP
= =
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Example 7A Matrix That Is Not
Diagonalizable Find a matrix P ( if possible ) that diagonalize thematrix
Solution.
The characteristic polynomial of A is
1 0 0
1 2 0
3 5 2
A
=
2
1 0 0
det( ) 1 2 0 ( 1)( 2)
3 5 2
I A
= =
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Example 7 (Cont.)A Matrix That Is Not Diagonalizable
so the characteristic equation is
(-1)(-2)2=0
Thus, the eigenvalues of A are =1 and =2. We can
easily show that bases for the eigenspaces are
Since A is a 33 matrix and there are only two basisvectors in total, A is not diagonalizable.
1 2
1/8 0
1: 1/ 8 2 : 01 1
= = = = p p
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Algebraic multiplicity of : It is defined as the number
of times the root occur in the characteristic equation
and is denoted by M
.
Geometric multiplicity of : It is defined as the number
of L.I. eigenvectors associated with and it denoted by
m
(= dimension of eigenspace of )
In general, m M
Defect of :
= M - m
Remark : (1) Defective matrices are notdia onalizable
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Application of Eigenvalues andEigenvectors
Computing Powers of a Matrix :There are numerous problems in applied mathematics
that require the computation of high powers of a square
matrix. We shall conclude this section by showing howdiagonalization can be used to simplify such
computations for diagonalizable matrices.
If A is an n n matrix and P is an invertible matrix, then(P-1AP)2 = (P-1AP)(P-1AP) = P-1AIAP=P-1A2P
More generally, for any positive integer k
(P-1
AP)k
= P-1
Ak
P
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Computing Powers of a Matrix(cont.)
It follows form this equation that if A is diagonalizable,
and P-1AP=D is diagonal matrix, then P-1AkP = ( P-1AP )
k = Dk
Solving this equation for Ak yield : Ak = PDkP-1
This last equation expresses the kth power of A in terms
of the kth power of the diagonal matrix D. But Dk is
easy to compute; for example, if
1 1
2 k 2
0 ... 0 0 ... 0
0 ... 0 0 ... 0, and D
: : : : : :
0 0 ... 0 0 ...
k
k
k
n n
d d
d dD
d d
= =
E l 8
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Example 8Power of a Matrix
Find A13, where
Solution.
We showed in Example 5 that the matrix A has three L.I.
eigenvectors and so the matrix A is diagonalized by
and that
0 0 2
1 2 1
1 0 3
A
=
1 0 2
0 1 1
1 0 1
P
=
1
2 0 0
0 2 0
0 0 1
D P AP
= =
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Example 8 ( Cont.)Power of a Matrix
Thus , we have13
13 13 1 13
13
1 0 2 2 0 0 1 0
0 1 1 0 2 0 1 11 0 1 0 0 1 1 0
8190 0 163828191 8192 8191
8191 0 16383
A P D P
= =
=
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Cayley-Hamilton Theorem
Every square matrix satisfy its characteristic polynomiali.e. If p( ) = det(A I) is the characteristic polynomial
of an n n matrix A, then p(A) = 0.
Application of Cayley-HamiltonTheoremThe Cayley-Hamilton Theorem can be used to find:
The Power of a matrix and
The Inverse of an n n matrix A, by expressingthese as polynomials in A of degree < n.
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The eigenvalue and eigenvector method of
mathematical analysis is useful in many fields
because it can be used to solve homogeneous linear
systems of differential equations with constantcoefficients. Furthermore, in chemical engineering
many models are formed on the basis of systems of
differential equations that are either linear or can belinearized and solved using the eigenvalue
eigenvector method. In general, most ODEs can be
linearized and therefore solved by this method
Use of eigenvectors and eigenvalues in solving
Linear differential equations
ow o so ve n a va ue pro em
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ow o so ve n a va ue pro emusing Eigenvalues and
Eigenvectors Solve the following Initial value problem:
111 1 12 2 1
221 1 22 2 2
= a + a + .........+ a
= a + a + .........+ a
........................................ ...............
........................................ ...............
= a
n n
n n
nn
duu u u
dt
du u u udt
du
dt1 1 2 2+ a + .........+ a
that u = b when t = 0 , for i = 1 , 2 , ...., n
n nn n
i i
u u u
given
w rite the abo ve system in the matrix form as:w e
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1 11 12 1
2 21 22 2
1 2
w rite the abo ve system in the matrix form as:
( ) ...
( ) ...
= Au , wh ere u(t) = , A = ,: ... ... ... ...
( ) ...
n
n
n n n n n
w e
u t a a a
u t a a ad u
d t
u t a a a
1 2
1 2
, , ......., be the eigenvalues ofmatrix A andX , X , ............, X be the corresponding eigenvectors.
Then the solutions of this L.D.E. is given by :
u = X e , where X is the co
n
n
t
Let
rresponding eigenvector of .
1
general solution is given by :
u(t) = in
t
i ii
The
c X e
= b x x
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1 1 1
2 2 2
1
, u(0) = , where X
: : :
solving these equations , we can findc , for i = 1,2,...,
So the solution of the given Initial
i i
n
i i
i i
i
n in in
i
b x x
b x x Now c
b x x
on
=
= =
1 1
2 2
1
value problem is :
( )
( )( ) = , On comparing , we get
: :
( )
u ( ) = , for i = 1 , 2 , ....., n .
i
i
i
n
i t
i
i
n in
t
i i ii
u t x
u t x
u t c e
u t x
t c x e
=
=
E l 9
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Example 9Solve the Initial Value Problem
Solve the Initial value problem
The problem is to find v(t) and w(t) for t > 0.
We write the system in the matrix form as :
As in EX. 1 , we see that eigenvalues of the matrix A are
= 3v ; v = 6 at t = 0
= 8v - w ; w = 5 at t = 0
dv
dt
dw
dt
( ) 3 0= Au , w here u(t) = , A = , u(0( ) 8 1
v td uw td t
1 23 , = =
The corresponding eigenvalues are1 0
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The corresponding eigenvalues are
, The solution to this L.D.E. is given
by
The general solution is given by
Thus ; the solution of the given Initial valueproblem is :
1 2
1 0= , X = .
2 1X
= Audu
dt
X e , where X is the eigenvector correspon ing to eigenvalutu =
1 2
1 1 2 2 1 2
u(t) = C X e + C X e , where C and C be arbitr ary constantt t
1 2 1 2
6 1 0So , u(0) = = C + C , that C 6 and C 7.
5 2 1implies
= =
3 -( ) 1 0
So we get u(t) = = 6 e -7 e( ) 2 1
t tv t
w t
3 3comparing , we get v(t) = 6e and w(t)= 12e 7et tO n
Eigenvectors and eigenvalues are used in
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Eigenvectors and eigenvalues are used instructural geology to determine the directionsof principal strain; the directions where angles
are not changing.In seismology,these are the directions of leastcompression (tension), the compression axis, andthe intermediate axis (in three dimensions).Some facts: The product of the eigenvalues = det|A|
The sum of the eigenvalues = trace(A)
The (x,y) values of A can be thought of asrepresenting points on an ellipse centered at(0.0). The eigenvectors are then in the directions
of the major and minor axes of the ellipse, and
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DB Consulting, 1999,all rights reserved
Eigenvalues & Eigenvectors
Example of uses:Structural analysis (vibrations).
Correlation analysis(statistical analysis and data mining).
We use the Jacobi method applicable tosymmetric matrices only.
A 2x2 rotation matrix is applied on the matrix toannihilate the largest off-diagonal element.
This process is repeated until all offdiagonalelements are negligible.
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Questions
?
Thanks