Eigenvalues And Eigenvectors And Their Applications

Eigenvalues and Eigenvectors and their Applications By Dr. P.K.Sharma Sr. Lecturer in Mathematics D.A.V. College Jalandhar. Email Id: [email protected]

• • • • • • • • •

The purpose of my lecture is to make 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 given eigenvalue exists ? What are the eigenvalues corresponding to special types of matrices like symmetric , skew symmetric , orthoganal and unitary matrices etc. Some important Theorems relating to eigenvalues Why eigenvectors and eigenvalues are important ? What are the application of eigenvectors and eigenvalues ?

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 changing both its magnitude and its direction. However, a matrix may act on certain vectors by changing only their magnitude, and leaving their direction unchanged (or possibly reversing it). These vectors are the eigenvectors of the matrix. A matrix acts on an eigenvector by multiplying its magnitude by a factor, which is positive if its direction is unchanged and negative if its direction is reversed. This factor is the eigenvalue associated with that eigenvector.

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 of x ; that is , A x = λ x , for some scalar λ. The scalar λ is called an eigenvalue of A , and x is called the eigenvector of A corresponding to the eigenvalue λ.

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

4

Example 1 Eigenvector of a 2×2 Matrix 1  • The vector x=   2   the matrix

is an eigenvector of

3 0  A=  8 − 1  

Corresponding to the eigenvalue λ=3, since

3 0   1   3  Ax =  =   = 3x    8 −1  2   6 

5

Example 2 X is not an Eigenvector of a 2×2 Matrix

• The vector

3 0  A=  8 − 1  

 2  is not eigenvector of the matrix x=   λ 3 , there  donot exist scalar such that

λ

 3 0  2  6  A x = = = λ x       Hence the vector x is not eigenvector of the matrix A 8 − 1  3  13   eigenvalues . Only square Note: Not all matrices have matrices have eigenvalues and eigenvectors.

6

Origin of Eigenvalues and Eigenvect Eigenvalues and eigenvectors have their origins in physics, in particular in problems where motion is involved, although their uses extend from solutions to stress and strain problems to differential equations and quantum mechanics. Recall from last class that we used matrices to deform 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 the eigenvector from the original length. The basic equation in eigenvalue problems

•

How to find eigenvalues of a square matrix of order n To find the eigenvalues of an n × n matrix A . we rewrite

Ax = λx

or equivalently,

as

Ax = λI x

(λI - A)x = 0

(1)

Equation (1) has a nonzero solution if and only if det (λI - 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 (λI - A) is a polynomial p in λ called the characteristic polynomial of A. The set of all eigenvalues of A is called the Spectrum of A .

Example 3 Eigenvalues of a 3×3 Matrix • Find the eigenvalues of

0 1 0 A =  0 0 1   4 −17 8 

Solution. The characteristic polynomial of A is 0   λ −1 det(λ I − A) = det  0 λ −1  = λ3 − 8λ 2 + 17λ − 4  −4 17 λ − 8 The eigenvalues of A must therefore satisfy the characteristic λ 3 − 8λ 2 equation + 17λ − 4 = 0 (2)

On solving we find λ = 4 , 2 + 3 , 2 - 3

•

Example 4 Eigenvalues of an Upper Triangular Matrix Find the eigenvalues of the upper triangular matrix

Solution.

a11 a12 a13 a14 0  a a a 22 23 24 A = 0 0 a33 a34   a44  0 0 0  

Recalling that the determinant of a triangular matrix is the product of the entries on the main diagonal , we obtain

−a12 −a13 a−14  λ − a11   0  λ − a − a − a 22 23 24  (= λa − )( aλ − )( aλ −)( det(λ I − A ) = det 22 33  0  11 0 λ − a33 −a34   0 0 0 λ − a 44   Thus, the characteristic equation is (λ-a 11 )(λ-a22 ) (λ-a33 ) (λ-a44 )=0 and the eigenvalues are λ=a11 , λ=a22 , λ=a33 , λ=a44 which are precisely the diagonal entries of A. 10

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Eigenvalues of special types of matrices Types of Matrices

T ( A =A ) • Symmetric T ( A • Skew Symmetric = -A ) • Orthogonal ( ATA = I ) • Hermitian ( Aθ= A ) θ • Skew Hermatian ( A = -A ) θ • Unitary (A A= I )

Nature of Eigenvalues • Reals • Purely Imaginary or Zero • Unit modulus • Reals • Purely imaginary or zero • Unit modulus

Theorem 1 If A is an n×n triangular matrix (upper triangular, lower triangular, or diagonal), then the eigenvalues of A are entries on the main diagonal of A. Theorem 2 If k is a positive integer, λ is an eigenvalue of a matrix A, and x is corresponding eigenvector, then λk is an eigenvalue of Ak and x is a corresponding eigenvector. Theorem 3 A square matrix A is invertible if and only if λ=0 is not an eigenvalue of A.

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Theorem 4 Equivalent Statements • a) b) c) d)

If A is an n × n matrix and λ is a real number, then the following are equivalent. λ is an eigenvalue of A. The system of equations (λI-A)x = 0 has nontrivial solutions. There is a nonzero vector x in Rn such that Ax=λx. λ is a solution of the characteristic equation det(λI-A)=0. 13

•

Finding Eigenvectors corresponding to given eigenvector (or Bases for The eigenvectors Eigenspaces) of A corresponding to an eigenvalue λ are the nonzero vectors x that satisfy Ax = λx.

• Equivalently, the eigenvectors corresponding to λ are the nonzero vectors in the solution space of (λI-A)x =0. We call this solution space the eigenspace of A corresponding to λ. • Are eigenvector x corresponding to eigenvalue λ unique ? No ; every scalar multiple of eigenvector x is also eigvevector corresponding to eigenvalue λ, for A(kx) = k(Ax) = k(λx) = λ(kx). • The set of L.I. eigenvectors forms the bases of the eigenspace. 14

•

Example 5 Finding Eigenvectors of a square matrix (or Bases for Eigenspaces) Find eigenvectors and hence the bases for the eigenspaces of

 0 0 −2  …… A = 1 2 1  (1) 1 0 3  Solution. The characteristic equation of matrix A is λ3-5λ2+8λ-4=0 or (λ-1)(λ-2)2=0 ; ------(2) Thus, the eigenvalues of A are λ=1 and λ=2, so we need to find the eigenvectors corresponding to these two distinct eigenvalues. 15

By definition,

 x1  x =  x2   x3 

Is an eigenvector of A corresponding to λ if and only if x is a nontrivial solution of (λI-A)x=0, that is, of

0 2   x1  0  λ  −1 λ − 2 −1   x  =  0    2    −1 0 λ − 3  x3   0  If λ=2, then (3) becomes  2 0 2  x1 −1 0 −  =  1 x    2  1  −1 0 −    x 3

0 0   0

(3)

Solving this system yield x1 = - s , x2 = t , x3 = s Thus, the eigenvectors of A corresponding to λ=2 are 16

s −s  − =  +0 =t x = t        s     s   

Since

 −1  0  0  and  1       1   0

− 0  s+    0  

 1 0 t    1

0 1 0

are linearly independent eigenvectors.

and so these vectors form a basis for e eigenspace th correspondingλ＝ to

Similarly, the eigenvectors corresponding to λ ＝ 1 are the nonzerovector so it form a basis for the eigenspace corresponding to λ ＝ 1. -2 1    1 Note: The number of L.I. eigenvector corresponding to eigenvalue

λ equal to n – rank(λI – A) , where n is the order of square matix A.

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. 18

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. 19

Theorem (5/3) Equivalent Statements p) A has rank n. q) A has nullity 0. r) The orthogonal complement of the nullspace of A is Rn. s) The orthogonal complement of the row space of A is {0}. t) ATA is invertible. u) λ= 0 is not eigenvalue of A. 20

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; the matrix P is said to diagonalize A. 21

Theorem 6 If v1, v2, … vk, are eigenvectors of A corresponding to distinct eigenvalues λ1, λ2, …, λk , then {v1, v2, … vk} is a linearly independent set. Theorem 7 If an n × n matrix A has n distinct eigenvalues, then A is diagonalizable.

22

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. 23

Procedure for Diagonalizing a Matrix • 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-1 AP will then be diagonal with λ1, λ2, …, λn as its successive diagonal entries, where λi is the eigenvalue corresponding to pi, for i=1, 2, …, n.

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Example 6 Finding a Matrix P That Diagonalizes a Matrix A

• Find a matrix P that diagonalizes

Solution.

 0 0 −2  A = 1 2 1  1 0 3 

From Example 5 of the preceding section we found the characteristic equation of A to be (λ-1)(λ-2)2=0 and we found the following bases for the eigenspaces:

 −1 0 λ = 2 : p1 =  0  , p 2 = 1   1  0 

 −2  λ =1: p 3 =  1   1 

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Example 6 (Cont.) Finding a Matrix P That Diagonalizes a Matrix A There are three basis vectors in total, so the matrix A is diagonalizable and

 −1 0 −2  P =  0 1 1   1 0 1 

diagonalizes A. As a check, the reader should verify that  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   −1 0 −1 1 0 3   1 0 1   0 0 1  26

•

Example 7 A Matrix That Is Not Diagonalizable Find a matrix P ( if possible ) that diagonalize the matrix

 1 0 0 A =  1 2 0   −3 5 2 

Solution. The characteristic polynomial of A is

λ −1 0 0 det(λ I − A) = −1 λ − 2 0 = (λ − 1)(λ − 2)2 3 −5 λ − 2 27

Example 7 (Cont.) A Matrix That Is Not Diagonalizable so the characteristic equation is (λ-1)(λ-2)2=0 Thus, the eigenvalues of A are λ=1 and λ=2. We can easily show that bases for the eigenspaces are  1/ 8  0  λ = 1: p1 =  −1/ 8 λ = 2 : p 2 =  0   1  1  Since A is a 3×3 matrix and there are only two basis vectors in total, A is not diagonalizable. 28

Algebraic multiplicity of λ : It is defined as the number of times the root λ occur in the characteristic equation and is denoted by Mλ . Geometric multiplicity of λ : It is defined as the number of L.I. eigenvectors associated with λ and it denoted by mλ (= dimension of eigenspace of λ ) In general, mλ Defect of λ : ∆

≤

λ

Mλ = Mλ - mλ

Remark : (1) Defective matrices are not diagonalizable

Application of Eigenvalues and Eigenvectors • 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 how diagonalization 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-1 AP)2 = (P-1 AP)(P-1 AP) = P-1 AIAP=P-1 A2P 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 k

P-1AkP = ( P-1AP )

= Dk

Solving this equation for Ak yield :

Ak = PDk P-1

This last equation expresses the kth power of A in terms d1 0 ... 0   d1k 0 ... 0   k of the kth power of the diagonal matrix D. But D is    k 0 d 2 ... 0  0 d 2 ... 0  k   D =compute; for example, , and D = if easy to  : : : : : :     k  0 0 ... d n   0 0 ... d n 

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Example 8 Power of a Matrix • Find A13, where

 0 0 −2  A = 1 2 1  1 0 3 

Solution. We showed in Example 5 that the matrix A has three L.I.  −1 0 −2 0 1 1 eigenvectors and so the matrix A P is=diagonalized by    1 0 1 

and that 1 D =P −AP =

2 0 0 0 2 0    0 0 1 

32

Example 8 ( Cont.) Power of a Matrix

Thus , we have

 1   1  1  0 16382 −  − 8190    =  8191 8192 8191   8191  0 16383

−1 0 13 13 − 1  A = PD P  0= 1  1 0

2−  1 1 

13

2 0   0

0 0 13 2 0 13 0 1−

33

0 1  0−

Cayley-Hamilton Theorem Every square matrix satisfy its characteristic polynomial i.e. If p(λ ) = det(A −λ I) is the characteristic polynomial of an n × n matrix A, then p(A) = 0.

Application of Cayley-HamiltonTheorem The Cayley-Hamilton Theorem can be used to find:  The Power of a matrix and  The Inverse of an n × n matrix A, by expressing these as polynomials in A of degree < n.

Use of eigenvectors and eigenvalues in solving Linear differential equations • 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 constant coefficients. Furthermore, in chemical engineering many models are formed on the basis of systems of differential equations that are either linear or can be linearized and solved using the eigenvalue eigenvector method. In general, most ODEs can be linearized and therefore solved by this method

How to solve Initial value problem using Eigenvalues and Eigenvectors • Solve the following Initial value problem: du1 = a 11u1 + a 12 u2 + .........+ a 1 un n dt du 2 = a 21u1 + a 22 u2 + .........+ a 2 un n dt ........................................ ............... ........................................ ............... du n = a n1u1 + a n2u2 + .........+ a nnu n dt given that u i= b i when t = 0 , for i = 1 , 2 ,

...., n

w e w rite the above system in the m atrix rm as: fo  u1 ( t)  u ( t) du = Au , w here u(t) =2  : dt   un ( t)

     

a11 a21 , A= ...  1an

a12 ...n a22 ...n ... ... ... 2an

a1 a2 ... ann

  ,u   

Let λ1 , λ2 , ......., λn be the eigenvalues ofmatrix A and X1 , X 2 , ............, Xn be the correspon ding eigenvectors. Then the solutions of this L.D.E. is giv en by : λ t

u = X e , where X is the corresponding eigenvector ofλ . The general solution is given by :

u(t) =

n

λi t c X e ∑i i i =1

 b1 b 2  Now , u(0) = :   bn

xi 1   xi 1   x    n x i 2  = ∑ c   , where X = i 2  i  i =1 i  :  :       xin   xin  on solving these equations , we can findci , for i = 1,2,...,n So the solution of the given Initial value problem is :  u1 ( t )  u ( t)  2  = u(t) =  :     un ( t )  u i (t ) = c i x iieλit

xi 1 x i 2  ic :  xin

  n λi t e , On comparing , we get ∑  i =1   , for i = 1 , 2 , ....., n .

Example 9 Solve the Initial Value Problem

• Solve the Initial value problem dv = 3v dt dw = 8v - w dt

; v = 6 at t = 0 ; w = 5 at t = 0

• The problem is to find v(t) and w(t) for t > 0. We write the system in the matrix form as :

 v( t )  du = Au , where u(t) =  dt  w( t) 

3 0 , A = 8 − 1

  , u(0) 

As in EX. 1 , we see that eigenvalues of the matrix λ = A3 are , 1

λ =1

2

0  The corresponding eigenvalues are 1  X1 =   , X 2 = .  1  2  du =Au , The solution to this L.D.E. is given dt byλ t

u= X e

, w here X is the eigenvectorsponing corre to eigenvalue λ

The general solution is given by λ1 t

u(t) = C1 X1 e

+ C2 X2 e 2

λt

, where C1 and C2 beary arbitr constants

 6  1  0 So , u(0) =   = C 1   + C 2   , implies that C 1 =6 and C 2 = −7.  5  2  1

 v(t )  1 3 t  0 So we get u(t) =  = 6  e -7   e   w(t )  2  1

-t

Thus ; the solution of the given Initial value 3t problem is : On com paring , w e get v(t) = 6e and =w12e (t)

−

− 7e 3t

Eigenvectors and eigenvalues are used in structural geology to determine the directions of principal strain; the directions where angles are not changing. In seismology, these are the directions of least compression (tension), the compression axis, and the 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 as representing 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

Eigenvalues & Eigenvectors • Example of uses: – Structural analysis (vibrations). – Correlation analysis (statistical analysis and data mining). • We use the Jacobi method applicable to symmetric matrices only. • A 2x2 rotation matrix is applied on the matrix to annihilate the largest off-diagonal element. • This process is repeated until all offdiagonal elements are negligible. ©DB Consulting, 1999, all rights reserved

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