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The world of Eigenvalues-eigenfunctionsAn operator    A   operates on a function and produces afunction.For every operator...
Therefore, the operator A has an eigenvalue ncorresponding to eigenfunction x n .  1. Eigenfunctions are not unique.  Supp...
2   −1  Let A = 2 5  be the matrix and we want to                compute its eigenvalues and eigenfunctions. Its  ch...
Similarly, for e-value of 4, the eigenfunction appears           1 to be e2 = − 2 .              6. Faddeev-Leverrie...
And one proceeds this way to get          p3 = 1728  The CA polynomial = ( −1 )3 [λ3 − 44λ2 + 564λ −1728]  The eigenvalues...
Suppose, the eigenfunction of                                                 P −1 AP          is     y     witheigenvalue...
. So, we can normalize each( x ( i ) )t x (   j)                        = 0 , ∀i ≠ j                                      ...
3                           3∑ a1 j = 1 + 1 + 0 = 2 ,   ∑ a2 j = 1 + 2 + 1 = 4j =1                       j =13∑ a3 j = 6j ...
1In the limit   k → ∞,                        2k                             →0   . Therefore,   A   is a convergentmatrix...
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Eigenvalues

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Eigenvalues

  1. 1. The world of Eigenvalues-eigenfunctionsAn operator A operates on a function and produces afunction.For every operator, there is a set of functions whichwhen operated by the operator produces the samefunction modified only multiplied by a constantfactor.Such a function is called the eigenfunction of theoperator, and the constant modifier is called itscorresponding eigenvalue. An eigenvalue is just anumber: Real or complex.A typical eigenvalue equation would look like Ax = λ xHere, the matrix or the operator A operates on avector (or a function) x producing an amplified orreduced vector λx . Here the eigenvalue λbelongs toeigenfunction x . dSuppose the operator is A = ( x dx ) . A operating on d nx n produces Ax = x x = nx . n n dx
  2. 2. Therefore, the operator A has an eigenvalue ncorresponding to eigenfunction x n . 1. Eigenfunctions are not unique. Suppose Ax = λ x . Define, another vector z = cx , where c is a constant. Now, Az = Acx = cAx = cλ x = λ cx = λ z Therefore, z is also an e-function (eigenfunction) of A. 2. If Ax = λ x is an eigenvalue equation (and we assume that x is not a zero vector), then Ax = λx ⇔ (A - λI)x = 0 ⇐⇒ det(A - λI) = 0 This leads to a characteristic polynomial in λ: p A = det( A − λ I ) λ is an e-value of A only if pA = 0. 3. Spectrum of an operator A is σ( A ) : set of all its e-values. 4. Spectral radius of an operator A is ρ ( A ) = max | λ | λ∈σ ( A ) = 1maxn | λi | ≤i ≤ 5. Computation of spectrum and spectral radius:
  3. 3. 2 −1 Let A = 2 5  be the matrix and we want to   compute its eigenvalues and eigenfunctions. Its characteristic equation (CE) is: 2 − λ −1  det  = 0 ⇐⇒ (2 - λ )(5 - λ ) + 2 = 0  2 5 − λ This gives λ2 − 7λ + 12 = 0 ⇐ ⇒ ( λ − 3 )( λ − 4 ) = 0Therefore, A has two eigenvalues: 3 and 4. x Let the eigenfunction be the vector x =  1  x2 corresponding to e-value 3. Then  2 − 1  x1   x1   3 x1   2 5   x  = 3 x  =  3 x    2   2   2 Therefore, we have 2 x1 − x2 = 3x1 yieldingx1 = − x2 . Also, we get 2 x1 + 5 x2 = 3 x2 which gives us no newresult. Therefore, we can arbitrarily take the 1 following solution: e1 = −1 corresponding to e-value 3  for the matrix A.
  4. 4. Similarly, for e-value of 4, the eigenfunction appears 1 to be e2 = − 2 .   6. Faddeev-Leverrier Method to get characteristic polynomial. Define a sequence of matrices P = A, p1 = trace( P ) 1 1 1 P2 = A[ P − p1I ] , p2 = trace( P2 ) 1 2 1 P3 = A[ P2 − p2 I ] , p3 = trace( P3 ) 3 … … 1 Pn = A[ Pn −1 − pn −1I ] , p n = trace( Pn ) n Then the characteristic polynomial P( λ ) is [ P( λ ) = ( −1 )n λn − p1λn −1 − p2 λn − 2 − ... − pn ] 12 6 − 6  6 16 2  e.g. A=   − 6 2  16   Define P = A, p1 = trace( A ) = 12 + 16 + 16 = 44 1 P2 = A( P − p1I ) = 1 12 6 − 6− 32 6 −6   6 16 2  6 − 28 2     − 6 2  16  − 6  2 − 28  − 312 −108 108  = −108 − 408 − 60 , p 2 = −564    108  − 60 − 408 
  5. 5. And one proceeds this way to get p3 = 1728 The CA polynomial = ( −1 )3 [λ3 − 44λ2 + 564λ −1728] The eigenvalues are next found solving [λ3 − 44λ2 + 564λ −1728] = 0 7. More facts about eigenvalues. Assume Ax = λ x . Therefore, λ is the eigenvalue of A with eigenvector x . a. A−1 has the same eigenvector as A and the corresponding eigenvalue is λ−1 . b. An has the same eigenvector as A with the eigenvalue λn . c. ( A + µI ) has the same eigenvector as A with the eigenvalue ( λ + µ ) . d. If A is symmetric, all its eigenvalues are real. e. If P is an invertible matrix then P −1 AP has the same eigenvalues as A .Proof of e.
  6. 6. Suppose, the eigenfunction of P −1 AP is y witheigenvalue k .Then, P − APy = ky 1 ⇐⇒ APy = Pky = kPyTherefore, Py = x and k must be equal to λ. Thereforethe eigenvalues of A and P −1 AP are identical and theeigenvector of one is a linear mapping of the otherone.If the eigenvalues of A , λ1 ,λ2 ,...,λn are all distinctthen there exists a similarity transformation such that λ1 0 0 .. 0  0 λ 0 .. 0   2  −1P AP = D =  0 0 λ3 .. 0   .. .. .. .. 0     0 0 0 .. λn  Let the eigenvectors of A be x ( 1 ) , x ( 2 ) ,..., x ( i ) ,...x ( n )such that we have Ax( i ) = λi x( i )Then the matrix P = [ x( 1 ) , x( 2 ) ,..., x( n ) ]Then AP = [ Ax( 1 ) , Ax( 2 ) ,..., Ax( n ) ] [ = λ1 x( 1 ) ,λ2 x( 2 ) ,..., λn x( n ) ] [ ][ = x ( 1 ) , x ( 2 ) ,..., x ( n ) λ1e( 1 ) ,λ 2 e( 2 ) ,..., λn e( n ) ]= PDTherefore, P −1 AP = DAlso, note the following. If A is symmetric, then
  7. 7. . So, we can normalize each( x ( i ) )t x ( j) = 0 , ∀i ≠ j (i ) x (i)eigenvector and obtain u = x so that the (i )matrix Q = [u ( 1 ) ,u ( 2 ) ,...,u ( n ) ] would be an orthogonal matrix.i.e. Q AQ = DtMatrix-norm.Computationally, the l 2 -norm of a matrix isdetermined as l 2 -norm of [ A =|| A ||2 = ρ( At A ) ]1 / 2 1 1 0e.g. A = 1 2 1   −1  1 2  1 1 −1 1 1 0  3 2 −1Then A A = 1 t 2 1  1 2 1 =  2 6 4      0  1 2 −1  1 2 −1   4 5The eigenvalues are: λ1 = 0, λ2 = 7 + 7 , λ3 = 7 − 7Therefore, A2 = ρ( At A ) = 7 + 7 ≈ 3.106 A ∞ = max ∑ aijThe l∞norm is defined as 1≤i ≤n j 1 1 0 e.g. A =1 2 1    −1  1 − 4 
  8. 8. 3 3∑ a1 j = 1 + 1 + 0 = 2 , ∑ a2 j = 1 + 2 + 1 = 4j =1 j =13∑ a3 j = 6j =1 Therefore, A ∞ = max( 2 ,4 ,6 ) = 6In computational matrix algebra, we would often beinterested about situations when A k becomes small(all the entrees become almost zero). In that case, A isconsidered convergent. is convergent if klim∞( A )ij = 0 ki.e. A → 1   0Example. Is A = 2 convergent? 1 1   4 2 1  1  1   4 0 8 0 16 0 A2 =  A3 =  A4 =  1 1 , 3 1 , 1 1 ,        4 4 16 8   8 16 It appears that  1   2k 0Ak =  k 1    2k + 1 2k   
  9. 9. 1In the limit k → ∞, 2k →0 . Therefore, A is a convergentmatrix.Note the following equivalent results: a. A is a convergent matrix k b1. klim∞ A 2 = 0 → k b2. klim∞ A ∞ = 0 → c. ρ( A ) < 1 k d. klim∞ A x = 0 ∀x →Condition number K( A ) of a non-singular matrix Ais computed as K ( A ) = A . A -1A matrix is well-behaved if its condition number isclose to 1. When K ( A ) of a matrix A is significantlylarger than 1, we call it an ill-behaved matrix.

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