Math 19b: Linear Algebra with Probability Oliver Knill, Spring 2011Lecture 28: EigenvaluesWe have seen that det(A) 6= 0 if and only if A is invertible.The polynomial fA(λ) = det(A − λIn) is called the characteristic polynomial ofA.The eigenvalues of A are the roots of the characteristic polynomial.Pro of. If Av = λv,then v is in the kernel of A − λIn. Consequently, A − λInis not invertible anddet(A − λIn) = 0 .1 For the matrix A ="2 14 −1#, the characteristic polynomial isdet(A − λI2) = det("2 − λ 14 −1 − λ#) = λ2− λ − 6 .This polynomial has the roots 3, −2.Let tr(A) denote the trace of a matrix, the sum of the diagonal elements of A.For the matrix A ="a bc d#, the characteristic polynomial isλ2− tr(A) + det(A) .We can see this directly by writing out the determinant of the matr ix A − λ I2. The trace isimpo r tant because it always appears in the characteristic polynomial, also if the matrix islarger:For any n × n matrix, t he characteristic polynomial is of the formfA(λ) = (−λ)n+ tr(A)(−λ)n−1+ · · · + det(A) .Pro of. The pattern, where all the entries are in the diagonal leads to a term (A11− λ) ·(A22− λ)...(Ann− λ) which is (−λn) + (A11+ ... + Ann)(−λ)n−1+ ... The rest of this as wellas the other pat t erns only give us terms which ar e of order λn−2or smaller.How many eigenvalues do we have? For real eigenvalues, it depends. A rotation in the planewith an angle different from 0 or π has no real eigenvector. The eigenvalues are complex inthat case:2 For a rotation A ="cos(t) sin(t)− sin(t) cos(t)#the characteristic polynomial is λ2− 2 cos(α) + 1which has the roots cos(α) ± i sin(α ) = eiα.Allowing complex eigenvalues is really a blessing. The structure is very simple:Fundamental theorem of algebra: For a n × n matrix A, the characteristicpolynomial has exactly n roots. There are therefore exactly n eigenva lues of A if wecount them with multiplicity.Pro of1One only has to show a polynomial p(z) = zn+ an−1zn−1+ · · · + a1z + a0always has a rootz0We can then factor out p(z) = (z − z0)g(z) where g(z) is a polynomial of degree (n − 1) anduse induction in n. Assume now that in contrary the polynomial p has no root. Cauchy’s integraltheorem then tellsZ|z|=r|dzzp(z)=2πip(0)6= 0 . (1)On the other hand, for all r,|Z|z|=r|dzzp(z)| ≤ 2πrmax|z|=r1|zp(z)|=2πmin|z|=rp(z). (2)The right hand side goes to 0 for r → ∞ because|p(z)| ≥ |z|n|(1 −|an−1||z|− · · · −|a0||z|n)which goes to infinity for r → ∞. The two equations (1) and (2) form a contradiction. Theassumption that p has no root was therefore not possible.If λ1, . . . , λnare the eigenvalues of A, thenfA(λ) = (λ1− λ)(λ2− λ) . . . (λn− λ) .Comparing coefficients, we know now the following important fact:The determinant of A is the product of the eigenvalues. The trace is the sum of theeigenvalues.We can therefore often compute the eigenvalues3 Find the eigenvalues of the matrixA ="3 75 5#Because each row adds up to 10, this is an eigenvalue: you can check that"11#. We can alsoread off the trace 8. Because the eigenvalues add up to 8 the other eigenvalue is −2. Thisexample seems special but it often occurs in textbooks. Try it out: what are the eigenvaluesofA ="11 10012 101#?1A. R. Schep. A Simple Complex Analysis and an Advanced Calculus Proof of the Fundamental theorem ofAlgebra. Mathematical Monthly, 116, p 67-68, 20094 Find the eigenvalues of the matrixA =1 2 3 4 50 2 3 4 50 0 3 4 50 0 0 4 50 0 0 0 5We can immediately compute the characteristic polynomial in t his case because A − λI5isstill upper triangular so that the determinant is the product of the diagonal entries. We seethat the eigenvalues are 1, 2, 3, 4, 5.The eigenvalues of an upper or lower triangular matrix are the diagonal entries ofthe matrix.5 How do we construct 2x2 matrices which have integer eigenvectors and integer eigenvalues?Just take an integer matrix for which the r ow vectors have the same sum. Then this sumis an eigenvalue to the eigenvector"11#. The other eigenvalue can be obtained by noticingthat the trace of the matrix is the sum of the eigenvalues. For example, the matrix"6 72 11#has the eigenvalue 13 and because the sum of t he eigenvalues is 18 a second eigenvalue 5.A matrix with nonnegative entries for which the sum of the columns entries add upto 1 is called a Markov matrix.Markov Matrices have a n eigenvalue 1.Pro of. The eigenvalues of A and ATare the same because they have the same characteristicpolynomial. The matrix AThas an eigenvector [1, 1, 1, 1, 1]T.6A =1/2 1/3 1/41/4 1/3 1/31/4 1/3 5/12This vector ~v defines an equilibrium point of the Markov pr ocess.7 If A ="1/3 1/22/3 1/2#. Then [3/7, 4/7] is the equilibrium eigenvector to the eigenvalue 1.Homework due April 13, 20111 a ) Find the characteristic polynomial and the eigenvalues of the matrixA =1 1 12 0 14 −1 0.b) Find the eigenvalues of A =100 1 1 1 11 100 1 1 11 1 100 1 11 1 1 100 11 1 1 1 100.2 a ) Verify that n × n matrix has a at least one real eigenvalue if n is odd.b) Find a 4 × 4 matrix, f or which there is no real eigenvalue.c) Verify that a symmetric 2 × 2 matrix has only real eigenvalues.3 a ) Verify that for a partitioned matr ixC ="A 00 B#,the union of the eigenvalues of A and B are the eigenvalues of C.b) Assume we have an eigenvalue ~v of A use this to find an eigenvector of C.Similarly, if ~w is an eigenvector of B, build an eigenvector of C.(*) Optional: Make some experiments with random matrices: The following Mathemat-ica code computes Eigenvalues of random matrices. You will observe Girko’s circularlaw.M=1000;A=Table [Random[] −1/2 ,{M} ,{M} ] ;e=Eigenvalues [A ] ;d=Table [Min[ Table [ I f [ i==j , 1 0 ,Abs [ e [ [ i ]] − e [ [ j ] ] ] ] , { j ,M} ] ] , { i ,M} ] ;a=Max[ d ] ; b=Min[ d ] ;Graphics [ Table [ {Hue [ ( d [ [ j ]] − a ) / (b−a ) ] ,Point [ {Re [ e [ [ j ] ] ] , Im [ e [ [ j ] ] ] } ] } , { j ,M} ] ]