DIAGONALIZATION Math 21b, O.KnillSUMMARY. A n × n matrix, A~v = λ~v with eigenvalue λ and eigenvector ~v. The eigenvalues are the roots ofthe characteristic polynomial fA(λ) = det(λ − A) = λn− tr(A)λn−1+ ... + (−1)ndet(A). The eigenvectors tothe eigenvalue λ are in ker(λ − A). The number of times, an eigenvalue λ occurs in the full list of n roots offA(λ) is called algebraic multiplicity. It is bigger or equal than the geometric multiplicity: dim(ker(λ − A).EXAMPLE. The eigenvalues ofa bc dare λ±= T /2 +pT2/4 − D, where T = a + d is the trace andD = ad − bc is the determinant of A. If c 6= 0, the eigenvectors are v±=λ±− dcif c = 0, then a, d areeigenvalues to the eigenvectorsa0and−ba − d. If a = d, then the second eigenvector is parallel to thefirst and the geometric multiplicity of the eigenvalue a = d is 1.EIGENBASIS. If A has n different eigenvalues, then A has an eigenbasis, consisting of eigenvectors of A.DIAGONALIZATION. How does the matrix A look in an eigenbasis? If S is the matrix with the eigenvectors ascolumns, then we knowB = S−1AS . We have S~ei= ~viand AS~ei= λi~viwe know S−1AS~ei= λi~ei. Therefore,B is diagonal with diagonal entries λi.EXAMPLE. A =2 31 2has the eigenvalues λ1= 2 +√3 with eigenvector ~v1= [√3, 1] and the eigenvaluesλ2= 2 −√3 with eigenvector ~v2= [−√3, 1]. Form S =√3 −√31 1and check S−1AS = D is diagonal.APPLICATION: FUNCTIONAL CALCULUS. Let A be the matrix in the above example. What is A100+A37−1? The trick is to diagonalize A: B = S−1AS, then Bk= S−1AkS and We can compute A100+ A37−1 =S(B100+ B37− 1)S−1.APPLICATION: SOLVING LINEAR SYSTEMS. x(t+1) = Ax(t) has the solution x(n) = Anx(0). To computeAn, we diagonalize A and get x(n) = SBnS−1x(0). This is an explicit formula.SIMILAR MATRICES HAVE THE SAME EIGENVALUES.One can see this in two ways:1) If B = S−1AS and ~v is an eigenvector of B to the eigenvalue λ, then S~v is an eigenvector of A to theeigenvalue λ.2) From det(S−1AS) = det(A), we know that the characteristic polynomials fB(λ) = det(λ − B) = det(λ −S−1AS) = det(S−1(λ − AS) = det((λ − A) = fA(λ) are the same.CONSEQUENCES.1) Because the characteristic polynomials of similar matrices agree, the trace tr(A) of similar matrices agrees.2) The trace is the sum of the eigenvalues of A. (Compare the trace of A with the trace of the diagonalizematrix.)THE CAYLEY HAMILTON THEOREM. If A is diagonalizable, then fA(A) = 0.PROOF. The DIAGONALIZATION B = S−1AS has the eigenvalues in the diagonal. So fA(B), whichcontains fA(λi) in the diagonal is zero. From fA(B) = 0 we get SfA(B)S−1= fA(A) = 0.By the way: the theorem holds for all matrices: the coefficients of a general matrix can be changed a tiny bit so that all eigenvalues are different. For any suchperturbations one has fA(A) = 0. Because the coefficients of fA(A) depend continuously on A, they are zero 0 in general.CRITERIA FOR SIMILARITY.• If A and B have the same characteristic polynomial and diagonalizable, then they are similar.• If A and B have a different determinant or trace, they are not similar.• If A has an eigenvalue which is not an eigenvalue of B, then they are not similar.WHY DO WE WANT TO DIAGONALIZE?1) FUNCTIONAL CALCULUS. If p(x) = 1 + x + x2+ x3/3! + x4/4! be a polynomial and A is a matrix, thenp(A) = 1 + A + A2/2! + A3/3! + A4/4! is a matrix. If B = S−1AS is diagonal with diagonal entries λi, thenp(B) is diagonal with diagonal entries p(λi). And p(A) = S p(B)S−1. This speeds up the calculation becausematrix multiplication costs much. The matrix p(A) can be written down with three matrix multiplications,because p(B) is diagonal.2) SOLVING LINEAR DIFFERENTIAL EQUATIONS. A differential equation˙~v = A~v is solved by~v(t) = eAt~v(0), where eAt= 1 + At + A2t2/2! + A3t3/3!...)x(0). (Differentiate this sum with respect to tto get AeAt~v(0) = A~x(t).) If we write this in an eigenbasis of A, then ~y(t) = eBt~y(0) with the diagonalmatrix B =λ10 . . . 00 λ2. . . 00 0 . . . λn. In other words, we have then explicit solutions yj(t) = eλjtyj(0). Lineardifferential equations later in this course. It is important motivation.3) STOCHASTIC DYNAMICS (i.e MARKOV PROCESSES). Complicated systems can be modeled by puttingprobabilities on each possible event and computing the probabilities that an event switches to any other event.This defines a transition matrix. Such a matrix always has an eigenvalue 1. The corresponding eigenvectoris the stable probability distribution on the states. If we want to understand, how fast things settle to thisequilibrium, we need to know the other eigenvalues and eigenvectors.MOLECULAR VIBRATIONS. While quantum mechanics describes the motion of atoms in molecules, thevibrations can be described classically, when treating the atoms as ”balls” connected with springs. Such ap-proximations are necessary when dealing with large atoms, where quantum mechanical computations would betoo costly. Examples of simple molecules are white phosphorus P4, which has tetrahedronal shape or methanCH4the simplest organic compound or freon, CF2Cl2which is used in refrigerants. Caffeine or aspirin formmore complicated molecules.Freon CF2Cl2Caffeine C8H10N4O2Aspirin C9H8O4WHITE PHOSPHORUS VIBRATIONS. (Differential equations appear later, the context is motivation at thisstage). Let x1, x2, x3, x4be the positions of the four phosphorus atoms (each of them is a 3-vector). The inter-atomar forces bonding the atoms is modeled by springs. The first atom feels a force x2−x1+ x3−x1+ x4−x1and is accelerated in the same amount. Let’s just chose units so that the force is equal to the acceleration. Then¨x1= (x2− x1) + (x3− x1) + (x4− x1)¨x2= (x3− x2) + (x4− x2) + (x1− x2)¨x3= (x4− x3) + (x1− x3) + (x2− x3)¨x4= (x1− x4) + (x2− x4) + (x3− x4)which has the form ¨x =Ax, where the 4 × 4 ma-trixA =−3 1 1 11 −3 1 11 1 −3 11 1 1 −3v1=1111, v2=−1001, v3=−1010, v4=−1100are the eigenvectors to the eigenvalues λ1= 0, λ2= −4, λ3= −4, λ3= −4. With S = [v1v2v3v4],the matrix B = S−1BS is diagonal with entries 0, −4, −4, −4. The coordinates yi= Sxisatisfy¨y1= 0, ¨y2= −4y2, ¨y3= −4y3, ¨y4= −4y4which we can solve y0which is the center of mass
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