Math 19b: Linear Algebra with Probability Oliver Knill, Spring 2011Lecture 30: DiagonalizationDiagonalizationTwo matrices are called similar if S−1AS. A matrix is called diagonalizable if itis similar to a diagonal matrix.A matrix is diagonalizable if and only if it has an eigenbasis, a basis consisting ofeigenvectors.Pro of. If we have an eigenbasis, we have a coordinate transformation matrix S which contains theeigenvectors vias column vectors. To see that the matrix S−1AS is diagonal, we checkS−1ASei− S−1Avi= S−1λivi= λiS−1vi= λiei.On the other hand if A is diagonalizable, then we have a matrix S for which S−1AS = B is diago-nal. The column vectors of S are eigenvectors because the k’th column of the equation AS = BSshows Avi= λivi.Are all matrices diagonalizable? No! We need to have an eigenbasis and therefore that thegeometric multiplicities all agree with the algebraic multiplicities. We have seen that the shearmatrixA ="1 10 1#has the eigenvalues 1 for which the geometric multiplicity is smaller than the algebraic one. Thismatrix is not diagonalizable.Simple spectrumA matrix has simple spectrum, if all eigenvalues have algebraic multiplicity 1.If a matrix ha s simple spectrum, then it is diagonalizable.Pro of. Because the algebraic multiplicity is 1 for each eigenvalue and the geometric multiplicityis always at least 1, we have an eigenvector for each eigenvalue and so n eigenvalues.1 We have computed the eigenvalues of the rotation matrixA ="cos(α) −sin(α)sin(α) cos(α)#We have seen that the eigenvalues are eiα= cos(α) + i sin(α), the eigenvectors are"±i1#.The eigenvectors are the same for every rotation-dilation matrix. WithA ="a −bb a#, S ="i −i1 1#we haveS−1AS ="a + ib 00 a − ib#.Functional calculus2 What is A100+ A37− 1 if A ="2 31 2#? The matrix has the eigenvalues λ1= 2 +√3 witheigenvector ~v1= [√3, 1] a nd the eigenvalues λ2= 2 −√3. with eigenvector ~v2= [−√3, 1].Form S ="√3 −√31 1#and check S−1AS = D is diagonal. Because Bk= S−1AkS caneasily be computed, we know A100+ A37− 1 = S(B100+ B37− 1)S−1.Establishing similarity3 Show that the matrices A ="3 52 6#B ="4 43 5#are similar. Proof. They have the sameeigenvalues 8, 9 as you can see by inspecting the sum of rows and the trace. Both matricesare therefore diagonalizable and similar to the matrix"8 00 9#.• If A and B have the same characteristic polynomial and diagonalizable, t hen they aresimilar.• 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.• If A and B have the same eigenvalues but different geometric multiplicities, then theyare not similar.Without proof we mention the following result which gives an if and only if result for simi-larity:If A and B have the same eigenvalues with geometric multiplicities which agree andthe same holds for all powers Akand Bk, then A is similar to B.Cayley Hamilton theoremFor any polynomial p,‘ we can form the ma t rix p(A). For example, for p(x) = x2+ 2x + 3,we have p(A) = A2+ 2A + 3.If fAis the characteristic polynomial, we can form fA(A)If A is diagonalizable, then fA(A) = 0.The matrix B = S−1AS has the eigenvalues in the diagonal. So fA(B), which containsfA(λi) in the diagonal is zero. From fA(B) = 0 we get SfA(B)S−1= fA(A) = 0.The theorem holds for all matrices: the coefficients of a general matrix can be changed atiny bit so that all eigenvalues are different. For any such perturbations one has fA(A) = 0.Because the coefficients of fA(A) depend continuously on A, they are zero in general.An application in chemistryWhile quantum mechanics describes the motion of atoms in molecules, the vibratio ns canbe described classically, when treating the atoms as ” balls” connected with springs. Suchapproximations are necessary when dealing with large atoms, where quantum mechanicalcomputations would be too costly. Examples of simple molecules are white phosphorus P4,which has tetrahedronal shape or methan CH4the simplest organic compound or freon,CF2Cl2which is used in refrigerants.Freon CF2Cl2Caffeine C8H10N4O2Aspirin C9H8O41Let 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 aforce x2− x1+ x3− x1+ x4− x1and is accelerated in the same amount. Let’s just choseunits 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 the4 × 4 ma trixA =−3 1 1 11 −3 1 11 1 −3 11 1 1 −3v1=1111, v2=−1001, v3=−1010, v4=−1100arethe 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 thecenter of mass satisfies y0= a + bt (move molecule with constant speed). The motionsyi= aicos(2t) + bisin(2t) of the other eigenvectors are oscillations, called normal modes.The general motion of the molecule is a superposition of these modes.Homework due April 20, 20111 What is the probability that an upper triangular 3 × 3 matrix with entries 0 and 1 isdiagonalizable?2 Which of the following matrices are similar?A =1 1 0 00 1 0 00 0 1 10 0 0 1, B =1 1 0 00 1 1 00 0 1 00 0 0 1, C =1 1 0 00 1 1 00 0 1 10 0 0 1, D =1 0 0 00 1 1 00 0 1 10 0 0 1.3 Diagonalize the following mat rix in the complex:A =2 −3 0 03 2 0 00 0 5 60 0 6 51We grabbed the pdb Molecule files from http : //www.sci.ouc.bc.ca, translated them with ”povchem” from.pdb to .pov rendered them under