determinant2_adeterminant2_bDETERMINANTOR II, (Judgement day) Math 21b, O.KnillHOMEWORK: Section 6.2: 6,8,16,42,18*,44* and 6.3: 12DETERMINANT AND VOLUME. If A is a n × n matrix, then |det(A)| is the volume of the n-dimensiona lparallelepip ed Enspanned by the n column vectors vjof A.Proof. Use the QR decomposition A = QR, where Q is orthogonal and R isupper triangular. From QQT= 1, we get 1 = det(Q)det(QT) = det(Q)2see that|det(Q)| = 1. Therefore, det(A) = ±det(R). The determinant of R is the productof the ||ui|| = ||vi− projVj−1vi|| which was the distance from vito Vj−1. Thevolume vol(Ej) of a j-dimensional parallelepiped Ejwith base Ej−1in Vj−1andheight ||ui|| is vol(Ej−1)||uj||. Inductively vol(Ej) = ||uj||vol(Ej−1) and thereforevol(En) =Qnj=1||uj|| = det(R).The volume of a k dimensional parallelepiped defined by the vectors v1, . . . , vkispdet(ATA).Proof. QTQ = Ingives ATA = (QR)T(QR) = RTQTQR = RTR. So, det(RTR) = det(R)2= (Qkj=1||uj||)2.(Note that A is a n × k matrix and that ATA = RTR and R are k × k matrices.)ORIENTATION. Determinants allow to define the orientation of n vectors in n-dimensional spac e. This is”handy” because there is no ”rig ht hand rule” in hyperspace... To do so, define the matrix A with columnvectors vjand define the or ie ntation as the sign of det(A). In three dimensions, this agrees with the right handrule: if v1is the thumb, v2is the pointing finger and v3is the middle finger, then their orientation is positive.bidetx det(A) = iCRAMER’S RULE. This is an explicit formula for the solution of A~x =~b. If Aidenotes the matrix, where the column ~viof A is replace d by~b, thenxi= det(Ai)/det(A)Proof. det(Ai) = det([v1, . . . , b, . . . , vn] = det([v1, . . . , (Ax), . . . , vn] =det([v1, . . . ,Pixivi, . . . , vn] = xidet([v1, . . . , vi, . . . , vn]) = xidet(A)EXAMPLE. Solve the system 5x+3y = 8, 8x+5y = 2 using Cramers rule. This linear system with A =5 38 5and b =82. We get x = det8 32 5= 3 4y = det5 88 2= −54.GABRIEL CRAMER. (1704-1752), born in Geneva, Switzerland, he worked on geometryand analysis. Cramer used the rule named after him in a book ”Introduction `a l’analysedes lignes courbes alg´ebraique”, where he solved like this a system of equations with 5unknowns. According to a short biography o f Cramer by J.J O’Connor and E F Robertson,the rule had however been used a lready before by other mathematicians. Solving systemswith Cramers formulas is slower than by Gaussian elimination. The rule is still important.For example, if A or b depends on a parameter t, and we want to see how x depends onthe parameter t one can find explicit fo rmulas for (d/dt)xi(t).THE INVERSE OF A MATRIX. Because the columns of A−1are solutions of A~x = ~ei, where ~ejare bas isvectors, Cramers rule together with the Laplace expansion gives the formula:[A−1]ij= (−1)i+jdet(Aji)/det(A)Bij= (−1)i+jdet(Aji) is called the classical adjoint of A. Note the change ij → ji. Don’t confuse theclassical adjoint with the transpose ATwhich is sometimes also called the adjoint.EXAMPLE. A =2 3 15 2 46 0 7has det(A) = −17 and we get A−1=14 −21 10−11 8 −3−12 18 −11/(−17):B11= (−1)1+1det2 40 7= 14. B12= (−1)1+2det3 10 7= −21. B13= (−1)1+3det3 12 4= 1 0.B21= (−1)2+1det5 46 7= −11. B22= (−1)2+2det2 16 7= 8 . B23= (−1)2+3det2 15 4= −3.B31= (−1)3+1det5 26 0= −12. B32= (−1)3+2det2 36 0= 1 8. B33= (−1)3+3det2 35 2= −11.THE ART OF CALCULATING DETERMINANTS. When confronted with a matrix, it is good to go througha checklist of methods to crack the determinant. Often, there are different possibilities to solve the problem, inmany cases the solution is particularly simple using one method.• Is it a 2 × 2 or 3 × 3 matrix?• Is it a upper or lower triangular matrix?• Is it a partitioned matrix?• Is it a trick: like A1000?• Does geometry imply noninvertibility?• Do you see duplicated columns or rows?• Can you r ow r educe to a triangular case?• Are there only a few nonzero patters?• Laplace expansion with some row or column?• Later: Can you see the eigenvalues of A − λIn?EXAMPLES.1) A =1 2 3 4 52 4 6 8 105 5 5 5 41 3 2 7 43 2 8 4 9Try row reduction.2) A =2 1 0 0 01 1 1 0 00 0 2 1 00 0 0 3 10 0 0 0 4Laplace expans ion.3) A =1 1 0 0 01 2 2 0 00 1 1 0 00 0 0 1 30 0 0 4 2Partitioned matrix.4) A =1 6 10 1 152 8 17 1 290 0 3 8 120 0 0 4 90 0 0 0 5Make it triangular.APPLICATION HOFSTADTER BUTTERFLY. In solid state physics, one is interested in the function f(E) =det(L − EIn), whereL =λ cos(α) 1 0 · 0 11 λ cos(2α) 1 · · 00 1 · · · ·· · · · 1 00 · · 1 λ cos((n − 1)α) 11 0 · 0 1 λ cos(nα)describes an electr on in a periodic crystal, E is the energy and α = 2π/n . The electron can move as a Blochwave whenever the determinant is negative. These intervals for m the spectrum of the quantum mechanicalsystem. A physicist is interested in the rate of change of f (E) or its dependence on λ when E is fixed. .-4 -2 2 4123456The graph to the left shows the function E 7→log(|det(L − EIn)|) in the case λ = 2 and n = 5.In the energy interva ls, where this function is zero,the electron can move, otherwise the crystal is an in-sulator. The picture to the right shows the spectrumof the crystal depending on α. It is called the ”Hofs-tadter butterfly” made po pular in the book ”G¨odel,Escher Bach” by Douglas Hofstadter.DETERMINANTOR II, (Judgement day) Math 21b, O.KnillHOMEWORK: Section 6.2: 6,8,16,42,18*,44* and 6.3: 12DETERMINANT AND VOLUME. If A is a n × n matrix, then |det(A)| is the volume of the n-dimensiona lparallelepip ed Enspanned by the n column vectors vjof A.Proof. Use the QR decomposition A = QR, where Q is orthogonal and R isupper triangular. From QQT= 1, we get 1 = det(Q)det(QT) = det(Q)2see that|det(Q)| = 1. Therefore, det(A) = ±det(R). The determinant of R is the productof the ||ui|| = ||vi− projVj−1vi|| which was the distance from vito Vj−1. Thevolume vol(Ej) of a j-dimensional parallelepiped Ejwith base Ej−1in Vj−1andheight ||ui|| is vol(Ej−1)||uj||. Inductively vol(Ej) = ||uj||vol(Ej−1) and thereforevol(En) =Qnj=1||uj|| = det(R).The volume of a k
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