10 Eigenvalue problems/complex numbersRead: Boas Ch. 3, sec. 10-12, Ch. 210.1 Eigenvalues and eigenvectorsFigure 1: Left: 3 masses for computation of inertia tensor. Right: imagine rotating cylinder aroundaxis of symmetry, or around one rotated by an angle.Let’s start with a physical example which illustrates the kind of math we needto develop. A set of masses or a rigid body with some mass distribution can becharacterized in its resistance to rotation about any axis by its inertia tensorIij≡Xkmkr2kδij−Xkmkxk,ixk,j, (1)where k is an index that runs over a set of assumed discrete particles with massmk. The indices i and j refer to Cartesian coordinates x1≡ x, x2≡ y, and x3≡ z,and finally rkis the distance of particle k from the origin,rk=qx2k,1+ x2k,2+ x2k,3(2)I need to convince you that this has something to do with what you learned inelementary physics about the moment of inertia associated with the rotation ofa rigid body about some axis. First imagine there is a point mass m located at(a,0,0). There’s only one mass so that the sum over k has only one term. Checkthat I11= m · a2− m · a2= m(a − a) = 0, I22= I33= m · a2− m · 0 · 0 = ma2.These correspond to the formulae you learned for the moment of inertia associatedwith rotating a point mass about x axis where it sits, a distance 0 away, or aboutthe y or z axes, a distance a away.1In a more general situation as shown in the figure, we may have many massesindexed by k in (1). For the situation shown, let’s calculate the Iijexplicitly:I11= (2 + 2 + 2) − (1 + 0 + 1) = 4 (3)I12= −(1 + 0 + 0) = −1, (4)etc. We find for the matrix or 2nd rank tensor IijI =4 −1 −1−1 4 −1−1 −1 4. (5)Note I is symmetric due to its definition.mechanics:|Li = I|ωi (6)|τi = I|αi, (7)where now the vectors |ωi and |αi have components which are angular velocitiesand accelerations, respectively, around each of three axes. Since there are off-diagonal elements of Iij, it’s hard to get much intuition for what it means; notefor example that Lx= Ixxωx+ Ixyωy+ Ixzωz. It is possible, however, to find abasis (rotate to a new set of coordinates) where the inertia tensor is diagonal, i.e.L0x= I0xxω0x, L0y= I0yyω0y, L0z= I0zzω0z. This should be intuitive to you as you areused to calculating the moments of inertia for highly symmetric situations, as theI for a cylinder rotated around its own axis, as shown in Fig. 1b. As also shownthere, however, you can calculate the same thing for a rotation around a z0axis,at a different angle with respect to the cylinder axis. The Iijis no longer diagonal,but some very general symmetric matrix for such coordinates. So the idea is: insome general case, let’s find a coordinate system where the tensor is diagonal. Theelements on the diagonal will be the moments of inertia with respect to the threeprincipal axes.We know the effect of a rotation R on both vectors and matrices now: |Ai →|A0i = R|Ai, and M → M0= R−1MR (similarity transform). Remember thatrotations are orthogonal, so I could write M → M0= RTM R, or RM0= MR(I’ll start dropping the underline for matrices, and hope that the context will makeclear what is meant). Let’s do this for the inertia tensor:I11I12I13I21I22I23I31I32I33R11R12R13R21R22R23R31R32R33=R11R12R13R21R22R23R31R32R33λ10 00 λ200 0 λ3, (8)2and now notice that if we consider R as a set of 3 column vectors~Riwe haveI|R1i = λ1|R1i ; I|R2i = λ2|R2i ; I|R3i = λ3|R3i. (9)In linear algebra a matrix equation M|vi = λ|vi is known as an eigenvalue problem(Eigen = “proper” or “own” in German). λ is the eigenvalue and |vi is called theeigenvector. In general for a matrix M of rank d, there are d eigenvalues and deigenvectors corresponding to them. It’s a very special situation: you’re asking fora particular value λ and a particular vector |vi such that the effect of the lineartransformation M on the vector only dilates it (multiplies it by a number), butdoes not change its direction.Now note that the linear equation is of the homogeneous type (homogeneousmeans: matrix times vector (x, y, z) =0, not some constant vector):I11I12I13I21I22I23I31I32I33v1v2v3= λv1v!v3, orI11− λ I12I13I21I22− λ I23I31I32I33− λv1v2v3= 0 (10)We solve such an equation by observing that it can only have a solution if itsdeterminant is zero (if you don’t buy this, try a general problem·a bc d¸·xy¸= 0,and you’ll see that you’ll arrive at a contradiction unless ad − bc = 0). So todiagonalize the matrix I we solve the secular equation det(I − λ1)=0, where 1 isthe identity matrix.For the particular case of the set of masses in the figure above, we have¯¯¯¯¯¯4 − λ −1 −1−1 4 − λ −1−1 −1 4 − λ¯¯¯¯¯¯= (4 − λ)¯¯¯¯4 − λ −1−1 4 − λ¯¯¯¯− (−1)¯¯¯¯−1 −1−1 4 − λ¯¯¯¯+(−1)¯¯¯¯−1 4 − λ−1 −1¯¯¯¯= 0⇒ (λ − 2)(λ − 5)2= 0, (11)so the eigenvalues of this problem are λ = 2, 5. λ =2 has multiplicity 1, and λ = 5multiplicity 2, since there are 2 roots. We say there are two eigenvectors of Icorresponding to λ = 5.3The next task is to find the eigenvectors corresponding to each eigenvalue. Let’sstart with λ = 2:4 − 2 −1 −1−1 4 − 2 −1−1 −1 4 − 2v1v2v3=000⇒ 2v1− v2− v3= 0−v1+ 2v2− v3= 0−v1− v2+ 2v3= 0⇒ v1= v2= v3≡ v. (12)So any vector of the form (v, v, v) works ∀v. This is natural because of thehomogeneous nature of the problem, but usually we get rid of the ambiguity byspecifying only the normalized eigenvector, i.e. the unit eigenvector v21+v22+v23= 1,so here we have v = 1/√3,|~v1i =1√3111(13)Now λ = 5. This is a little funny, because now every element of (I − λ1) is-1! So the 2 equations which determine the remaining eigenvectors are the same,v1+ v2+ v3= 0. Even if we impose a normalization condition h~v|~vi = 1, we do nothave enough equations to fix v1, v2, v3; there are infinitely many choices. We couldtake|~v2i =1√210−1, (14)which satisfies our equation. Then any other one which has eigenvalue 5 and islinearly independent of the one we just found is a good answer to the question, whatare the eigenvectors of I? However it’s conventional to look for an orthonormal set,eigenvectors which are normalized to 1 and mutually orthogonal.