The Generalized Eigenvalue-Eigenvector Problem or How to Simultaneously Diagonalize Two Quadratic orms The following records deal with the generalized eigenvalue problem of two positive definite matrices G and C which are connected with two quadratic forms Q and Q. This problem arises in the study of map projections when one aims at learning about the distortion of the mapping of one manifold onto another manifold. In particular, Tissot has shown that during the mapping process a circle of infinitesimal radius in the one manifold becomes an ellipse in the map the semi axes of which point towards the minimal and maximal distortion. or this reason the Tissot ellipse is also called ellipse of distortion. The starting point of the following statements is the Cauchy-Green deformation tensor (matrix) C connected with the metric tensor (matrix) G. Because both C and G are positive definite, they geometrically represent ellipses which are neither parallel to each other nor aligned with the coordinate axes, in general. When we talk about ellipses G and C in the subsequent derivations we always mean the ellipses represented by matrices G and C. In the simple example starting from the bottom, however, it is assumed that ellipse G is already aligned with the coordinate axes. This does not mean any serious limitation because one of the quadratic forms can always be easily aligned with the coordinate axis by a simple rotation of the coordinate system. Use the MATLAB-files gevp.m and plotell.m to compute and plot more general examples. Given two matrices C and G the generalized eigenvalues, and eigenvectors [, ] are determined from the equations det(c G) 0 (eigenvalues) and C G (eigenvectors, eigendirections). In short, the eigenvalues and eigenvectors are i i i given by, tr(cg ) tr(cg ) 4det(CG ) (C G ) G (C G )(C G )G (C G ) G (C G ) G (C G )(C G )G (C G ) G C G (C G ) (C G ) C G
The eigenvectors are normalized in such a way that using :, the postulates 'G I and 'C : Diag(, ) are satisfied. These requirements as well as size, shape and orientation of the Tissot ellipses will not be influenced by a change of sign within matrix. Postmultiplying by one of the matrices SDiag(,), S Diag(, ), S3 Diag(, ) will only cause a change of the mutual orientation of the eigendirections: det det(s 3) 0, det(s ) det(s ) 0.
3 Example: Q: x + 4y = x' G x =, Q:.5x - 3xy +.5y = x' C x =, G 0 0 4 C.5.5.5.5 ) In the first step we would like to plot ellipse G connected with Q and that's why we perform an eigenvector-eigenvalue decomposition of matrix G : det G I 0 4 0, 4 eigenvalues G I 0 eigenvector semi-major axis: a semi-minor axis: b 0.5 0 0 0 3 0 3 0 0 G I 0 0 0 eigenvector y 0.5 b a x
4 ) The same procedure is applied for matrix C in order to plot ellipse C connected with Q:.5.5 det C I 0.5.5 0.5.5, 4 (eigenvalues) semi-major axis: a semi-minor axis: b 0.5.5.5 C I 0.5.5 0.5.5 put s s.5.5 C I 0.5.5 0.5.5 (eigenvector ) put s s (eigenvector ) y 0.5 a b x
5 3) Now we solve the generalized eigenvalue problem which can be geometrically interpreted as a simultaneous diagonalization of Q and Q. Without loss of generality we state the problem in such a way that G will become a circle [ det(c G) 0 ]. In order to perform a comparison of the ellipses the problem can also be formulated the other way round [C becomes a circle: det(g C) 0 ]..5.5 9 det(c G) 0.5.5 4 0.5.5 4 4 5 369 6 0.369,.763 semi-major axis: a3.663 semi-minor axis: b3 0.608 C G 0.5.5.38.5.5.054.38.5 0.5.5 4 put s.454s s Normalization will be done later.454 C G 0.5.5 0.36.5.5 8.554 0.36.5 0.5.5 4 put t 0.754t t Normalization will be done later 0.754
6 s 0,.454 0.754 0 t The eigendirections of the generalized eigenvalue problem are normalized so that 'G I and 'C : Diag(, ) is fulfilled. s 0.454 0 s 0 'G 0 t 0.754 0 4.454 0.754 0 t 9.7s 0 0.3t I s 0.330, t 0.94363 0.330 0.9436, 0.478 0.655 0.330 0.9436 0.478 0.655 G after simultaneous diagonalization with C y a3 b3 x Direction of Direction of C after simultaneous diagonalization with G
7 Check: 'C : Diag(, )? 0.330 0.478.5.5 0.330 0.9436 0.369 0 0.9436 0.655.5.5 0.478 0.655 0.769, e, e cos 54.9477, cos 9.9480 R tan det G tan R R tan G G tan tan tan 70.6699 R tan det G tan R R tan G G tan tan tan 9.330 where Gij are the elements of G and Rij are the elements of the upper triangular matrix R as being derived from the Cholesky decomposition of G R ' R. Clearly, 90 is satisfied. 4) The whole procedure can also be done using elementary coordinate transformations: Q: x + 4y = versus Q:.5x - 3xy +.5y = a) Scale x and y so that Q becomes a unit circle, i.e. put x'=x, y'=y y=0.5y' Q': x' + y' = versus Q':.5x' -.5x'y' + 0.65y' = b) Rotate x' and y' so that Q' is aligned with the coordinate axis and the mixed term in x'y' vanishes. The rotation matrix is obtained through the solution of the eigenvector-eigenvalue decomposition of M..5 0.75 x ' x ' x ' ' Q : x ',y' x ',y' M x ',y' ' 0.75 0.65 y' y' y' eigenvalues eigendirections= rotation matrix
8 Put x" x ' y" y' and diagonalize M in order to find. Diagonalization of M : Eigenvalues.5 0.75 5 369 0 0.75 0.65 6 0.369,.763 Diagonalization of M : Eigendirections.5 0.75 0 s.85078 0.330 s.85078 0.94363.5 0.75 0 t 0.35078 0.94363 t 0.35078 0.330 0.330 0.94363 0.94363 0.330, cos sin sin cos 70.6699 cos sin x ' x" x ' x"cos y"sin sin cos y' y" y' x"sin y"cos c) Apply the rotation Q : x y " " " Q : 0.369x.7630y x y " " " " "
9 Remarks: A few general statements follow associated with the treatment of det(g C) 0 instead of det(c G) 0 (i) det(g C) 0 det(c G) 0 (ii) 'G (iii) 'C I det(g C) 0 (iv) det(g C) 0 det(c G I) 0 (v) eig( G,C ) = eig(inv( C )* G,eye()) (MATLAB code) In order to show this start from det(g C) 0, =0.369, =.763 G C 0 0.0955 0.5485 0 s 0.7546 G C 0 5.90775 4.4465 0 t.4539 s 0, 0.7546.4539 0 t 'C I s 0.56766, t 0.550 0.56766 0.550 0.0996 0.7848 'G
0 Summary: Two quadratic forms Q, Q with positive definite matrices G and C and ellipses. G and C, which are not necessarily parallel to each other nor aligned with the coordinate axes Q b a Q a eigenvectors need not be normalized b Step : Transform Q to become G a circle of radius ; Q and C will change correspondingly Q ' Step : Eigenvectors of Q ' can be rotated into the eigendirections of Q ' Q ' Step 3: Now the eigenvalues can be compared. If necessary both quadratic forms can simultaneously be aligned with the coordinate axes Q " " ' Q Q