Mathematics 307|December 6, 1995
Generalized eigenvalues and conservative systems
Assume M positive denite. To solve
Mx00 + Kx = 0
we convert it into a rst order system by setting
y = Mx
and getting
0 x = 0 M ,1 x
y
,K 0
y
or
v0 = Av; v = eAt v0
We apply Gauss elimination to factor
M = L tL; M ,1 = tL,1 L,1 ; tL M ,1 L = I
and then multiply
t
L 0
0 M ,1 tL,1 0 =
0
I
0 L,1 ,K 0
0
L
,L,1 K tL,1 0
The matrix
K = L,1 K tL,1
is still symmetric. We can nd an orthogonal X such that
X ,1 K X = D
where D is diagonal. So X ,1 0
0 I X 0 = 0 I
0
X ,1 ,K 0 0 X
,D 0
Finally we use
p
! 0
0 1
if ! = k. So we set
0 1
,k 0
Thus if
0
0 = 0 !
1
,! 0
2!
= 64
and get
,1
!
0
0 I
0 I
,D 0
0 0 3
0 !2 0 75
0 0 !3 1
,1
0
0
0 I = ,
0 = C
,1
t
0
X
0
L
0
Y= 0 I 0
X ,1 0 L,1
,1 tL
X
0
=
0
X ,1 L,1
t ,1
,1 0 L
X
,
1
Y =
0
LX
Solving the generalized eigenvalue problem
then
Finally
2
Y A Y ,1 = ,
0 0 ; A = Y ,1 AY; eAt = Y ,1eCt Y
!it , sin !i t
eCt = cos
sin !i t cos !i t
Lower triangular systems
We want to nd x such that Lx = c. We nd x0, x1 etc.
`0;0x0 = c0
`i;0x0 + `i;1x1 + + `i;i xi = ci
x0 = c0 =`0;0
xi = ci , `i;0 x0 , `i;1 x1 , We will call this with a parameter n, the dimension, so we can use partial vectors anmd matrices.
Cholesky factorization
We have
which we solve by induction.
0
` t` = ta
0 t
a A
2 = ` = a
` = ,1 a
` t` + t = A
t = A , ` t`
Note that ` t` is a square matrix with entries `i `j .
For i = 0 to n ,p2
= `i;i := mi;i
for j = i + 1 to n , 1
`j;i := mj;i=
for j = i + 1 to n , 1
for k = i + 1 to j
pmj;k := mj;k , `j;i `k;i
`n,1;n,1 := mn,1;n,1
Lower triangular inverses
To solve
or
0
x X
0 =I
` = 1; x + X` = 0
is the easiest way to do it inductively. Note that we should also have a routine for nding Lx where L is
lower triangular.