avatar
C*a
1
Given two matrices M and K and a vector,f
system [M+{\delta}K]u^(n+1)=Mu^(n)
it's required that ||u^(n+1)||< ||u^(n)||.
Please show that this condition leads to
||u^(n+1)|| < |\lembda|_max ||u^(n)||,where \lembda_max is the
biggest eigenvalue of the generized eigenvalue problem of form
Xv= \lembda Yv, X,Y are two matrices. For the specific system they
are M+{\delta}K and M,respectively.
那位稍微指点一下。。。线性代数实在是忘了差不多了
avatar
g*i
2
is there any condition for M and M+delta K?
If M is invertible, and M+delta K is diagonalizable,
then it works out

【在 C******a 的大作中提到】
: Given two matrices M and K and a vector,f
: system [M+{\delta}K]u^(n+1)=Mu^(n)
: it's required that ||u^(n+1)||< ||u^(n)||.
: Please show that this condition leads to
: ||u^(n+1)|| < |\lembda|_max ||u^(n)||,where \lembda_max is the
: biggest eigenvalue of the generized eigenvalue problem of form
: Xv= \lembda Yv, X,Y are two matrices. For the specific system they
: are M+{\delta}K and M,respectively.
: 那位稍微指点一下。。。线性代数实在是忘了差不多了

avatar
C*a
3
there's a error in typing:the system should be
M u^(n+1)= [M+{\delta}K] u^(n)
M is SPD, K is -Laplacian.

【在 g***i 的大作中提到】
: is there any condition for M and M+delta K?
: If M is invertible, and M+delta K is diagonalizable,
: then it works out

avatar
g*i
4
then u can try the eigendecomposition of
M+delta K?

【在 C******a 的大作中提到】
: there's a error in typing:the system should be
: M u^(n+1)= [M+{\delta}K] u^(n)
: M is SPD, K is -Laplacian.

avatar
C*a
5
i don't understand, please show me...

【在 g***i 的大作中提到】
: then u can try the eigendecomposition of
: M+delta K?

avatar
g*i
6
( M+\delta K ) V = M V \Lambda
here \Lambda is the diagonal matrix with
diagonal entries to be the eigenvalues
so
M V \Lambda V' = M+\delta K
...
may or may not work. pls check

【在 C******a 的大作中提到】
: i don't understand, please show me...
avatar
C*a
7

from here,we get Mu^(n+1)=M V\Lambda V^(-1) u^(n)
can we say u^(n+1)=V \lambda V^(-1)?
if so, ||u^(n+1)||i don't even need the constraints of ||u^(n+1)||
【在 g***i 的大作中提到】
: ( M+\delta K ) V = M V \Lambda
: here \Lambda is the diagonal matrix with
: diagonal entries to be the eigenvalues
: so
: M V \Lambda V' = M+\delta K
: ...
: may or may not work. pls check

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