j**********7
发帖数: 4 | 1 Consider a matrix $J$. every eigenvalue of $J$ has positive real part.
Now let us consider infinite product of (1-J/n). Does
infinite product converges? and how about the converge rate?
Is there anyone can give some advice? thanks! | p*****k
发帖数: 318 | 2 for powers of matrix, you normally want to diagonalize the matrix first:
say P'JP = diag{j1,j2,....jm}, where P'=P^(-1), and j's are the eigenvalues of J.
if i understand your question correctly, you want to compute the limit of (I-J/n)^n when n->infty (where I is the identity matrix)?
insert bunch of PP'(=I) in the product:
(I-J/n)^n = PP'(I-J/n)PP'(I-J/n)...PP'(I-J/n)PP'=P{[P'(I-J/n)P]^n}P'
=P[diag{(1-j1/n)^n,(1-j2/n)^n,...,(1-jm/n)^n}]P'
so seems to me the limit when n->infty is:
P[diag{e^(-j | j**********7
发帖数: 4 | 3 Many thanks! It is quite clear and correct.
You are so kind :)
eigenvalues of J.
(I-J/n)^n when n->infty (where I is the identity matrix)?
J is diagonalizable http://en.wikipedia.org/wiki/Diagonalizable, the above procedure should work, no?
【在 p*****k 的大作中提到】
: for powers of matrix, you normally want to diagonalize the matrix first:
: say P'JP = diag{j1,j2,....jm}, where P'=P^(-1), and j's are the eigenvalues of J.
: if i understand your question correctly, you want to compute the limit of (I-J/n)^n when n->infty (where I is the identity matrix)?
: insert bunch of PP'(=I) in the product:
: (I-J/n)^n = PP'(I-J/n)PP'(I-J/n)...PP'(I-J/n)PP'=P{[P'(I-J/n)P]^n}P'
: =P[diag{(1-j1/n)^n,(1-j2/n)^n,...,(1-jm/n)^n}]P'
: so seems to me the limit when n->infty is:
: P[diag{e^(-j
| i****e
发帖数: 78 | 4 maybe the question is (1-J/n)^m as m-> infty? then positive real parts may
be related to the convergence and rate.
eigenvalues of J.
(I-J/n)^n when n->infty (where I is the identity matrix)?
J is diagonalizable http://en.wikipedia.org/wiki/Diagonalizable, the above procedure should work, no?
【在 p*****k 的大作中提到】
: for powers of matrix, you normally want to diagonalize the matrix first:
: say P'JP = diag{j1,j2,....jm}, where P'=P^(-1), and j's are the eigenvalues of J.
: if i understand your question correctly, you want to compute the limit of (I-J/n)^n when n->infty (where I is the identity matrix)?
: insert bunch of PP'(=I) in the product:
: (I-J/n)^n = PP'(I-J/n)PP'(I-J/n)...PP'(I-J/n)PP'=P{[P'(I-J/n)P]^n}P'
: =P[diag{(1-j1/n)^n,(1-j2/n)^n,...,(1-jm/n)^n}]P'
: so seems to me the limit when n->infty is:
: P[diag{e^(-j
| H****y
发帖数: 19 | 5 It's not (I-J/n)^n when n->infty, it should be Prod_n^infty (I-J/n)
If you take a ln, then the i^th diagonal element becomes Sum_n ln(1-Ji/n).
If you approximate ln(1-Ji/n) by -Ji/n, then it becomes Sum_n (-Ji/n), which
does not converge.
One can confirm this argument with Mathematica, using Ji=1, and calculate
Sum_n ln(1-1/n). You'll find the sum does not converge. | p*****k
发帖数: 318 | 6 thanks. the question makes sense now.
but you sure it does not converge? seems to me you are saying the log of
this infinite product is going to -\infty, so the product converges to zero,
no? |
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