s**p
发帖数: 207 | 1 不好意思，脑子没转过来
如何把一个矩阵的转置用初等变换实现，如果矩阵很大，希望这个变幻是稀疏的
比如说,矩阵
[1 2 3
A= 4 5 6
7 8 9]
转置成
[1 4 7
2 5 8
3 6 9] =A'
发现行变换列变换都不成
[ 0 0 1;] [ 3 2 1 ]
[ 0 1 0;] A= [ 6 5 4 ]
[ 1 0 0 ] [ 9 8 7 ]
或者
[ 0 0 1; ] [ 0 0 1;] [ 9 8 7;]
[ 0 1 0; ] A [ 0 1 0;] = [ 6 5 4;]
[ 1 0 0 ] [ 1 0 0 ] [ 3 2 1 ]
but none of them are transpose. | B********e
发帖数: 10014 | 2 I guess you are right
you can't complete it by elementary transform
actually 'elementary transfrom' is used to keep the equivalence relation bet
ween matrix when we solve the equation system. 'transpose' doesnt do that e
ven though it's also linear transform on real field.
in some sense 'transpose' is as 'elementary' as 'elementary transform'.
and i'm wondering why you want to carry out transpose by 'elementary transfo
rm'? after all, transpose is a simpler operation, you can do it by nothing
【在 s**p 的大作中提到】
: 不好意思，脑子没转过来
: 如何把一个矩阵的转置用初等变换实现，如果矩阵很大，希望这个变幻是稀疏的
: 比如说,矩阵
: [1 2 3
: A= 4 5 6
: 7 8 9]
: 转置成
: [1 4 7
: 2 5 8
: 3 6 9] =A'
| s**p
发帖数: 207 | 3 ok,
i have a symmetric matrix A and I want to solve Ax=b
if I can get its upper left matrix, let's say B, (the diagonal term is 1/2
Aii)
then B+B'=A; if I have a sparse transformation matrix T: B'=TB
Ax=b=> (B+B')x=b => (I+T)Bx=b
Bx=(I+T)^-1 b
(I+T)^-1 is easy. because T is sparse
so x is obtained in O(N^2), inversion free.
i guess T must be dense, otherwise, too simple to be true
bet
e
transfo
nothing
【在 B********e 的大作中提到】
: I guess you are right
: you can't complete it by elementary transform
: actually 'elementary transfrom' is used to keep the equivalence relation bet
: ween matrix when we solve the equation system. 'transpose' doesnt do that e
: ven though it's also linear transform on real field.
: in some sense 'transpose' is as 'elementary' as 'elementary transform'.
: and i'm wondering why you want to carry out transpose by 'elementary transfo
: rm'? after all, transpose is a simpler operation, you can do it by nothing
| c******m
发帖数: 599 | 4 B'=TB
so T=B'*B^{-1}
你觉得T会是怎么样的?
【在 s**p 的大作中提到】
: ok,
: i have a symmetric matrix A and I want to solve Ax=b
: if I can get its upper left matrix, let's say B, (the diagonal term is 1/2
: Aii)
: then B+B'=A; if I have a sparse transformation matrix T: B'=TB
: Ax=b=> (B+B')x=b => (I+T)Bx=b
: Bx=(I+T)^-1 b
: (I+T)^-1 is easy. because T is sparse
: so x is obtained in O(N^2), inversion free.
: i guess T must be dense, otherwise, too simple to be true
| B********e
发帖数: 10014 | 5 got you
不过如你所言，根据能量守恒，不可能是真的，呵呵
【在 s**p 的大作中提到】
: ok,
: i have a symmetric matrix A and I want to solve Ax=b
: if I can get its upper left matrix, let's say B, (the diagonal term is 1/2
: Aii)
: then B+B'=A; if I have a sparse transformation matrix T: B'=TB
: Ax=b=> (B+B')x=b => (I+T)Bx=b
: Bx=(I+T)^-1 b
: (I+T)^-1 is easy. because T is sparse
: so x is obtained in O(N^2), inversion free.
: i guess T must be dense, otherwise, too simple to be true
| s**p
发帖数: 207 | 6 犀利
【在 c******m 的大作中提到】
: B'=TB
: so T=B'*B^{-1}
: 你觉得T会是怎么样的?
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