c*******h
发帖数: 1096 | 1 if lambda is eigenvalue, 2-lambda is also eigenvalue
if 2-lambda is eigenvalue, 4-lambda is also eigenvalue
if 4-lambda is eigenvalue, 8-lambda is also eigenvalue
if 8-lambda is eigenvalue, 16-lambda is also eigenvalue
...
how many eigenvalues do you have? |
|
n*s
发帖数: 752 | 2 【 以下文字转载自 Mathematics 讨论区 】
发信人: nos (redemption), 信区: Mathematics
标 题: matrix eigenvalue question
发信站: BBS 未名空间站 (Tue Jun 3 09:54:44 2008), 转信
I have a matrix and it turns out
if lambda is eigenvalue, 2-lambda is also eigenvalue
is this a kind of symmetry? does it help solving the eigenvalue problem?
for example, does this property help in decomposing the matrix? |
|
n*******l
发帖数: 2911 | 3 Two cases.
1. If t=0 is an eignevalue of AB, then AB is singular and BA is singular too.
Hence t=0 is an eigenvalue of BA.
2. If t is a nonzero eigenvalue of AB, then there is a nonzero eigenvector x
such that ABx=tx and Bx can not be zero (if Bx=0, then ABx=0, but tx is not
zero, contradiction). Thus Bx is a nonzero eigenvector for BA with
eigenvalue t. |
|
n*s
发帖数: 752 | 4 I have a matrix and it turns out
if lambda is eigenvalue, 2-lambda is also eigenvalue
is this a kind of symmetry? does it help solving the eigenvalue problem?
for example, does this property help in decomposing the matrix? |
|
w*********r
发帖数: 488 | 5 雪天跪求高手赐教。随机生成几个数排到矩阵里，但是总不能保证矩阵的eigenvalues
全小于1.我现在可以设
一个条件如果不满足小于1的条件就重新生成，但是觉得这样不是很efficient，没有办
法保证一定会出现满
足条件的矩阵。不知道有没有可能先设N个eigenvalues 给N -dimension的矩阵，然后
根据这些
eigenvalues生成矩阵，而且矩阵不一定是对角阵。
谢了先～～～～～～～ |
|
c*******h
发帖数: 1096 | 6 compute the eigenvalues and scale the matrix by the largest eigenvalue
eigenvalues |
|
W*****k
发帖数: 158 | 7 考虑 下面这个eigenvalue 问题，c表示eigenvalue
a(x) f'(x)+b(x)^2/2 f''(x)= -c f(x)
满足混合边界条件
f'(0)=0 和 f(1)=0
假设存在一个eigenvalue序列 0
那么在什么条件下，或者说对a(x)和b(x)加上什么条件可以使得
c_1对应的eigenfunction f_1(x) 在区间(0,1)上恒是正的
有什么reference可以推荐吗？ |
|
A*******s
发帖数: 3942 | 8 感谢啊，包子随后奉上。明白了为什么overestimate first few eigenvalues-->
underestimate last few eigenvalues. 不过还是没有彻底明白为啥co-linearity->
overestimate first few eigenvalues. 能推荐几篇paper么？ |
|
g**********t
发帖数: 475 | 9 【 以下文字转载自 Computation 讨论区 】
发信人: geneticdrift (不懂微积分), 信区: Computation
标 题: 如何用CUDA同时计算几百个实对称矩阵的eigenvalues/eigenvecot
发信站: BBS 未名空间站 (Mon Jul 2 02:38:51 2012, 美东)
我有一个程序要反复计算几百个（约500个）64 x 64的实对称矩阵的所有的
eigenvalues/eigenvectors。自己用CUDA实现了一个Jacobi algorithm with chess
tournament ordering。具体来说，每个block(含有32个threads)处理一个矩阵，这32
个threads并行消去一个矩阵中的32个off-diagonal elements，直到算法收敛。结果无
误，计算单个矩阵所花的时间也和最近的一篇paper里的数据接近。但是这个算法和CPU
上的library比没有太大的优势。在同时处理这500个矩阵的情况下，和GSL里面高度优
化的函数比较(用单CPU)，用GPU仅仅快了一倍。我觉得主要是... 阅读全帖 |
|
g**********t
发帖数: 475 | 10 【 以下文字转载自 Computation 讨论区 】
发信人: geneticdrift (不懂微积分), 信区: Computation
标 题: 如何用CUDA同时计算几百个实对称矩阵的eigenvalues/eigenvecot
发信站: BBS 未名空间站 (Mon Jul 2 02:38:51 2012, 美东)
我有一个程序要反复计算几百个（约500个）64 x 64的实对称矩阵的所有的
eigenvalues/eigenvectors。自己用CUDA实现了一个Jacobi algorithm with chess
tournament ordering。具体来说，每个block(含有32个threads)处理一个矩阵，这32
个threads并行消去一个矩阵中的32个off-diagonal elements，直到算法收敛。结果无
误，计算单个矩阵所花的时间也和最近的一篇paper里的数据接近。但是这个算法和CPU
上的library比没有太大的优势。在同时处理这500个矩阵的情况下，和GSL里面高度优
化的函数比较(用单CPU)，用GPU仅仅快了一倍。我觉得主要是... 阅读全帖 |
|
g**********t
发帖数: 475 | 11 【 以下文字转载自 Computation 讨论区 】
发信人: geneticdrift (不懂微积分), 信区: Computation
标 题: 如何用CUDA同时计算几百个实对称矩阵的eigenvalues/eigenvecot
发信站: BBS 未名空间站 (Mon Jul 2 02:38:51 2012, 美东)
我有一个程序要反复计算几百个（约500个）64 x 64的实对称矩阵的所有的
eigenvalues/eigenvectors。自己用CUDA实现了一个Jacobi algorithm with chess
tournament ordering。具体来说，每个block(含有32个threads)处理一个矩阵，这32
个threads并行消去一个矩阵中的32个off-diagonal elements，直到算法收敛。结果无
误，计算单个矩阵所花的时间也和最近的一篇paper里的数据接近。但是这个算法和CPU
上的library比没有太大的优势。在同时处理这500个矩阵的情况下，和GSL里面高度优
化的函数比较(用单CPU)，用GPU仅仅快了一倍。我觉得主要是... 阅读全帖 |
|
d**s
发帖数: 920 | 12 Folks:
A naive question. What is the relationship between Singular Value and
Eigenvalue ?
Does every matrix has singular values ?
Does every matrix has Eigenvalues ?
When two of them are same ?
Thanks. |
|
d**s
发帖数: 920 | 13 Thanks, does any matrix has a eigenvalue ?
I think any matrix has singular value, if the matrix also have eigenvalue,
when
two of them are same ? |
|
p*****r
发帖数: 30 | 14 only n*n (square) matrix has eigenvalues.
It has n eigenvalues,although some may be the same. |
|
w*********r
发帖数: 488 | 15 我的这个问题比较tricky，实际上是要求两个矩阵A和B相加之后的新矩阵的
eigenvalues全小于1.回去翻了
一下论文，说A和B 可以是对角阵，于是随机生成1组N个绝对值小于1的数，形成对角阵
A，再生成N个
【0，1】之间的数去scale对角阵I-A，形成B，这样A+B是一个对角阵，对角线上的所有
元素的绝对值都小于
1，也就是eigenvalues的绝对值都小于1。 我想这样应该是对的。 |
|
h**********c
发帖数: 4120 | 16 1。不能用matlab
2。可以用fortran c/c++
比方 x^2 + m x + n = 0
矩阵
| m/2 sqrt(m^2/4 -n)|
|sqrt(m^2/4 -n) m/2 |
的eigenvalues 应该就是 x^2 + m x + n = 0 的 eigenvalues.
如果polynomial 的维数高了，怎么得到这样矩阵，因为矩阵式是numerical friendly
的。 |
|
d*z
发帖数: 150 | 17
Matrix include unknow variable? Then human being maybe the
best tool.
Suppose the polynomial is f(x), and the matrix is f(A)
if lamda(i) is a eigenvalue of A then f(lamda(i)) is
eigenvalue of f(A). |
|
a********s
发帖数: 188 | 18 I am not a big bull, but I think, the covariance matrix ("Sigma") can be
eigendecomposed into
Sigma = FDt(F)
where, D is the diagonal matrix of eigenvalues. The larger the eigenvalue it
is, the higher the variation of covariance matrix. |
|
A*******s
发帖数: 3942 | 19 sorry i didn't state my question clearly.
say we want to estimate population eigenvetors and eigenvalues based on an
observed sample. we rank our estimated eigenvalues or lambda_hats from largest
to smallest. Then for larger lambda_hats, say lambda_hat_1, it is biased
high, in other words, E[lambda_hat_1] > lambda_1. It is biased low for
smaller lambda_hats, say E[lambda_hat_n] < lambda_n
Why? or it is just my misunderstanding?
it |
|
A*******s
发帖数: 3942 | 20 No. i saw it on Freidman's paper "Regularized Discriminant Analysis". Right
after the formula (18) you can see his description of this problem.
"This shrinkage has the effect of decreasing the larger eigenvalues and
increasing the smaller ones, thereby counteracting the biasing inherent in
sample based estimation of eigenvalues."
I recall I saw similar statements before but I don't know where it is from.
re |
|
s*****9
发帖数: 108 | 21 这方面的paper还挺多的。说说我的想法:
比方说做PCA, 如果sample size n和参数空间维数p相比不算太大的话。那么估计出来
的头几个PC所代表的variation会比其实际的要大一些吧（co-linearity等等因素），
因此最大的几个eigenvalue就会被over-estimated。总的variation的估计不会偏差多
少，那么最后几个PC所在的variation就会被压缩的很小。最小的那个eigenvalue经常
会被shrink到0，导致covariance matrix不可逆。要是n >> p ，应该不存在这个问题
吧。 |
|
s***0
发帖数: 117 | 22 What's the model of the GPU?
And, did you try using the NVIDIA SDK eigenvalue code? |
|
g**********t
发帖数: 475 | 23 显卡用的是tesla M2070 (Fermi generation)。已经看过了cuda SDK里面的例子，但是
那个是bisection algorithm，貌似这个算法只能算eigenvalues，不能算eigenvectors? |
|
T****s
发帖数: 915 | 24 【 以下文字转载自 Joke 讨论区 】
发信人: daemonself (mit行为艺术专业博士后导师), 信区: Joke
标 题: Re: 如何用CUDA同时计算几百个实对称矩阵的eigenvalues/eigenvecot
发信站: BBS 未名空间站 (Mon Jul 2 08:59:55 2012, 美东)
http://saahpc.ncsa.illinois.edu/10/papers/paper_19.pdf
based on this paper the speed-up should be significant
32
CPU |
|
w**d
发帖数: 2334 | 25 Anybody knows some more recent upbound for the
absolute eigenvalues of a real matrix? Searched a lit bit without
finding anything.
Thanks in advance. |
|
l******e
发帖数: 55 | 26 我需要遍一个小程序算出3*3 matrix 的eigenvalues and eigenvectors.
哪位大侠能提供一个类似的程序啊?
多谢了!! |
|
g**********t
发帖数: 475 | 27 我有一个程序要反复计算几百个（约500个）64 x 64的实对称矩阵的所有的
eigenvalues/eigenvectors。自己用CUDA实现了一个Jacobi algorithm with chess
tournament ordering。具体来说，每个block(含有32个threads)处理一个矩阵，这32
个threads并行消去一个矩阵中的32个off-diagonal elements，直到算法收敛。结果无
误，计算单个矩阵所花的时间也和最近的一篇paper里的数据接近。但是这个算法和CPU
上的library比没有太大的优势。在同时处理这500个矩阵的情况下，和GSL里面高度优
化的函数比较(用单CPU)，用GPU仅仅快了一倍。我觉得主要是Jacobi algorithm对于这
个大小的矩阵效率太差，而GSL里面的函数用的好像是QR decomposition，虽然只有一
个thread但是效率很高。有没有比较适合我的问题的能在GPU上高效执行的算法？有没
有什么paper/code可以参考的？先谢谢了。 |
|
s***r
发帖数: 52 | 28 suppose a matrix A.
singular value is the eigenvalue of A*A'. |
|
x******r
发帖数: 367 | 29 Let x=(sin(a),sin(2a),...,sin(na))T
If a=j*pi/(n+1),pi-3.1415926...
then x is an eigenvector of A corresponding to the eigenvalue 2-2cos(a).
The proof can be found on page 621
of Numerical analysis: Mathematics of Scientific Computing(third Edition).
The authors are David Kincaid & Ward Cheney.
Hope that it will be helpful. |
|
w**********m
发帖数: 82 | 30 rt
对于一般的Hermitian矩阵，
求它的eigenvalue decomposition，
传统的求法有cyclic Jacobi algorithm，
但是复杂度相对较大，有没有相对compleixty更小的算法？
thanks |
|
h***o
发帖数: 26 | 31 Is there a fast way to obtain the largest/smallest eigenvalues of a
symmetric matrix? or the upper bound for the largest and lower bound for the
smallest?
The matrix is large and sparse.
Thanks a lot |
|
c*******h
发帖数: 1096 | 32 ok, to be serious.
there are numerous ways.
the best way i prefer is to apply a few steps of lanczos iterations
and then solve the largest eigenvalue of a tridiagonal matrix.
i beleive matlab implements this in eigs(), since it uses arpack.
the
a |
|
s******t
发帖数: 150 | 33 我想如果你彻底了解你自己的问题，那么你必然是已经得到了一个ｅｉｇｅｎｅｑｕａ
ｔｉｏｎ
ＡＸ＝＼ｌａｍｄａＸ
因此如果你要求Maximum Eigenvalue，只要求解Ａ－＼ｌａｍｄａＩ＝０的最大特征根。 |
|
m*********e
发帖数: 277 | 34 Gerschgorin's theorem
eigenvalues |
|
c*******h
发帖数: 1096 | 35 i did a small 2x2 example and find that the norm of the matrix can
not be upper bounded. in fact, depending on the eigenvectors, the
norm of the matrix can be arbitrarily large.
btw, i found a lower bound from a problem in Horn's book. the f-norm
of the matrix is larger than or equal to the root of the squared sum
of the eigenvalues. equality holds iff the matrix is normal. |
|
d**e
发帖数: 2420 | 36 Given two 2x2 matrices
A=diag(a1,a2), B=[a1+b1+p1,-p2;-p1,a2+b2+p2]
Let BI=the inverse of B, that is, B^(-1)
and C=A*BI
Assume all parameters are positive,i.e., a1,a2,b1,b2,p1,p2>0.
If a2,b1,b2,p1,p2 are fixed, then the principal eigenvalue of C is a
function of a1, denoted by, f(a1).
Prove or disprove that function f is monotone with respect to a1.
Any comment or suggestion is welcome. Thanks a lot. |
|
n********s
发帖数: 150 | 37 if f(x)=exp(ix), and there is an operator -i(d/dx), where x is angle,
then the eigenvalue of the operator should be 0,1,2,3.........
is it correct?
Why or why not?
Thanks a lot. |
|
e**********n
发帖数: 359 | 38 You need to identify the Hilbert space (of the wavefunctions) first, which
is all periodic functions with period 2pi in this case. The operator -i(d/dx
) is an operator defined on this Hilbert space. Its eigenvalues are all
integers and the eigenfunctions are exp(i nx) with n \in Z. |
|
g*****g
发帖数: 18 | 39 -i(d/dx)f(x)=Af(x), A is eigenvalue
thus f(x)=exp(iAx),
since x is angle
exp(iAx) should no difference with exp(iA(x+2pi))
therefore, exp(i2piA)=1, then A should be 0, 1, 2,...
-1, -2,....
( I am not sure it is right.) |
|
s*******m
发帖数: 50 | 40 This question is too general. From what you describe, I can‘t see this
property of eigenvalue would help. |
|
r******s
发帖数: 2155 | 41 【 以下文字转载自 Sociology 讨论区,原文如下 】
发信人: fdstephen (kiwilo), 信区: Sociology
标 题: [转]心理学的故事--第十三章 社会心理学 eigenvalue
发信站: Unknown Space - 未名空间 (Sun Apr 18 00:52:33 2004), 站内信件
发信人: Smille (冰冷如火), 信区: Thoughts
标 题: [转]心理学的故事--第十三章 社会心理学家
发信站: The unknown SPACE (Wed Mar 22 11:24:43 2000), 转信
心理学的故事 上页 下页 素心学苑
第十三章 社会心理学家
无人之境
问：现代心理学中什么领域极为忙碌和高产而又没有明确的身
份，甚至没有一个普遍接受的定义？
答：社会心理学。与其说它是一个领域，倒不如说它是心理学和
社会学之间的一片无人之境，它与社会学彼此重叠，还对好几种其它
的社会科学造成影响。自从社会心理学诞生之后 |
|
b****d
发帖数: 30 | 42 say, A is matrix, X is vector, e is a number. there's solution that
AX = eX
such X are called eigenvectors, e are called eigenvalues. |
|
A*******s
发帖数: 3942 | 43 I saw some conclusions many times like:
sample estimations of larger eigenvalues (of covariance matrix) are biased
high, and those of smaller are biased low.
Could any big bull tell me why? thanks in advance and baozi will be sent
later |
|
I*****a
发帖数: 5425 | 44 Yes this is intuitively understandable. I think when the sample size is so
large that the data are not sparse in any dimension, then the estimation is
not problematic.
I just checked Friedman's paper, and he said the biases are most severe espe
cially when the eigenvalues are close to each other. I don't quite get that.
Do you know the reason for this ? |
|
s*****9
发帖数: 108 | 45 客气了，我google了这个topic，也没找到理论上的公式来说明这个现象。倒是有很多
paper里包括了
相关的simulation的结果。
这个paper：“A bootstrap approach to eigenvalue correction”
的第一页最后和第二页开头提了一些paper，不知道有没有帮助。
我猜可以通过定义eigen-value作为一个优化问题的解，进而写出eigen-value的一个近
似表达式。
应该能得到consistence和rate的一些结果。 |
|
s*****e
发帖数: 115 | 46
-1
(a
0}
center
From which textbook did you find that you can check the multiplicity of the
eigenvalue of YOUR problem like what you just did?
Everything you've tried so far (i.e., checking the algebraic multiplicity of
an eigenvalue, constructing generalized vector) is for the following
eigenvalue problem:
(a*I - M ) x = 0, where M is a CONSTANT matrix. In this case, eigenvalues
are roots of the characteristic POLYNOMIAL.
Does your problem look the same?
Without any context, I also don't see w... 阅读全帖 |
|
h********3
发帖数: 2075 | 47 你是说power iteration这个算法会converge吧。
你看看它的矩阵等式。
Perron-Frobenius Theorem: If M is a positive, column stochastic matrix, then:
1 is an eigenvalue of multiplicity one.
1 is the largest eigenvalue: all the other eigenvalues are in modulus
smaller than 1.
the eigenvector corresponding to eigenvalue 1 has all entries positive. In
particular, for the eigenvalue 1 there exists a unique eigenvector with the
sum of its entries equal to 1.
它其实是用largest eigenvector去近
似原来的矩阵，不断拉大largest eigenvector和其他eigenvect... 阅读全帖 |
|
G*********l
发帖数: 2 | 48 这位朋友, 这个matrix的eigenvalues是{4，0，0，0}。你可以用MATLAB算下。
Simple check: Sum of all eigenvalues should equal to the 他race, which is
the sum of the entries on the main diagonal, i.e. 1,1,1,1.
In fact，做个简单的row elimination, 矩阵变为
1 1 1 1
0 0 0 0
0 0 0 0
0 0 0 0
Elimination告诉我们原来的矩阵有一个正的Pivot（1），三个Pivots = 0.
据此，即使不去计算具体的eigenvalues, 我们也可以知道该矩阵必然有1个正
eigenvalue, 3个repeated eigenvalues = 0. |
|
f*****k
发帖数: 353 | 49 depends on A's eigenvalues,
if A has two real non-equal eigenvalues r1, r2, and the eigenvalue is v1, v2
, then
u = C*(e^(r1 t) v1 + e^(r2 t) v2);
if A has two conjugate complex eigenvalue r1, r2, then
then pick up the real and the imaginary part of e^(r1 t) v1, the linear
combination is the solution.
if A has double eigenvalue r, then find the two linear independent solution
v1, v2 of
(A - r I )^2 v = 0,
vv1 = (A - rI)*v1, vv2 = (A - rI)*v2;
solution u = C (e^(rt)*(v1 + vv1*x) + e^(rt)*(v2 + v |
|
s***e
发帖数: 911 | 50
If we use the spectral norm, which is defined as the square root of the
maximum eigenvalue of A^(+)A, where A^(+) is the adjoint matrix of A. In
real space, A^(+)=A^(T).
Then:
||A||=Max[Eigenvalue(A^(T)A)]
||P^(T)AP||=Max[Eigenvalue(P^(T)A^(T)PP^(T)AP)]
=Max[Eigenvalue(P^(T)A^(T)AP]
=Max[Eigenvalue(A^(T)A)]
于是得证.
以上用到:
1. 正交矩阵A^(T)=A^(-1);
2. Det[B^(-1)AB]=Det[A] |
|
