s***n
发帖数: 9499 | 1 Minimize Trace((A'A)^(-1)),
Where variable A is an n by n real matrix, and every element of the matrix A
is between 0 and 1.
A' denotes the transpose of A.
^(-1) denotes the inverse of a matrix.
Any suggestions or references are highly appreciated! |
|
t***o
发帖数: 4265 | 2 S是一个 PxK tall real matrix（行数P>列数K）,它的每个列向量都为单位向量，
并且各列线性无关：
S=[S_1 S_2 ... S_K],
S_i'*S_i=1, where ' is transpose;
S_i and S_j are independent for any unequal i and j.
M = S'*S
容易知道 M 是个 KxK 对称矩阵，并且主对角线元素M(n,n)都为1。
我用matlab试验发现, K越大，M^(-1)的主对角线元素M^(-1)(n,n)也越大。
比如，M^(-1)(1,1)随着K增大而则增大。K
猜想，M^(-1)(n,n) monotonically increases with K.
请问能从理论上证明这个结论吗？ |
|
e***n
发帖数: 286 | 3 【 以下文字转载自 Computation 讨论区 】
发信人: erain (红花会大老板), 信区: Computation
标 题: 紧急求问: 是否可以将一个对称不定矩阵 A 分解为 A = B * B'
发信站: BBS 未名空间站 (Sat Apr 21 16:47:52 2007)
Urgent!
For any symmetric definite matrix, for sure we can factor it with Cholesky
method. How about symmetric indefinite matrix? I need factor such a matrix
A exactly into the product form
A = B * B'
where B' is the transpose of B and B is some n x n matrix ( not necessarily
to be triangular).
I know we can factor it with a LDLT method and fur |
|
w**a
发帖数: 1024 | 4 let vector v=[x(t) y(t)]', ' is transpose
A=[a b]
[c d]; assume a,b,c,d are const.
dv/dt = A *v
you can do a coordinate rotation then diagonalize matrix A.
let C be a coordinate transformation matrix
(1) dv/dt = A*C'*C*v
multiply (1) by C
(2)d(C*v) = C*A*C'*(C*v)
let vector u = C*v and matrix E = C*A*C', because E is diagonal matrix,
E11 = e1; E12=E21=0; E22 = e2;
(3) then we have,
du1/dt = e1*u1
du2/dt = e2*u2
the solution is obvious now.
bifurcation might occur. |
|
i*******d
发帖数: 6 | 5 Can anyone prove that the 2-norm of the following matrix is equal to 1?
i.e. || I - Mxx'M/(x'M'Mx) ||2 = 1
M is an any n by n matrix and x is a vector having n elements; ' indicats
transpose.
Thank you very much! |
|
a*****t
发帖数: 822 | 6 Let B=(A+A')/2. A' is the conjugate transpose of A.
Then sufficient and necessary both hold. |
|
s*******n
发帖数: 101 | 7 【 以下文字转载自 EE 讨论区 】
发信人: sensorman (所谓成长), 信区: EE
标 题: 请教一个矩阵的问题
发信站: BBS 未名空间站 (Tue Jul 24 12:23:30 2007), 转信
想计算I+CDC^T的逆，i.e.(I+CDC^T)^(-1),C^T是C的transpose, D是diagonal的, 如果
C是向量，可以用matrix inversion lemma,但如果C是矩阵，有什么公式可以化简吗？
3x |
|
s**p
发帖数: 207 | 8 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*****n
发帖数: 78 | 9 实数变量 x, 矩阵A, B:
Max f(x) = (x^T *A *x)/(x^T *B *x)
where ^T dedotes Transpose. Martrix A and B are symmetric positive-definite.
谢谢 |
|
c****r
发帖数: 576 | 10 这样的方程：J*C + C*J' = D,其中J' = transpose(J)，J，C，D都是带有符号变量的
矩阵，求符号解。 |
|
s**********2
发帖数: 61 | 11 Given a matrix A=(1.5,0.5,0)
(0.5,1.5,0)
(0, 0, 3)
Find a matrix S & a diagonal matrix D such that A=SDS^-1, where S^-1 is the
inverse of S. Also, choose S such that S^-1 = S' (S' is the transpose of S)
I know that D is a similar matrix to A, & have found it, but I am having
trouble finding S. I'm trying to find the basis w/ eigenvectors, but I find
that route gives me an uninvertible matrix. Any suggestions as to what I am
doing wrong? |
|
b*********n
发帖数: 56 | 12 "where S^-1 is the inverse of S. Also, choose S such that S^-1 = S' (S' is
the transpose of S)"
This requires that S be orthonormal.
About what you get:
Firstly, z seems wrong, because it is not orthogonal to x or y;
Secondly, after you get x, y, z right, you need to normalize them, that is,
to make each of them be of norm 1. |
|
|
f***e
发帖数: 43 | 14 用不着软件，这个问题很简单, linear regression
写成一个objective function, minimize it, find the coefficient
y=, w, x a vector, <> inner product
minimize J=sum_i(y_i-)^2, i is the index of sample
take partial j/partial w =0,
w=(x^T x)^(-1)x^T y (T is the transpose)
(x^Tx)^(-1)x^T is also called pseudo inverse of x
多给包子,我给你发一个详细的求解过程。 |
|
l******3
发帖数: 6 | 15 Guess your question is to solve
minimize f(D) over all D with eigenvalues equal to b1, b2,...bn.
Seems not a trivial problem.
You may try out the following method:
Let D0=diag[b1, b2,...,bn]. Your problem can then be rewritten as
minimize norm (R^T*D0*R*A*R^T*D0*R - a* Id) over all real number a,
rotation matrix R, where Id is the identity matrix and R^T is the transpose.
Then you may try to use the usual first-order condition...
The algebraic calculations are tedious but worth trying if this i |
|
c******m
发帖数: 41 | 16 Let v be a non-zero column vector in R^(n)( i.e n维列向量空间） and v ^(T) (
i.e v 的转置）be its transpose.
(a) Show that the n*n matrix A defined by v*v^(T) ( i.e v乘以v的转置） is of
rank one and is a diagonalizable matrix.
(b) If u is a unit vector in R^(n), show that the matrix H=I-2u*u^(T) is an
invertible matrix. Here I is the identity matrix. ( Hint: Note that u^(T)*v
is a scalar for any column vector v. You may show that the null space of H
is trivial)
(c) If v and w are any two nonzero orthogonal vecto |
|
k**a
发帖数: 121 | 17 这个似乎和LP本身一样是NP-hard问题，也就是说除了直接解，没有更好的方法。当然
，针对具体问题，可以试试用Farkas's Lemma:
[Farkas's Lemma] Let A be an m × n matrix and b an m-dimensional vector.
Then, exactly one of the following two statements is true:
1. There exists an x ∈ R^n such that Ax = b and x ≥ 0.
2. There exists a y ∈ R^m such that A'y ≥ 0 and b'y < 0.
where ' denotes transpose.
或者试试dual。如果 optimal dual solution unbounded，那么原本的LP就是
infeasible。
求dual和Farkas's Lemma道理差不多，都是把系数矩阵转置过来，如果你的问题把系数
矩阵转置过来比较好看，可以试试。不过说来说去对于一般的LP，这些都是NP- |
|
h***r
发帖数: 726 | 18 after the ellipsoid paper, Karmarkar developed the interior point method for
LP.
Does anyone understand why the canonical form, on the first page, is the
following? (WHY Z, not R?)
min x' x
s.t. Ax = 0
\sum xi = 1
x >= 0
x \in R^n, c \in Z^n, A \in Z^(mn)
x' means x transpose
R^n means R to the nth
So, Why c, A in Z, not R???
Thanks! |
|
n*********y
发帖数: 54 | 19 WEI ZHANG TO RECEIVE 2010 SASTRA RAMANUJAN PRIZE
http://www.math.ufl.edu/sastra-prize/2010.html
The 2010 SASTRA Ramanujan Prize will be awarded to Wei Zhang, who is now a
Benjamin Pierce Instructor at the Department of Mathematics, Harvard
University, USA. This annual prize which was established in 2005, is for
outstanding contributions by very young mathematicians to areas influenced
by the genius Srinivasa Ramanujan. The age limit for the prize has been set
at 32 because Ramanujan achieved so ... 阅读全帖 |
|
f*******g
发帖数: 55 | 20 Suppose A, B, and C are real, symmetric, and positive semi-definite matrices
. Is the matrix
ABC + C'B'A'
also positive semi-definite? (Here ' means transpose)
Thanks! |
|
b***k
发帖数: 2673 | 21 Given an arbitrary matrix A with dimension m x n,
what is the difference or relationship of A'*A and A*A'?
here A' is transpose of A.
obviousely both of them are symmetric, and they have identical eigenvalues.
but their eigenvectors are different which means eigenspace of both matrix
are distinct, right? |
|
x********i
发帖数: 905 | 22 http://arxiv.org/abs/1509.05576
The Riemann hypothesis is, and will hopefully remain for a long time, a
great motivation to uncover and explore new parts of the mathematical world.
After reviewing its impact on the development of algebraic geometry we
discuss three strategies, working concretely at the level of the explicit
formulas. The first strategy is "analytic" and is based on Riemannian spaces
and Selberg's work on the trace formula and its comparison with the
explicit formulas. The second... 阅读全帖 |
|
s***n
发帖数: 821 | 23
Not true, you even can define a strain measure yourself. Also you can apply
your strain measure at any strain level. They are not limited to large strains.
Some of the strain measures are:
Green deformation tensor (Right Cauchy-Green tensor) C = transpose(F)F
Left Cauchy-Green tensor b= F tran(F)
C is a material strain tensor,
and b is a spatial strain |
|
J*******3
发帖数: 1651 | 24 Probably it is my misunderstanding. when I read the manual, I got
impression that the matrix of supercell is the transpose of Monkhorst matrix
, is this correct? |
|
a**u
发帖数: 17 | 25 有100个被试,每个人都有6个VARIABLES: A1, A2, A3, B1, B2, B3
我想要做的是,给每个人计算两个SLOPE, 其实就是根据三个点来一个LINEAR
REGRESSION,根据A1,A2,A3计算一个SLOPE,再根据B1,B2,B3计算一个SLOPE.
请问是不是需要TRANSPOSE或者RESTRUCTURE将数据变形,然后再用REGRESSION 后者CURVE
ESTEMATED得到SLOPE,也就是BETA, 然后再怎么保存成一列变量哪?
怎么都想不出来怎么办,好急啊,老板又着急要结果.请帮忙!!! |
|
a**u
发帖数: 17 | 26 好吧,其实是一个简单的问题,
做LINEAR REGRESSION的时候,怎么在DATA EDITOR中保存BETA?
一般来说,我们都是针对一个变量,也就是一个COLUMN来做REGRESSION(这里的自变量我用
时间),但是现在我要对横着的三个变量做REGRESSION. 也就是说,这三个变量是在三个时
间点测量的SCORE, 我希望能找到他们变化的斜率.
当然,首先我想到的就是将数据TRANSPOSE,这样使这三个变量成纵向排列,仿佛是一个VARI
ABLE,这样我可以得到SLOPE,但是问题是如何将这些BETA保存在DATA EDITOR里面,因为里
面的被试数太多,我不可能一个一个从OUTPUT里面COPY到DATA窗口中.
另外,这纯粹是一个SPSS的使用技术问题,我猜想叙说变量名什么的没有必要,所以采用A,B
代替.这样叙述的清楚吗? |
|
r******s
发帖数: 2155 | 27 beta for what?
If for each variable, you have 6 IVs (??) in total, just write them down.
用
时
当然,首先我想到的就是将数据TRANSPOSE,这样使这三个变量成纵向排列,仿佛是一个VARI
里
另外,这纯粹是一个SPSS的使用技术问题,我猜想叙说变量名什么的没有必要,所以采用A,B |
|
a**u
发帖数: 17 | 28 I just want to make it simple. Actually there are about 30 variables. Most
important thing is that there are about 100 subjects. I need to get slope for
EACH subject, I can't write them one by one.
我
个
当然,首先我想到的就是将数据TRANSPOSE,这样使这三个变量成纵向排列,仿佛是一个VARI
为
另外,这纯粹是一个SPSS的使用技术问题,我猜想叙说变量名什么的没有必要,所以采用A,B |
|
s*****a
发帖数: 56 | 29 I would agree to create two stacked variables for both A and B, and get the
change trajectories for A and B. However, you can just obtain two beta values
(slopes for A and B). As you set them as new variables, they would be
constants and are not really variables. Maybe I misunderstand your
question...
用
时
当然,首先我想到的就是将数据TRANSPOSE,这样使这三个变量成纵向排列,仿佛是一个VARI
里
另外,这纯粹是一个SPSS的使用技术问题,我猜想叙说变量名什么的没有必要,所以采用A,B |
|
s*****t
发帖数: 9 | 30 what's wrong with astrology? why cant modern humans have some entertainment
in peace? astrology is just like religion, its own purpose and position in
society have long ago been transposed beyond what narrow-minded tunnel-
visioned science-crammed self-righteous modern humans can possibly imagine.
the
0
4 |
|
b***k
发帖数: 2673 | 31 也许这个程序用c实现比C++来实现要容易的多。呵呵，我是c出身的。
这个话题似乎转到了如何学习C++的问题。
以前litaihei在板上说过一个方法，我觉得很好。
就是写个例子程序来练习C++的各种简单功能
比如写个matrix class，包括
constructor (default, copy)
destructor
copy assignment
operator overloading(including +,-,*,transpose, inverse, [],etc)
member function (eigenvalue)
output/input overloading
inheritance (bandmatrix, sparse matrix)
template(generize the matrix class)
这个matrix class写完善了后，可以再编写一点小main程序来测试它。
这个过程中也可以理解const，inline，extern等等的keyword
通过这个实践，可以初步理解C++中一些最基本的但是也是最重要的概念。 |
|
l***y
发帖数: 126 | 32
Assume the matrix is a[i][j] with 1<=i<=m and 1<=j<=n, m>=n (otherwise,
transpose a first), and the rows and columns are sorted in ascending order,
then the following should print out all entries of the matrix in ascending
order. The correctness is left as an exercise :). The time complexity is
O(mn log(n)), but I suspect it is actually better than that, although it is not obvious and certainly not easy to prove.
1. Initialize an array b[i]=0 for all i between 1 and n
2. set b[1]=1
3. Initializ |
|
x******a
发帖数: 6336 | 33 A'= the transpose of A?
the null spaces should be the same?
Ax=0 => A'Ax=0 => Null(A)\subset Null (A'A)
A'Ay=0 => y'A'Ay=0, i.e, (Ay, Ay)=0 => Ay=0
=> Null(A'A)\subset Null(A). |
|
J**Y
发帖数: 34 | 34 In economics, solving M is usually done with Cholesky decomposition as
I said before, because the var-cov matrix of Z: V=MM', is a symmetric
positive definite matrix. So V can be written as the product of a lower
triangle matrix and its transpose. |
|
t***a
发帖数: 4 | 35 ||A||^2=Aij*Aij
=A_ij*A~_ji
=Tr(A*A~)
=Tr(P~*A*P*P~*A~*P)
=||P~*A*P||^2
here X~=transpose(X)
sum over repeated indices |
|
h**********a
发帖数: 40 | 36 transpose?? just ' bah. |
|
B***y
发帖数: 83 | 37
Notice that it is a second order iteration equation, thus first write it as a
vector equation:
Y_(n+1) = A(n) * Y_(n) ,
where Y_n = (a(2n-1), a(2n))^t, where "t" means "transpose".
Easy to check that A(n) asymptotically behaves like (b 0
0 b )
Thus if |b| > 1, then the series Y_n will diverge, if |b|< 1 then the series
will converge. |
|
c*****t
发帖数: 1879 | 38
别忘了 chiasmata, virus, transposible elements, improper segregation,
chemicals/radiations, aging, etc... 所以概率远大于一. 特别是 chiasmata
的作用最大. |
|
b****d
发帖数: 1311 | 39
(1) rk(A) = rk(QR) =< min(rk(Q), rk(R)) . You can use linearly dependence of
row or column vector to prove this.
Also, any invertible matrix can be presented as products of fundamental
transformation matrixes. so if diagonal entries of R are nonzero, it's just
doing some basic transformation on column vectors of Q. so it's rk n.
(2) rk(A) = rk(Q) >=Q . first we can know rk(A) =< min(rk(Q),rk(R))=rk(Q) .
second, Let Q' be the transpose of Q. rk(A) >= rk(Q'A) = rk(Q'QR) = rk(DR)=
rk(Q), D is a nx |
|
f****n
发帖数: 355 | 40 OK, I looked at my textbook, and here's what it says:
in case of white gaussian noise, you can use the "square-law" detector,
which is
r'r ~ gamma
gamma is the threshold you choose based the desired power. r is the column
vector of input. r' is its transpose, and ~ means comparison.
For non-white gaussian noise, if K is the covariance matrix, the square-law
detector becomes
r'* K^-1 * r ~ gamma
K^-1 is the inverse of K.
My textbook is written by our teacher and probably can't be found in
booksto |
|
G**7
发帖数: 391 | 41 Possibly, you may use "proc transpose" to do this. |
|
p********a
发帖数: 5352 | 42 ☆─────────────────────────────────────☆
honestman (snail) 于 (Fri Jan 12 15:26:03 2007) 提到:
怎么将
1 14
1 25
1 26
2 35
2 43
2 45
改成
1 14 25 26
2 35 43 45
由于数量比较大, 不能手工改, 请版上的同学帮帮忙,谢谢!
☆─────────────────────────────────────☆
Bighappy (快乐大大大) 于 (Fri Jan 12 15:59:32 2007) 提到:
hehe, you can read the data using SAS, then transfer them using PROC
TRANSPOSE, then save the dataset as excel file...
☆─────────────────────────────────────☆
papertigra (长工胖头猪) 于 |
|
a********g
发帖数: 651 | 43 昨天终于考完ADV了，前后复习了1个多月，从本版得到了很大的帮助。前面的几个总结
性的帖子很管用，特别是pricematch 的帖子提到的，我基本上都考到了。
下面具体的说说我考的题目。
SQL：
各种的join都有涉及，和merge，set a, set b 的情况比较。
SQL insert 的格式
IC的使用
SQL中index 和view的基本用法。
Dictionary.table
MACRO:
很多%在一起的结果，有2题
Automacro 中sysday and sysdate的区别
特别要注意的一个考点，这个是我没有在online tutor 中见到的：
Call symput 的用法，其输出什么情况下在local，什么情况在global。
Symput and symputx 的区别，这个自己瞎蒙的。
％let在什么地方，是否影响symput
Efficient:
前人已经说得很全面了，我只想补充几点。
Transpose 的用法
By groupformat
By x notsorted
Array的基本用法
If and select比较
hash object |
|
a********a
发帖数: 346 | 44 Thanks, actually what I want is more complete than I gave above in my real
data. I want to keep transpose function in my code. And could you tell me
why the sum is not working in my code?
|
|
x*******u
发帖数: 500 | 45 只想出这个笨办法， 有没有更简单的
data a;
input x y z m n;
cards;
1 2 3 . 4
2 3 4 5 6
1 . 3 5 6
1 3 4 . .
;
run;
proc contents data=a out=out(keep=name); run;
proc sql noprint;
select name into: name separated by ' ' from out;
run;
proc transpose data=a out=b;
run;
data c; set b;
missnum=nmiss(of col1-col4);
run;
data d(where=(missnum=0))
e(where=(missnum>0));
set c;
run; |
|
|
j*****e
发帖数: 182 | 47 Use proc transpose (by id) to put all the 'aa' responses from the same id
into one row.
In the data step, generate an indicator that eqals one if one of the 'aa'
responses is 3(use array statement).
Then, use proc freq. tables trt*indicator. |
|
d*****n
发帖数: 65 | 48 proc summary data=one;
var x y z;
output out=temp1(drop=_type_ _freq_) mean= ;
proc transpose data=temp1 out=temp2 prefix=mean;
proc summary data=temp2;
var mean1;
output out=data(drop=_type_ _freq_) mean= ;
run; |
|
d*****n
发帖数: 65 | 49 sure。
you can look at the "proc transpose".
mean |
|
p********a
发帖数: 5352 | 50 data a;
input year loc $ id $;
datalines;
2007 NY a
2007 NY b
2008 CA c
2008 CA d
2008 CA e
2008 FL f
run;
proc sort data=a;
by year loc;
proc transpose data=a out=b(drop=_name_) prefix=all_id;
by year loc;
var id;
proc print;
run; |
|
