R********n
发帖数: 519 | 1 虽然从CS角度看过一些manifold learning文章，但是其实数学上对manifold其没什么
认识:).
目前只知道the most simple case is topological manifold,
如果需要calculus operation on manifold, then it is differentiable manifolds
(包括smooth and analytic manifold),
如果需要度量角度，长度，内积什么的, 就是Riemannian manifold,似乎这个最广泛?
The subspaces of a Euclidean space也是.
这些认识对吗,大牛能不能讲讲呢,还有什么比较基本的？
另外目前一些学习方法，我觉得Lapacian Eigenmaps比较本质, 因为它是从连续
manifold的到离散的角度做的,Lapacian operator比较本质, 在manifold上的它的特征
算子就是sin/cos, 不知这么理解对吗?
谢谢谢谢:) |
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G******i
发帖数: 163 | 2 E:=Banach space C[-pi,pi]
(f,g): the inner product in L^2 [-pi,pi].
For a subset A of {1,cos t, sin t, cos 2t, sin 2t, cos 3t, sin 3t, ....},
define
X(A):={ f in E: (f,g)=0 for g in A},
Y(A):={ f in E: (f,g)=0 for g not in A}.
Question: Do we have X(A)+Y(A)=E ?
It's easy to know:
(1) X(A) and Y(A) are closed subspaces of E, with intersection 0.
(2) X(A)+Y(A) is dense in E.
(3) If A is finite, X(A)+Y(A)=E.
(4) If A={1,cos t,cos 2t, cos 3t, ....}, X(A)+Y(A)=E.
For a general A, I would think X(A)+ |
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o**a
发帖数: 86 | 4 I should have said traceless and symmetrical. traceless means the diagonal
elements sum to zero. Symmetrical means Aij = Aji. 5 comes from 2 degrees of
freedom on the diagonal and 3 from off-diagonal. Talking about 3x3 matrix
here.
If A is such, so is U.A.U(-1). Thus Q is defined on a subspace with only 5
base vectors.
My feeling is that Q is still unitary here, but I should verify somehow.
Just wondering if there is any obvious way based on certain fundermental
properties of matrix operation.
T |
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c*******h
发帖数: 1096 | 5 When you consider A as a 9D vector, the traceless symmetric A's form a
subspace that has dimension 5. The transformation is still the 9-by-9 matrix.
If you do a linear transformation f:R^9 -> R^5 for the 9D vectors first,
effectively you are asking for a 5-by-5 matrix Q such that f(UAU') = Q*f(A)
for all A that are traceless symmetric. Let f have the matrix form P, where
P is
5-by-9 and has full row rank. Then Q equals to P * (U @ U) * P+, where P+
means the pseudo inverse of P. Unless the rows |
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p******n
发帖数: 46 | 6 2)中的 2nd countable必要吗？为什么？
metrizable space是normal space.因此必然是 completely regular space. 一个
completely regular space is homeomorphic to a subspace of [0,1]^J.
为什么还要 2nd countable呢？我的推理有问题吗？
second |
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r********e
发帖数: 103 | 7 metrizable是completely regular Hausdorff space 或者说是Tychonoff space没错，
但是这个只保证homeomorphic to a subspace of the cube [0,1]^I, rather than
the Hilbert cube [0,1]^N. Notice that the cube may not be metrizable since
the index set I may not be countable (think about the cantor cube).
Actually from Urysohn Metrization Theorem, we have the following:
For a Hausdorff space X, TFAE:
1. X can be embedded in the Hilbert cube
2. X is separable metrizable space
3. X is regular and second countable
You can see |
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l******e
发帖数: 4 | 8 来自主题: Mathematics版 - 代数群问题 Say the group is G, and the closed subset is X. Consider the right regular
representation of G in K[G]. Take a finite dimensional G-invariant subspace
V of K[G] such that W=V\cap I(X) generates I(X). Let H be the stabilizer of
W in G. Then H is a subgroup. It is not difficult to show that H=X. |
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c******m
发帖数: 599 | 9 比如A和B都是SPD(symmetric positive definite)
对于一般的R^n上的内积, BA不是SPD,
但是, 重新定义R^n上的内积<\cdot,\cdot>= (A\cdot, \cdot)
那么BA在这个<,> 内积下就还是SPD的
推导自己根据定义自己推一下
ref.
Xu, J. Iterative Methods by Space Decomposition and Subspace Correction
SIAM Review, Vol. 34, No. 4, (Dec., 1992), pp. 581-613. |
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s*****s
发帖数: 1559 | 10 第一个就是subpace.
第二个是说subspace平移以后的一个东西。
也许要看行文，才能知道作者到底指哪个。 一般指第二个，但是这个词也很少出现在
数学书里，没有太大意思。数学书多用plane, hyperplane 更直观的说法。 |
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h********0
发帖数: 12056 | 11 it seems the subset you described is not a subspace
since the subset is obtained from a set of nonlinear constraints. |
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e****z
发帖数: 119 | 12 Oh, yes, they may not define a subspace... sorry I have forgot a lot of
basic mathematical concepts... So can I explicitly formulate these
constraints after the transformation? Thanks a lot. |
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s*****e
发帖数: 115 | 13 The point is that even if A(t) if smooth, the rank of A(t) may change
between n-1 and 0 as t changes. The (n-1)-dimensional subspace that contains
the range of A(t) may change
abruptly even A(t) is C^{infinity} in t.
to |
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C**********n
发帖数: 164 | 14 Every finite dimensional subspace Y
of a normed space X is complete. In particular, every finite dimensional
normed space is complete.
求证明！ |
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z***c
发帖数: 102 | 15 OK I was not being precise. But I think the OP was asking about the same
fixed point.
The concept of transversality in differential topology is the following: Two
submanifolds M, N of the manifold W are transversal at x if
TxM + TxN = TxW
the sum is in the sense of vector subspace. This is somewhat different from
the naive sense of transversality. If two submanifolds intersects at at a
point, then the transversality condition is the same as having "nonzero
angle".
As I explained, the intersec... 阅读全帖 |
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q********y
发帖数: 162 | 16 线性代数和矩阵都是给工程和其他学科学着玩的。数学系的应该高中就学过了。
你要会选择基构造线性空间，然后用算子对基作用 (很多情况下你更本不知道explicit
form of your matrix)。
我最得意的一件事就是我导师想到一个数学结构，非常有创造性，但他以为是一个群，
我思考了一晚上后告诉他是不变子空间（invariant subspace). |
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q********y
发帖数: 162 | 17 线性代数和矩阵都是给工程和其他学科学着玩的。数学系的应该高中就学过了。
你要会选择基构造线性空间，然后用算子对基作用 (很多情况下你更本不知道explicit
form of your matrix)。
我最得意的一件事就是我导师想到一个数学结构，非常有创造性，但他以为是一个群，
我思考了一晚上后告诉他是不变子空间（invariant subspace). |
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x********i
发帖数: 905 | 18 http://iccm.mcm.ac.cn/dct/page/1
Invited Lectures
Group 1
Fan Qin: Cluster algebras and monoidal categorification
Fang Li: Positivity of acyclic sign-skew-symmetric cluster algebras via
unfolding method and some related topics
Cheng-Chiang Tsai: An attempt for affine Springer theory
Li Cai: The Gross-Zagier formula: arithmetic applications
Ming-Hsuan Kang: Geometric zeta functions on reductive groups over non-
archimedean local fields
Huanchen Bao: Canonical bases arising... 阅读全帖 |
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i****a
发帖数: 88 | 19 这里有没有做Model order reduction的？ 用ANSYS 或 Femlab建模得到系统矩阵，然
后Arnoldi method 得到 Krylov
matrix来project from n to r subspace.网上free的 mor4anays好像找不到了。。。 |
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i****a
发帖数: 88 | 20 这里有没有做Model order reduction的？ 用ANSYS 或 Femlab建模得到系统矩阵，然
后Arnoldi method 得到 Krylov
matrix来project from n to r subspace.网上free的 mor4anays好像找不到了。。。 |
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a********e
发帖数: 508 | 21 can't see what's wrong with 3-dimension
L2 is of infinite dimension, but the subspace spaned by 3 vectors are of
dimension 3
}{ |
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C***m
发帖数: 120 | 22 perp 是指垂直
vec(1) 应该是一列1vector 作为base的subspace
你说这道题就是考怎么证明correlation 不变，那么直接带入correlation 公式就可以
了？ |
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b***k
发帖数: 2673 | 23 U*D*U^-1=D
the column vectors of U are in the subspace with basis vectors as
the eigenvectors of D.
but so what? |
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y******6
发帖数: 61 | 24 你也说过 U 根 D 只是 eigen space 一样， 可是 这只说明 U的 eigen space 根 I
一样。。。
所以 U is block diag for general case. 比如 D=(1,1,2) , 这个时候， U 只要
满足 [u(1,1),u(1,2), u(2,1), u(2,2)]
是可逆的， u(3,3) 非0 ，然后其他元素都是0 就 ok. 这样 对应于 1的eigen space
是 2-dim 的子空间， 你的错误在于选定了一个特殊的基，其实任何基都ok。本质上只
要能代表这个 subspace 的 grassman manifold 就ok。
matrix
can |
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h******x
发帖数: 13 | 25 Further Advances In Quantum Computing
Posted by Hemos on Wednesday November 01, @12:15AM
from the a-quiet-place-in-subspace dept.
Porfiry writes: "Scientists at the U.S. Department of Energy's Los
Alamos National Laboratory have taken another step forward in
the quest for a quantum-based computer by demonstrating the
existence of a physical state immune to certain types of
information-corrupting "noise," which could otherwise disrupt |
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S*********g
发帖数: 5298 | 26 If x is A's eigenvector, i.e., Ax=ax
Then Bx saitisfies, A(Bx)=B(Ax)=a(Bx).
If x is nondegenerate, then Bx=bx, i.e., x is B's eigenvector.
If x is degenerate, then Bx=\sum c_i x_i where Ax_i=a x_i.
We can find linearly independent combinations of x_i, say X_i, that satisfies
B X_i= b_i X_i.
In rephrase, this means we can diaognalize B in the subspace
of eigenvalue a of A. |
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r****y
发帖数: 1437 | 27 I think one big problem with IsoMap is how to define
"neighbor". My experience shows the result can be totally
different upto how large you define the "neighbor". As a result,
it makes your interpretation of your result more difficult.
There was a paper on Science recently, talking about this IsoMap
and howt to reduce to a subspace. We tried their code, the "neighbor"
problem still exists.
I also saw one paper dealing with nonlinear EOF with
neural network method, I cannot see th |
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s*****n
发帖数: 2174 | 28 这个就是典型的回归问题啊. 设前5各点为V1,..,V5. 后面的两个点叫V6, V7.
分别用V6和V7对V1和V5的组合做回归, 得到两个residual向量, 就是S2空间的两个基.
回归(y=ax+e)的本质, 就是找y在X的列向量生成的空间中的投影, 其剩余部分e就是y里
面垂直于x空间的的部分.
如果用代数, 就把5各点列向量合并成矩阵X, 两个点的列向量合并成矩阵Y. S2空间其实
就是
Y - X(X^TX)^{-1}X^TY = (I - X(X^TX)^{-1}X^T)Y
那个X(X^TX)^{-1}X^T也叫hat矩阵, 是高维空间向子空间的投影矩阵. 而 I-hat 就是子
空间对高维空间的补空间的投影矩阵. |
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d*****t
发帖数: 7903 | 29 一句话点醒梦中人，你这个法子好，我也重新更新了一遍对回归问题的理解。
我中午正在琢磨着能否用QR decomposition (Gram-Schmidt process),现在想来原理是
大同小异的。Gram-Schmidt is nothing but a series of regressions.
多谢多谢！
.
其实
是子 |
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h******a
发帖数: 198 | 32 也可以把这7个点组成一个矩阵，计算特征值，算单位化正交特征向量 也可以 |
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R********n
发帖数: 519 | 33 对啊，呵呵，那些人脸上面的各种linear DR 或者近似linear DR，还有什么linear
subspace的方法，实在有点太多了，大同小异 |
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