A*******r
发帖数: 768 | 1 inverse eigenvalue problem
很多文章的，飘过
存在 |
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i*********s
发帖数: 159 | 2 S. Moskow and M. Vogelius, First order corrections to the homogenized
eigenvalues of a
periodic composite medium. A convergence proof, Proc. Roy. Soc.
Edinburgh Sect. A. 127
(1997), 1263–1299. CMP 98:06
Please sent it to m*********[email protected]
Thanks. |
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l******3
发帖数: 6 | 3 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 |
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c*******h
发帖数: 1096 | 4 what can we say about the norm of the matrix?
assume that the matrix is unsymmetric but diagonalizable. |
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s**e
发帖数: 1834 | 6 (1) If M=N.
If B is non-degenerate, then AB = B^(-1)(BA)B is similar to BA. So AB and BA
have the same eigenvalues.
If B is degenerate (det(B)=0), disturb B a little bit to make it non-
degenerate, then the above conclusion still holds.
(2) If M>N.
Extend A and B to MxM matricess by filling 0's. Then following (1). |
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s**e
发帖数: 1834 | 7 You are right. e.g. A = [1,0;0,1], Q=[cos(t),sin(t);-sin(t),cos(t)].
Then AQ=Q has eigenvalues exp(i*t) and exp(-i*t).
But you can still get that "AQ's eightvalue's absolute value is between [0,1]". |
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t**o
发帖数: 2618 | 8 was wrong. I think you can only get eigenvalue between -1 and 1 even if
det Q=1.
如果是rotation，我想AQ的特征值的符号不会变 |
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d**e
发帖数: 2420 | 9 Given two positive nxn square matrices A and B with A>=B but A not equal to
B.
Here positive means each entry in A or B is positive.
then whether A's principal eigenvalue is strictly greater than B's?
非常感谢。 |
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G********n
发帖数: 615 | 10 If the matrix is analytic, it is known that eigenvalues and eigenvectors are
analytic too. |
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G********n
发帖数: 615 | 11 隐函数定理只对siple eigenvalue有用
如果有重数就不行了 |
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g****t
发帖数: 31659 | 13 隐函数定理本来就是个判断local连续或者可微的定理.
如果要找global性质.当然不行了.
如果要看多项式防城对系数的全局性变化,我记得是很难的事情.
如果系数是单参数决定的.可以画根轨迹.
多参数的,那就没办法了.
隐函数定理只对simple eigenvalue有用
对多重特征值就不行了 |
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R********n
发帖数: 519 | 14 请问一类特殊的矩阵的名词，想知道是否有约定的名词来统称
矩阵是M Non-negative definite，M的eigenvalue只有2个值，c 或者0，c是一个非负
实数，可以把c就看做1
M = c * \sum_1^k (e_i)(e_i)^T, e_1, e_2 .. .e_k 是R^D空间的k个正交向量，k<=D
一种特殊情况是k=D,这样M = c*I, I就是D by D的单位阵
形象地说，M是由一组正交向量张成的sphere，内在的rank是k，如果k是D的话，M就是
一个D维超球
这样的M，可以叫做什么呢？谢谢 |
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d**e
发帖数: 2420 | 15 这里我不希望用直接求解主特征值，然后求导的方法，因为那样也蛮复杂的。
更重要的是对于2阶以上矩阵，类似问题求解就更不好使了。
Observation: B是一个主角占优，不可约的M矩阵，它的逆BI是一个正矩阵。
从而C是一个正矩阵，它的主特征值是正实数，而且小于1。
我用matlab模拟了一下，似乎f关于a1单调增加。
请大家帮忙想想有什么巧妙的方法证明主特征值关于a1单调or not.
非常感谢。 |
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h**********c
发帖数: 4120 | 16 我记得线形代数好象说过，polynomial最后都能归到求eigenvalue.
上次回家把书扔家乐，还望那位线形代数ap出来更正一下。 |
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M******n
发帖数: 508 | 18 dX1dt=I-k1×X1
dX2dt=2k1×X1-k2×X2
⋮
dXidt=2ki-1×Xi-1-ki×Xi
⋮
dXndt=2kn-1×Xn-1-kn×Xn
I为常数，Ki为常数，n=1-10,
怎么在matlab里列矩阵用eigenvalues and eigenvectors来解这个题? |
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h**********c
发帖数: 4120 | 19 是我没搜到吗？
spline, eigenvalue(这个太常用了），echelon,diagonlizability |
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L******9
发帖数: 78 | 22 Mehta M.L. - Random matrices - AP 1991 2ed
COLL 45 - Katz, Sarnak - Random Matrices, Frobenius Eigenvalues, and
Monodromy
An Introduction to Random Matrices Greg W. Anderson |
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b*******t
发帖数: 33714 | 23 有什么比较快的算法吗？
有人用过hypre么？ |
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s*****e
发帖数: 115 | 24 Still not able to see the point of the question. For the boundary value problem you have, associated with each eigenvalue there are infinitely many eigenfunctions, one differing from another just by a scalar multiple. So for each $\lambda$ you can always pick $g_{\lambda}$ such that its sup-norm is exactly 1, unless you have to pick $g_{\lambda}$ in curtain special way that is not described here. |
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l******r
发帖数: 18699 | 25 Thanks for nice solution!
Just have a quick question. Is the Sobolev Imbedding theorem saying
||f||_sup<=constant||f||_soblev_norm ?
The sobolev norm of the eigenfunctions are usually depending on the
eigenvalues, which is an increasing sequence. So, it seems that directly
using Sobolev embedding thm cannot give us the desired result.
I agree with your first suggestion, mimick the SL theory to get the
conclusion. Thx! |
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s*****e
发帖数: 115 | 26 Not so familiar with the Green's function approach, so I cannot comment on that.
However, I believe your conjecture is true for the following reasons:
What you want to prove is equivalent to that if the eigenfunctions are normalized with respect to sup-norm then their L2 norms are always bounded away from zero.
Clearly, this not true for general functions. For example, both exp(-kt) and exp(-k)*exp(kt) (k>0) on [0,1] have L2 norms approaching to zero as k-->+ infinity. On the other hand, one can... 阅读全帖 |
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s*******e
发帖数: 432 | 28 2. 二维空间的旋转算子是不是没有特征值？
it has imaginary eigenvalue but not real value |
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p***c
发帖数: 2403 | 29 任何一个n乘n矩阵它的特征值都是存在的啊
这不就是代数学基本定理么，
eigenvalues |
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b******v
发帖数: 1493 | 30 determinant是0表示0是其中一个特征值
eigenvalues |
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J*******4
发帖数: 110 | 31 特征矩阵：
P(a) = a*I - (I + [ 0 1 0; 0 0 1; -1 0 0]*e^{-a})
where a is the eigenvalue, I is the 3x3 identity matrix, [...] above is
another 3x3 matrix written in matlab format.
容易验证 a=0 是特征多项式（det(P(a))）的一个二重根，并且 [1; -1; 1] （column
vector written in matlab format） 是属于 a=0 的本征矢。然而在求广义本征矢的
时候，却发现无法找到线性独立于本征矢的广义本征矢。求教版上前辈，我的方法有何
错误？
我用的方法是：
P(0)*V_1 = 0, 解出本征矢 V_1，然后，试图用 P(0) * V_2 = V_1 却发现方程组
inconsistent。
我查了Jack Hale的Introduction to Functional Differential Equations一书的7.2
节，也没有看出我的计算方法有什么问题。不... 阅读全帖 |
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J*******4
发帖数: 110 | 32 Thanks to Schiele for commenting!
The characteristic equation is derived from the following linear system of
delay differential equations:
\dot{z_1}(t) = z_1(t) + z_2(t-1)
\dot{z_2}(t) = z_2(t) + z_3(t-1)
\dot{z_3}(t) = z_3(t) - z_1(t-1)
The method I used to determine the double zeros of the characteristic
equation is from several papers (for example, you can find it from this
paper: http://iopscience.iop.org/0951-7715/21/11/010/). Meanwhile, it is also one of the conditions to determine the dou... 阅读全帖 |
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i****g
发帖数: 3896 | 33 http://blog.sina.com.cn/s/blog_c24597bf0101b871.html
致谢：I would like to thank Prof. Shing-Tung Yau for suggesting the title of
this article, Prof. William Dunham for information on the history of the
Twin Prime Conjecture, Prof. Liming Ge for biographic information about
Yitang Zhang, Prof. Shiu-Yuen Cheng for pointing out the paper of
Soundararajan cited in this article, Prof. Lo Yang for information about
Chengbiao Pan quoted below, and Prof. Yuan Wang for detailed information on
result... 阅读全帖 |
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x********i
发帖数: 905 | 34 Sarnak Awarded Wolf Prize
Peter Sarnak has been awarded the 2014 Wolf Prize in Mathematics. A
mathematician of extremely broad spectrum and far-reaching vision, Sarnak
has influenced the development of several mathematical fields, often by
uncovering deep and unsuspected connections. "By his insights and his
readiness to share ideas he has inspired the work of students and fellow
researchers in many areas of mathematics," the Wolf Foundation said. Sarnak
is the Eugene Higgins Professor of Mathem... 阅读全帖 |
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c******x
发帖数: 350 | 35 假设a11 和a12是第一个eigen vector 的两个分量，那还需要乘eigenvalue的平方根,
比如说是sqrt(d1)，
x1= sqrt(d1)[a11*G1 + a12*G2]
不知理解对吗？ |
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b***k
发帖数: 2673 | 36 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? |
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f*******g
发帖数: 55 | 38 Consider a dynamic process { Q[t] } evolving on the set of M-by-M positive
semi-definite matrices. In particular, { Q[t] } follows the recursive
equation:
Q[t+1] = Q[t] - Q[t] a[t] a[t]' Q[t] / (a[t]' (c I + Q[t] ) a[t]) + d I
where:
a[t] is an M-by-1 vector, which can be adjusted;
c and d are fixed positive real numbers; and
I is an M-by-M identity matrix.
Note that by normalization, we can consider a[t] to be a unit vector without
loss of generality.
The "total reduction" at time t
... 阅读全帖 |
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F****z
发帖数: 12 | 39 In fact, M may always have a real eigenvalue which is >=3. However I don't
know how to prove it. |
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t*****r
发帖数: 42 | 40 谢谢，原问题有个小错误，事实上
x=(1/a+1/b+1/c)/3-1,而非(1/a+1/b+1/c)-1.
当然如果x=(1/a+1/b+1/c)/3-1，的确有“M has a real eigenvalue which is >=3.” |
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t*****r
发帖数: 42 | 41 谢谢，事实上假设y=(1/a+1/b+1/c)/3，则x=y-1.
det(M)=abc(1+x^3)=abc(1+(y-1)^3)=abcy(y^2-3y+3)>0.
然后呢？
eigenvalues |
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e**********n
发帖数: 359 | 42 cyclic is another way of saying periodic boundary condition, here Q is
identified with Q+2 pi.
Similar to crystal momentum, when the variable is periodic, its conjugate
momentum, which is the generator of its translation, has integer eigenvalues
. You probably need to familiarize yourself with Luttinger liquid theory in
1D before going to 3D solitons. There are lots of good tutorials on
Luttinger liquid on the web. |
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n********s
发帖数: 150 | 43 I think you are right, thanks a lot, guys. |
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v**g
发帖数: 20 | 44 I just know that for a bipartite graph, such a property holds for its
normalized laplacian matrix... |
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x********g
发帖数: 595 | 45 请把"时间演化算符"翻译成英文。
The operators are Hermitian because we want to have real eigenvalues. The
operators are to qm what the physics quantities are to classical mechanics. |
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a*****e
发帖数: 4577 | 46 正在看楼上推荐的那本书
多谢各位建议
现在有几点难处在于
1.计算的体系是loss material，半导体或者金属
2.计算的体系是3维的柱状阵列
3.要考虑不同频谱的情况，包括可见光和红外
不知道什么软件比较合适？
如果自己写code的话，大家觉得工作量大概多大？（一千行？）
如果仅仅是解eigenvalue problem不知道能否解决问题，还是需要用FEFD，或FDTD? |
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c****e
发帖数: 2097 | 47 nod. this is the right (although sloppy) statement.
(generating) vectors in Hilbert space are labeled by eigenvalues of such
commuting operators. :D so wordy.
。 |
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Q*T
发帖数: 263 | 48 Just recall how you prove that momentum is the generator of
translation group.
You can do similar things to A or B to show that their
eigenvalues are not bounded. |
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d****i
发帖数: 123 | 50 损耗或增益 eigenvalues are in general complex. they can be real in case of
PT symmetry |
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