s**e
发帖数: 1834 | 1 Find a D such that DAD=I, so the answer is 1. |
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l******3
发帖数: 6 | 2 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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b****d
发帖数: 1311 | 3 两对称正定矩阵乘积未必对称更未必正定。但是所有特征值都为正实数。
所以在某种内积下（比如选特征向量组成的一组基为正交基）可以是正定的。 |
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s**e
发帖数: 1834 | 4 (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 | 6 Only for M=N.
B^(-1) is the inverse of B if B is non-degenerate.
When B is degenerate, B^(-1) is not defined. But we disturb B a little bit
to B_0 such that B_0 is non-degenerate. Then AB_0=B_0^(-1)(B_0A)B_0
is similar to B_0A, so AB_0 and B_0A have the same eigenvalues. Since this
is true for any small disturbance B_0, take limits, then AB and BA have the
same eigenvalues. |
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n*****g
发帖数: 1199 | 7 考虑分块矩阵 $A & I \\ I & B$. 用两种不同的方法求其行列式，得到AB和BA特征多
项式间的关系。具体参见北大高等代数教材课后习题。 |
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s**e
发帖数: 1834 | 9 I think the proof is fairly "严格", though might not be as elegant or "clean" as some proof that is more algebraic.
Because when B_0 --> B we have
det(x*I-AB_0) --> det(x*I-AB), so "AB_0 and B_0A have the same eigenvalues" implies "AB and BA have the same eigenvalues".
Btw, when both A and B are degenerate square matrices, the proof only claims
that "AB and BA have the same eigenvalues", but does not claim that "AB and
BA are similar". |
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h*******e
发帖数: 225 | 10
Why? Your reasoning is too "intuitive". Unless there is an explicit analysis
of the limit, your claim is not necessarily true. B0 -> B does not mean B0
has the same or similar properties as B. There could be a drastic change in
properties at the limiting point. |
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s**e
发帖数: 1834 | 11 You are right that such an "intuitive" argument might not be true when
taking limit for a general "function". In this case, we are only dealing
with "polynomials", so the argument is true.
For (fixed) n-degree POLYNOMIALS f_i (i=1,2,3,...), if f_i --> f, then the "
roots of f_i" --> "roots of f".
When f_i are general functions, it might not be true.
Since det(x*I-AB_0) and det(x*I-AB) are M-degree polynomials of x, so the
roots of det(x*I-AB_0) also converge to the roots of det(x*I-AB).
analysi |
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n******d
发帖数: 244 | 12 You need to require that Q be special 正交矩阵, i.e. det Q=1.
Otherewise you only get ＡＱ的特征值在【-1，１】之间.
You can prove it easily with the observation of its geometry. |
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t**o
发帖数: 2618 | 13 我想符号 是一定的
由于rotation是单联通的， 所以 特征值不会穿越0这个点。（A正定是对的，如果A半
正定，估计也对）
只是 此题我始终没看出来
如何用几何 直接把它说清楚 |
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t**o
发帖数: 2618 | 15 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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R********n
发帖数: 519 | 16 我想通过一个联系，把算法中用到的向量用矩阵来表示。比如有一个向量v，希望表达成矩阵M，可以通过M包含V的所有信息。
问题是，一般的做法都是用M的特征向量来表示v的方向，M的特征值来表示v的norm，但是特征向量是可以乘-1还是，这样方向就变了，所以存在一个方向模糊的问题。
就是说分辨不出来v or -v
比如M = v*v'，这样通过M，我只能得到v or -v，但是不知道具体是哪一个
有什么方法可以把v的正负极性在矩阵中保留呢？谢谢~~~ |
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d**e
发帖数: 2420 | 18 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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d**e
发帖数: 2420 | 20 this is not a counterexample. But still thanks. |
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q********e
发帖数: 1255 | 22 ok,貌似answer is yes at least for symmetric real matrix，屡败屡战哈
根据singular value 的性质，只需要证明：
if A>B, then ||A||_2 >||B||_2. (这里A>0,B>0,A>B all according to your
definition, ||.||_2 是矩阵的2 norm).
证明这个只需要证明
sup_x {||Ax||_2/||x||_2}=sup_{x>0}{||Ax||_2/||x||_2}
貌似不难，因为A>0. |
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d**e
发帖数: 2420 | 23 ≥由Gelfand's formula易得，如何证明严格＞？苦恼呀。
谢谢你的热心帮助，多谢了！！ |
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x******i
发帖数: 3022 | 24
I think the following is true:
suppose A and B are both symmetric and positive definite,
with B strictly positive definite, and suppose C= A+B
then if we list A's eigenvalues as
a_1>=a_2>=a_3>= ... >= a_n
and C's eigenvalues as
c_1>=c_2>=c_3>= ... >= c_n
then
c_1 > a_1
c_2 > a_2
c_3 > a_3
...
c_n > a_n
This can be easily proved by recognizing that the ellipsoid
corresponding to x*C*x = 1 lies strictly within the ellipsoid
corresponding to x*A*x = 1.
Suppose the eigenvectors corresponding to a_ |
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d**e
发帖数: 2420 | 25 谢谢你认真仔细的回复。如果在对称正定矩阵，也许可以通过
同时对角化证明。再次感谢。 |
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G******i
发帖数: 163 | 26 A>=B but A not equal to B, A>0, B>0
=> A^4 > B^4 (i.e., every entry of A^4 -B^4 is positive)
=> (The principal eigenvalue of A^4) > (the principal eigenvalue of B^4)
=> (The principal eigenvalue of A) > (The principal eigenvalue of B)
to |
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d**e
发帖数: 2420 | 27
You are right, it seems 3 is engouh such that A^3>B^3.
how do you get this result? thank you very much. |
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d**e
发帖数: 2420 | 28 great, that's right.
given A>B>0, we can choose sufficiently small d>0 such that
A-dI_n>B>0, then \rho(A-dI_n)>=\rho(B)
Note that \rho(A)=d+\rho(A-dI_n)>\rho(B).
The proof is complete.
非常感谢诸位的帮忙，实在让人敬佩。 |
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N**D
发帖数: 10322 | 29 depend on definition of matrix smooth,
my wildest guess. |
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G********n
发帖数: 615 | 30 By smooth I mean
A(x_1, ... ,x_n) = ( a_{ij}(x_1, ..., x_n) )
where a_{ij}(x_1, ..., x_n) are smooth functions on x_1, ..., x_n. |
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G********n
发帖数: 615 | 31 If the matrix is analytic, it is known that eigenvalues and eigenvectors are
analytic too. |
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B****n
发帖数: 11290 | 32 大部分情況下肯定是平滑的 比方說特徵值是多項式的解 係數是矩陣元素的平滑函數
根據隱函數定理 大致上可知在非singular的情況下 解是平滑的 |
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H*********r
发帖数: 659 | 33 where to learn this concept? I only learned numerical linear algebra
are |
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G********n
发帖数: 615 | 34 Thank you.
But I need a result for ALL symmetric matrices...
根據隱函數定理 大致上可知在非singular的情況下 解是平滑的 |
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g****t
发帖数: 31659 | 35 就是隐含数定理啊,找本matrxi analysis书就可以看到了.
Thank you.
But I need a result for ALL symmetric matrices...
根據隱函數定理 大致上可知在非singular的情況下 解是平滑的 |
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G********n
发帖数: 615 | 36 隐函数定理只对siple eigenvalue有用
如果有重数就不行了 |
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z***c
发帖数: 102 | 37 连续性没问题，不过光滑性就不一定了。比方说考虑x^2=a这个方程，当a从正数变为负
数的时候根从实
轴跑到虚轴拐了一个直角的弯。我觉得没重根的时候是光滑的，有重根就不光滑了。 |
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l********e
发帖数: 3632 | 38 不一定differentible
但是连续性正如你自己所说：simple root自然连续。多重的话，都重复了，当然还是
连续啦。 |
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G********n
发帖数: 615 | 39 考虑一个2*2矩阵
a_{11} = -a_{22} = exp(-1/t^2)cos(2/t).
a_{12} = a_{21} = exp(-1/t^2)sin(2/t).
A(t)关于t是光滑的，但是在t=0时特征向量不连续... |
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c*******h
发帖数: 1096 | 40 重根嘛。重根是会出问题的。特征子空间维度大于一，随便一组正交基旋转一下还是正
交基 |
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a****t
发帖数: 720 | 42 有重根的对称方程codimension 2， measure 0，可以assume 无重根。 |
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p**o
发帖数: 3409 | 43 我的D矩阵取了绝对值的。关于你的"反例"：
>> A=[1.0753, 2.6961, -2.6924;
2.6961, 0.6375, -0.9651;
-2.6924, -0.9651, 7.1568];
>> D=diag(sum(abs(A)))
D =
6.4638 0 0
0 4.2987 0
0 0 10.8143
>> Ap=D^(-1/2)*A*D^(-1/2)
Ap =
0.1664 0.5115 -0.3220
0.5115 0.1483 -0.1415
-0.3220 -0.1415 0.6618
>> [U,L]=eig(Ap)
U =
0.7240 0.4198 -0.5473
-0.6769 0.5853 -0.4463
0.1330 0.6936 0.7080
L =
-0.3710 0 |
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z*******9
发帖数: 167 | 44 Sorry。没看到那个绝对值符号。你原先的结论是对的。但是证明是
非常类似的。
见附件。（其中L是奇异的，因为它的每一列都可以用其他列线性表达。） |
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z*******9
发帖数: 167 | 46 Another little mistake.
See attachment. |
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z*******9
发帖数: 167 | 48 还是有点错
L=D-A不一定是奇异的。
当且仅当sign（A）的rank为1的时候，等号才成立, see attachment.
否则等号不成立。一个例子是：
A =
5.5389 -0.6245 2.8300
-0.6245 -0.1261 0.5906
2.8300 0.5906 2.9794
>> d = sum(abs(A));Dh = diag(1./sqrt(d));D = diag(d);[U,S]=eig(Dh*A*Dh)
U =
0.3456 -0.5269 0.7765
0.8444 0.5355 -0.0125
-0.4093 0.6600 0.6299
S =
-0.2653 0 0
0 0.3313 0
0 0 0.9214
>> rank(sign(A))
ans =
3
=========================== |
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