f********0
发帖数: 73 | 1 assume both BA and AB are conformable. |
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H****h
发帖数: 1037 | 2 det(tI+AB)=det(A(tA^{-1}+B))=det((tA^{-1}+B)A)=det(tI+BA) |
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x*****d
发帖数: 427 | 4 if ABx = tx, then BABx = tBx, i.e. Bx is an eigenvector of BA
belonging to t. Similar argument for the converse. |
|
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H****h
发帖数: 1037 | 6 还可以证明特种根的重数也是相同的。
too.
x
not |
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x*****d
发帖数: 427 | 7 本来我以为对非方阵也对，结果找到反例 (1,0), (1,0)^T,
所以还是你的方法比较先进 |
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g******a
发帖数: 69 | 8 [I B][0 0 ][I -B] = [BA 0]
[0 I][A AB][0 I ] [A 0] |
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H****h
发帖数: 1037 | 9 非方阵的情况可以通过补零形成方阵。所以AB和BA中阶数较高的方阵
多出来的特征值都是零。
如果A或者B任意一个是可逆的，那么AB和BA是相似的，于是有相同的若当标准型。
在一般情况下，AB和BA未必相似，比如二阶方阵A=e_{11}, B=e_{12}。
但可以证明AB和BA的非零特征值对应的若当块组合是完全一样的。
现在的一个问题是，如果AB和BA都只有零特征值，AB和BA的若当型有没有
任何联系，或者任何一对零特征值若当型都可以分别属于AB和BA？ |
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m****n
发帖数: 45 | 10 Sorry, I meant a matrix in SO(3).
So 1 must be an eigenvalue |
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g******a
发帖数: 69 | 11 this is because all the eigenvalues are
on the unit circle, and appear conjugately. |
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b****d
发帖数: 1311 | 12
good
Show him [ 0 1] which has eigenvalues i and -i
[-1 0] |
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r****y
发帖数: 1437 | 13 why? Just solve the eigenvalues and eigenvectors for
covariance matrix iteratively. There is numerical routine ready for this,
the first one is always 1st PC, etc.
In matlab, they have built-in PCA command, try princomp. |
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r*******y
发帖数: 1081 | 14 I also think so. And I-xx'/x'x is symmetric and 2-幂等矩阵 which has
only 1 and 0 as eigenvalues. |
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i*******d
发帖数: 6 | 15 I think it is easy for me to prove that the matrix is idempotent and an
idempotent matrix has only 1 and 0 as eigenvalues.
Thank you both! |
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c*******h
发帖数: 1096 | 16 just to remind you that 2-norm of a matrix A does not relate to the
eigenvalues
of A. you are just lucky here that the matrix in question is symmetric and
diagonalizable. |
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i*******d
发帖数: 6 | 17 I am confused. I found that ||A||_2 = sqrt(max(eigenvalues of A)) on
Internet. |
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i*******d
发帖数: 6 | 18 Yes, you are right for general matrix A. But for idempotent matrix A, i.e. A
^2=A, the eigenvalues of A is the same as ones of A'*A, right? |
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H****h
发帖数: 1037 | 19 If X and Y are square matrices, then XY and YX have the same
eigenvalues with same multiplicities. (det(xI-XY)=det(xI-YX).)
In your cases, you may expand X and Y to be square matrices by
adding 0 entries to below or right. Then XY and YX are both
M*M matrices. Since they have the same rank, and XY is essentially
an N*N matrix, so YX has rank no more than N. |
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M********d
发帖数: 91 | 20 Then you'll need to so some work, such as SVD or eigenvalue analysis |
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c*******h
发帖数: 1096 | 21 use eigs() in matlab. very fast... |
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B********e
发帖数: 10014 | 22 i suggest that go to Trefethen's book to find the
solution.
I learned it long time ago and forgot the details but I guess there is the a
nswer. |
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r*****f
发帖数: 247 | 23 根据定理，如果已知向量x和y，且 x majorize y, 那么肯定存在一个矩阵A，x的元素
是A的eigenvalues，y的元素是A的对角线。
那么这个A怎么根据x，y求？ 有现成的matlab算法么？
有人写了个paper给了算法，有点复杂。想找现成的。不知道哪位大侠有。 |
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s**c
发帖数: 1247 | 24 A,B n*n，symmetric, AB=BA,A的eigenvalue的multiplicity都是1
求证：A,B能simultaneously diagonalized
i.e. 存在orthogonal的矩阵P，A=PDP',B=PEP',D,E是diagonal的矩阵
3x |
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i********e
发帖数: 31 | 25 "A and B can be simultaneously diagonalized"
is sufficient for "A and B commute" but
not necessary.
Note that the original problem in this post
requires that A has "distinct" eigenvalues!
One of the necessary conditions is
that A and B have at least one common eigenvector. |
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s*n
发帖数: 245 | 26 matrix is symmetric.
element of matrix is A_{i,j}=2*delta_{i,j}-delta_{i,i+1}-delta_{i+1,i}.
就是主队角元素是2,离主对角元最近的副对角元是-1.其它元素都是0。
怎末求本征值。 |
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i********e
发帖数: 31 | 27 2-2*cos(k*pi/(n+1))
k = 1,2,..,n |
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c*******h
发帖数: 1096 | 29 泊松方程的解
土一点可以用暴力解法，特征多项式求根 |
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e**********n
发帖数: 359 | 30 Nearest tightbinding model in physics.
Assume the eigenvector is e^{i p j}, sqrt(-1) = i, j = 0,1,2,...n-1, solve
the linear equations that the eigenvectors satisfy to get all possible value
for p. |
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c*******h
发帖数: 1096 | 31 x is the largest eigenvector of the generalized eigenvalue problem
Ax = aBx.
or simply speaking, let B have cholesky B=GG' and let y=G'x, then
f(x) = y' * (G^-1*A*G^-T) * y / y'y
so y is the largest eigenvector of G^-1*A*G^-T.
or more simply speaking, the maximum is attained when x is the
largest eigenvector of A*B^-1 or B^-1*A
definite. |
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A*******r
发帖数: 768 | 32 relation b/w spectral norm and eigenvalues |
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j******n
发帖数: 21 | 33 题目是
Can λ=0 be an eigenvalue of a Sturm Liouville Problem?
Can y=0 be an eigenfunction of a Sturm Liouville Problem?
Explain your answer.
只是会用几个special case去证明第一问。但是觉得不是充分解释。
请问大家有没有什么好的办法说明这两个问题吗？ |
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b*********n
发帖数: 56 | 34 That's impossible. A has three distinct eigenvalues, and each e-val has one
eigenvector. So these three eigenvectors forms S. This S must be invertible.
Here is the result from matlab:
S =
-0.7071 0.7071 0
0.7071 0.7071 0
0 0 1.0000
D =
1 0 0
0 2 0
0 0 3 |
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t****y
发帖数: 576 | 35 两个同样size n by n的对称矩阵，如果A矩阵的每一个entry都比B矩阵大,是不是A矩阵
的eigenvalue就会比B矩阵大？ |
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a*******h
发帖数: 123 | 36 Eigenvalues get scaled by c as well. |
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c*******h
发帖数: 1096 | 37 i might be no better than an underg math major.
so please tell me more about multidimensional arrays and eigenvalues. |
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B********e
发帖数: 10014 | 38 呵呵，貌似不trivial，所以我就不try了
上网搜一下，于千万篇中掕一篇出来
Title:
Eigenvalues and invariants of tensors
Authors:
Qi, Liqun
Publication:
Journal of Mathematical Analysis and Applications, Vol. 325, No. 2, p. 1363
-1377
no offense,其实我只是想说，这个问题不是很有意思
on |
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r*****f
发帖数: 247 | 39 F. Zhang and Q. Zhang, “Eigenvalue inequalities for matrix product,”
IEEE Trans. Autom. Control, vol. 51, pp. 1506-1509, Sept. 2006.
J. R. Magnus and H. Neudecker, Matrix Differential Calculus with
Applications in Statistics and Econometrics, New York: Wiley, 1999. |
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n*s
发帖数: 752 | 40 if 2-lambda is e-value, 2-(2-lambda)=lambda is also e-value |
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i********e
发帖数: 31 | 41 Yes.
Matrix Analysis: Roger A. Horn, Charles R. Johnson - Google Books Result
by Roger A. Horn, Charles R. Johnson - 1985 - Science - 561 pages
"The following result, an immediate corollary of Weyl's theorem known as the
monotonicity theorem, says that all the eigenvalues of a Hermitian matrix .
.."
books.google.com/books?isbn=0521386322... |
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c*******h
发帖数: 1096 | 42 if alpha and beta are linearly independent, the two nonzero eigenvalues
are
/----------------------
(ax+by) + / (ax-by)^2 + 4(az)(bz)
-/ |
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c*******h
发帖数: 1096 | 43 by the definition of eigenvalues and eigenvectors,
and note that the corresponding eigenvectors are linear combinations of
alpha and beta. |
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c*******h
发帖数: 1096 | 44 if you want the first order term, a natural way is using (I+B)*exp(A)
or exp(A)*(I+B), or even exp(B)*exp(A) or exp(A)*exp(B).
you still can derive some bounds by help of some eigenvalue sensitivity
results. |
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s**********n
发帖数: 1485 | 46 QR iteration. If you mean numerically. |
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j**********7
发帖数: 4 | 47 An interview question I met.
"Consider a matrix $J$. every eigenvalue of $J$ has positive real part.
Now let us consider infinite product of (1-J/n). Do you think this
infinite product converges? and how about the converge rate?"
Is there anyone can give some advice. thanks! |
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m****i
发帖数: 875 | 48 不是有个啥×××circle的定理说可以的么？ |
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