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location Manchester, United Kingdom
age 37
visits member for 1 year, 7 months
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A research visitor at the University of Manchester.

Research interests: NLA, especially indefinite scalar products and matrix polynomials.


Dec
9
awarded  Caucus
Dec
2
comment A connection between the eigenvectors of $W$ and $D^{-1/2}WD^{-1/2}$
I don't think there is one, apart from the obvious: $(D^{-1/2})^* = D^{-1/2}$, so this is a congruence (which keeps the inertia).
Nov
18
reviewed Approve Function Relations
Nov
18
answered Difference between complex symmetric and hermitian matrices
Nov
17
comment Diagonalizing zero matrix
@EhBabay What do the eigenvalues of a similarity matrix $S$ have to do with this? The eigenvalues of $A = 0$ matter here, and they are zero, as Chrispy wrote in his reply to you.
Nov
16
revised Diagonalizing zero matrix
added 32 characters in body
Nov
16
comment Diagonalizing zero matrix
@EhBabay What?!
Nov
15
comment Properties of LU decomposition
$PA = LU$ is usually considered an LU factorization with pivoting, while an LU factorization (without the mention of pivoting) is usually just $A = LU$. What is the exact text of your problem?
Nov
13
comment Are Eigen vectors unique?
The answer is: yes, you can, if and only if $A$ and $B$ commute.
Nov
12
answered Determine the existence of a function
Nov
10
comment Determine the existence of a function
Good point. Did you find any others? Maybe that could give us some idea of the solution. Also, where did this problem come from?
Nov
10
comment Determine the existence of a function
I am pretty sure that $f$ that satisfies the conditions 1-3 must be an arithmetic mean of the elements of $x$, i.e., $f(x) = \frac{1}{n} \sum\limits_{k=1}^n x_k$, but I don't know how to prove it.
Nov
9
revised Matrices similar only to themselves
added 42 characters in body; edited tags; edited title
Nov
9
answered Matrices similar only to themselves
Nov
4
reviewed Approve A generalization to this pattern?
Nov
3
comment Connection between results of two SVDs
Ref. this last edit: while all of that is correct, there is no reason to use SVD (as the OP wants), which is quite unnatural for the functions of matrices. If the eigenvalue decomposition was used instead, all would remain the same, except we could ditch the condition $f(D) \ge 0$. If that one is really needed, it can be achieved by putting signs in the left or the right orthogonal matrix (but not both!) in the last step, and working with $|f(D)|$ after it. This way, all symmetric/Hermitian matrices such that $f(\lambda)$ exists for all eigenvalues $\lambda$ can be taken into account.
Nov
3
comment Connection between results of two SVDs
@Daniel It is possible, of course: $h(\Lambda)$ can be a symmetric permutation of the elements of $f(\Lambda)$. However, if you choose the elements in decomposition to be in certain (i.e., descending) order, then it is unique, as these are the eigenvalues and we know that if $\lambda$ is an eigenvalue of $X$, then $f(\lambda)$ is an eigenvalue of $f(X)$. All of this, of course, under the assumptions from the answer we're commenting.
Nov
3
comment Connection between results of two SVDs
@Daniel In that case, SVD is just an eigenvalue decomposition, so you have $f(U \Lambda U^T) = U f(\Lambda) U^T$, where $f(\Lambda)$ is computed elementwise because $\Lambda$ is a diagonal matrix. This is how you can define a function of any (orthogonally) diagonalizable matrix; nothing SVD-specific here.
Nov
2
comment Connection between results of two SVDs
My point was a bit different. Let's say that there is some neat relation, so something like $\Sigma' = \ln \Sigma$. There is no reason for it to work on some matrices and not on others (because $\Sigma \ge 0$ for all $M$, so $\ln \Sigma$ exists for all nonsingular $M$), and we do know that it cannot work on some of them (because $\ln M$ may not exist). My point here is that SVD "hides" the existence of the function. In his "Functions of Matrices", Higham mentions SVD only in Appendix B and problems' solution, which is another reason I don't expect there to be much of a connection (if any).
Nov
2
comment Connection between results of two SVDs
Given that the SVD of $M$ always exists, $\Sigma$ has only nonnegative elements, and $\ln M$ may not exist (at least not in $\mathbb{R}^{n \times n}$), I wouldn't hold my hopes up, at least not for something similar to what you get with the eigenvalue/Schur/Jordan decomposition.