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1932
bio website vsego.org
location Manchester, United Kingdom
age 37
visits member for 1 year, 8 months
seen Jan 25 at 3:12

A research visitor at the University of Manchester.

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


Jan
23
comment Is LU decomposition of matrices efficient for today's standards?
This also depends on the pivoting, i.e., if you are considering the LU factiorization with the complete pivoting, with the partial pivoting, or without any pivoting at all. The last one doesn't even exist for all matrices (might not be important for some applications), and the first one is computationally quite expensive.
Jan
16
answered Is $A$ diagonalisable if its unitary?
Jan
5
comment Diagonalisability of Self-Adjoint Operators for Non-Symmetric Metrics
Can you provide a reference for the diagonalizability of self-adjoint operators in symmetric bilinear forms? I'm having a problem with that, so I want to get all the terms right.
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
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.