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Questions about the properties of vector spaces and linear transformations, including linear systems in general.

7 votes

Finding eigenvalues of an 'almost-tridiagonal' circulant matrix

If you make an even-odd permutation, your matrix becomes $$ \begin{bmatrix} aI & I+Z\\ I+Z^{-1} & -aI \end{bmatrix}, $$ where $Z$ is the generator of the circulant algebra. Let $Z=FDF^{-1}$ be its eig …
Federico Poloni's user avatar
1 vote

Matrix equation solving guidelines

If you are solving this numerically on a computer, you should use a variant of the Bartels-Stewart method; see for instance http://dl.acm.org/citation.cfm?id=146929 (Gardiner, Laub, Amato, Moler). Th …
Federico Poloni's user avatar
2 votes

Adjusting matrix in generalized eigenvalue problem for the design of eigenfunctions

(Probably this is not what you are asking for, but it might be a cue to modify your question in the right direction.) If you fix $A$, $\gamma$, $\phi$, then what you get is a linear system in the en …
Federico Poloni's user avatar
8 votes
Accepted

Are the off-diagonal elements of exp(At) log-concave in t, for nonnegative matrices?

The following Matlab code (unless I coded something wrong, which is well possible) finds a few random counterexamples each time I run it, even when restricted to off-diagonal entries: n = 3; for trie …
Federico Poloni's user avatar
1 vote
Accepted

update SVD of linear combination after changing scalar weight

I don't think there is any computational advantage in general. For any pair of symmetric $M$, $N$ and given $c\neq d$, one can find symmetric $A, B$ such that $M=A+cB$, $N=A+dB$, so essentially you ar …
Federico Poloni's user avatar
2 votes
Accepted

singular value decomposition

Let $A_2A_1^{-1}=U_A DV_A$ and $B_2B_1^{-1}=U_B DV_B$ be the two SVDs with the same $D$. Set $U_2'=U_AU_B'$, $U_1=V_B'V_A$, $V=B_1^{-1} U_1 A_1$. The first equality is clear. The second one is prove …
Federico Poloni's user avatar
3 votes

Canonical form of a general Bilinear Form

People in matrix analysis would call this a "canonical form under congruence". Take a look at http://arxiv.org/abs/0709.2473; the solution is stated there.
Federico Poloni's user avatar
0 votes
Accepted

Strictly diagonally dominant hermitian matrices eigenvalues sign

You do not need to show that the discs are disjoints. In fact, this won't hold for most diagonally dominant matrices, unlike the main result that you wish to prove. What you need is a stronger form o …
Federico Poloni's user avatar
3 votes

naming for the map $T = x \mapsto a x b$

You might also be interested in checking out the representation of these operators with the Kronecker product and the vectorization map: $$ \operatorname{vec(AXB)}=(B^T \otimes A)\operatorname{vec}(X) …
Federico Poloni's user avatar
1 vote

fast merging of orthogonal bases

You may think to the situation as a Gram-Schmidt orthonormalization partially performed (the columns of $U_1$ are already orthonormal), and complete it. The matrix $V=U_2-U_1U_1^TU_2$ has columns ort …
Federico Poloni's user avatar
2 votes

Estimating the spectral radius of a matrix, noniteratively

Why are you trying to avoid eigenvalue calculations in the first place? I think Arnoldi methods (such as Arpack, used e.g. in Matlab's eigs) would do a respectable job, and maybe even the power method …
Federico Poloni's user avatar
0 votes

matrices self-adjoint with respect to some inner product

Take a look at: Indefinite linear algebra and applications --- Israel Gohberg,Peter Lancaster,L. Rodman, section 4.2.
Federico Poloni's user avatar
1 vote

Update to SVD by changing 2 row vectors

From Golub and Van Loan, Matrix Computations 3rd Edition, Section 12.5 ("updating the ULV decomposition": ...$O(n^3)$ flops are required to recompute the SVD of a matrix that has undergone a unit …
Federico Poloni's user avatar
3 votes
Accepted

Are there matrices $C,D,E$ such that $AXA^T = B \iff CX^{-1}D = E?$

Converting previous comments into an answer, because I noticed they can fully answer the question. First of all, one can change basis to assume $A = [I\,\, 0]$, and partition $$ X = \begin{bmatrix}X_{ …
Federico Poloni's user avatar
3 votes

FFT based algorithm for special matrices

Not exactly my field, but I think that the same techniques used for Vandermonde's determinant work for Moore's and on finite rings (with the added complication of using FFTs over rings). A good refer …
Federico Poloni's user avatar

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