Reputation
5,304
Top tag
Next privilege 10,000 Rep.
Access moderator tools
Badges
10 22
Newest
 Pundit
Impact
~49k people reached

16h
comment convex optimization?
I suppose $A$ is a vector?
1d
answered Fastest way to perform this multiplication expansion?
1d
comment Eigenvalues of Moore–Penrose Pseudo-Inverse of a Symmetric Matrix
@HosseinK.Mousavi - yes, I was referring to the Moore Penrose pseudoinverse.
1d
comment Eigenvalues of Moore–Penrose Pseudo-Inverse of a Symmetric Matrix
$\lambda_i \ne 0$ turns into $\lambda_i^{-1}$, $\lambda_i = 0$ turns into $0$.
2d
comment Increasing a singular value
If it's an extreme or nearly extreme singular value, this will change the condition of the matrix, with the usual effects on computation.
Aug
22
answered Simpson's rule with precision?
Aug
22
answered Let $(X,\mathcal{F},\mu)$ be a measure space and let $g\in L^1((X,\mathcal{F},\mu))$.
Aug
22
answered Analytic approximations of the step function
Aug
21
awarded  Pundit
Aug
21
comment The sphere $S^2$ is not contractible
It's supposed to be a map $f:I \times S^2 \to S^2$.
Aug
21
answered Integral should not be continuous.
Aug
19
answered Solving Frobenius minimization with linear algebra
Aug
19
answered Does $\mathbb{R}^n$ have a real vector space structure with dimension other than $n$?
Aug
19
comment A class of sparse matrices whose inverse is also sparse?
Sparse low rank perturbations of diagonal matrices will also have sparse inverses, by the Woodbury formula.
Aug
19
comment Reduce variance of peak values of function
Multiply everything with 0, that will flatten all peaks. Or is that nort what you want?
Aug
19
comment Reduce variance of peak values of function
What is $V$? What is $x$ in the graph that you drew? What is $\rho$ for?
Aug
16
comment Product of matrices with complex dimensions
It works iff $N = 2$.
Aug
15
comment Proof if $I+AB$ invertible then $I+BA$ invertible and $(I+BA)^{-1}=I-B(I+AB)^{-1}A$
This is true even if $A$ and $B$ are non-square matrices such that $A$ and $B^T$ have the same dimensions. If $A$ is a row vector and $B$ is a column vector, this is is known as the Sherman-Morrison-Woodbury fotmula.
Aug
15
answered Find such $g(x)$ for which $(b-a)\sqrt{f(a)f(b)}=g(b)f(b)-g(a)f(a)$ holds
Aug
15
comment Find such $g(x)$ for which $(b-a)\sqrt{f(a)f(b)}=g(b)f(b)-g(a)f(a)$ holds
Do you want the same $g$ to work for many different $f$? That will most likely not work. Or do you want a different $g$ for each $f$? Then you have to find a new $g$ for each $f$ which may be more work than just computing the left hand side.