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May
18
accepted Is every element contained in a smallest measurable set?
May
18
comment Is every element contained in a smallest measurable set?
Okay, my wording was a little bit misleading, I edited my post, hopefully it is clear now? I do not mean that the set of all $\{ E \}$ with $x \in E$ is countable, I mean that you can select some countable family of sets whose intersection is equal to the intersection of all the $E$'s.
May
18
revised Is every element contained in a smallest measurable set?
added 78 characters in body
May
18
comment Is every element contained in a smallest measurable set?
No, I am saying that the intersection I have written above is equal to an at most countable intersection (not that the resulting set is itself countable or that the measurable sets which contain $x$ itself form a countable family of sets). For a proof, where this set intersection is rewritten such that the sets over which is intersected are selected in accordance with elements from $X$ (and thereby at most countable if $X$ is), see: math.stackexchange.com/questions/931744/…
May
18
asked Is every element contained in a smallest measurable set?
May
17
asked Risk-neutral (i.e. martingale) measure if density is given for a single random variable (i.e. asset)
May
11
accepted Matrix identites, Derivative, Determinant, and a kind of Duality involving Traces
May
7
asked Matrix identites, Derivative, Determinant, and a kind of Duality involving Traces
May
6
comment A special argument to derive the derivative of the determinant
Ah okay, now I understand, the author assumes $t$ fixed in advance, guess he should have stating something like "Assume $\Phi(t) = I$ for some fixed $t$", then everything would work fine. Thank you!
May
6
comment A special argument to derive the derivative of the determinant
Ok, thank you, never thought of that. Do you understand the rest? I try to follow it, but as I see it the final formula $$ \frac{d}{d t} \mbox{det} \Phi(t) = \mbox{det}(\Phi(t)) \mbox{tr}(\Phi(t)^{-1} \dot{\Phi}(t)) $$ is also valid just for $t = 0$, but what sense has such a formula then?
May
6
asked A special argument to derive the derivative of the determinant
May
5
comment Differntiating matrix functions $f : \mathbb R^{n\times m} \to \mathbb R^{p\times q}$
In the usual sense of vectors $e_i$ that could be interpreted as slope in particular direction, of course derivatives taken in the direction of matrices are possible, but what is a natural analogon of the Jacobian?
May
5
asked Differntiating matrix functions $f : \mathbb R^{n\times m} \to \mathbb R^{p\times q}$
May
5
comment Matrix Calculus and Matrix Derivatives
As I understand copper's answer this matrix could not reasonably interpreted as a representation matrix (in general, so also in case we are not looking at the identity transformation), or am I wrong?
May
5
comment Matrix Calculus and Matrix Derivatives
But as a reprentation matrix of an endomorphism with a fixed base it does not represent the identity transformation...
May
5
accepted Matrix Calculus and Matrix Derivatives
May
4
comment Matrix Calculus and Matrix Derivatives
Okay, the matrix as written above has something to do with the projection maps $\phi(X) = [X]_{kl}$, i.e. how they behave on $\{ E_{ij} \}$, but then why should writing it as a row vector is more natural... guess in some sense I got the point, but it stil feels so artificial and arbitrary, if you have some motivational remarks as why to consider such matrix derivatives I would be thankeful. But nevertheless my question is answered in some sense.
May
4
comment Matrix Calculus and Matrix Derivatives
If I got your answer correctly it says that in this case, the collection of the partial derivatives (which you mention in your last sentence) did not yield in any natural way the representation matrix, in sharp contrast to the case $f : \mathbb R^n \to \mathbb R^m$, where the Jacobian (the collection of the partial derivatives) is exactly the representation matrix w.r.t. the standard basis? And for your remark "The matrix above doesn't make sense as a derivative", could it be meaningfully interpreted in any other way if it has nothing to do with the derivative?
May
4
comment Matrix Calculus and Matrix Derivatives
Thanks for your answer, but can you explain a little bit more. As I see it, if $f = id$, then $df$ is the identity of $L(R^{2\times 2}, R^{2\times 2})$, and by fixing the indices it should have the representation matrix $$ \begin{pmatrix} 1&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&1\end{pmatrix} $$but if I fix a basis the identity mapping always has as representation matrix the identity matrix?? In any basis, so why not here...
May
4
asked Matrix Calculus and Matrix Derivatives