bio | website | |
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location | Argentina | |
age | ||
visits | member for | 4 years, 2 months |
seen | Apr 15 at 1:42 | |
stats | profile views | 612 |
email: [the shortest string of letters containing my first name and last name as substrings]@gmail.com
Dec 16 |
awarded | Nice Answer |
Jul 29 |
awarded | Nice Question |
Jun 25 |
awarded | Excavator |
Apr 19 |
comment |
Giving $Top(X,Y)$ an appropriate topology
For the non-Hausdorff case, see ncatlab.org/nlab/show/exponential+law+for+spaces. |
Mar 8 |
comment |
Topological spaces determined by generalized metric spaces
Notice that this Arens' space is slightly different from the one given in Wikipedia (en.wikipedia.org/wiki/Arens%E2%80%93Fort_space), since in this space the set $\{c\}\cup\{a_{n,m}:n,m\in\omega\}$ is not open. |
Feb 20 |
awarded | Yearling |
Jan 25 |
asked | Nontrivial copies of SO(r) in SO(n) |
Jan 24 |
comment |
Suggestions for good notation
@Ben, the index in the coordinate expression $\frac{\partial f}{\partial x^j}$ for the 1-form $df$ is clearly in the low position! In fact, this is the main reason that I see for having to put the indexes of the coordinates in the high position as we do, instead of doing everything in the opposite way, which would be better in some way: we could write $f=x_1^2+x_3$ instead of $f=(x^1)^2+x^3$. |
Jan 24 |
comment |
Suggestions for good notation
Regarding differential geometry: If $f:M\to\mathbb R$ is a smooth function on a manifold and $x:M\to\mathbb R^n$ is a chart, I prefer $\left(\frac{\partial f}{\partial x}\right)_j$ or $\left(\frac\partial{\partial x}\right)_j f$ (or even $\partial_j f$ if the choice of the particular chart is clear or irrelevant). Because the notation $\frac{\partial f}{\partial x^j}$ suggests that $\frac{\partial f}{\partial g}$ could be defined using only $g$, and in fact you need to know that you are restricting to the curve along which the other coordinates $x^i$ are constant. |
Jan 21 |
revised |
Why is a topology made up of 'open' sets?
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Jan 21 |
revised |
Why is a topology made up of 'open' sets?
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Jan 14 |
comment |
real symmetric matrix has real eigenvalues - elementary proof
I think that the main difference is that Alexander extremises $x^tAx$ and I extremise $y^tAx$. That the two situations are not trivially equal is the subject of p.32 of Conway. |
Jan 14 |
revised |
real symmetric matrix has real eigenvalues - elementary proof
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Jan 13 |
revised |
real symmetric matrix has real eigenvalues - elementary proof
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Jan 13 |
comment |
real symmetric matrix has real eigenvalues - elementary proof
Alexander, when you said that the minimum is an eigenvalue, did you mean to prove it by applying the Lagrange multiplier equation to the function $f(x)=x^tAx$ restricted to a level set of $g(x)=x^tx$, or did you have a different idea in mind? |
Jan 13 |
comment |
real symmetric matrix has real eigenvalues - elementary proof
But Lagrange multipliers is, in my opinion, different from the argument above, which in fact was originally designed to deal with bounded operators, as explained in the comment. Can Lagrange multiplier be used to prove that $\pm\|T\|$ is an approximate eigenvalue of a bounded operator T? If not, is this enough to conclude that the proofs are different? |
Jan 13 |
comment |
real symmetric matrix has real eigenvalues - elementary proof
I don't understand Alexander's answer. How do you prove that if $R(x)=\frac{x^tAx}{x^tx}$ is maximum, then $x$ is an eigenvector? I got nowhere by derivating $R$, and the only easy way that I see to complete his proof is to normalize $x$ to get a maximum of $x^tAx$ in the unit sphere, and then write the Lagrange multipliers equation that tells you that $x$ is an eigenvector. |
Jan 13 |
awarded | Nice Answer |
Jan 13 |
comment |
real symmetric matrix has real eigenvalues - elementary proof
I meant $\frac 12\sum a_{ij}^2$. |
Jan 13 |
revised |
real symmetric matrix has real eigenvalues - elementary proof
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