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| bio | website | bruno.stonek.com |
|---|---|---|
| location | Montevideo, Uruguay | |
| age | 23 | |
| visits | member for | 2 years, 7 months |
| seen | yesterday | |
| stats | profile views | 717 |
Math student in Universidad de la República, Montevideo, Uruguay.
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May 17 |
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When quotient map is open? Another sufficient condition: that the map be a surjective submersion of smooth manifolds. |
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May 10 |
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Evaluate $\int_0^{\pi \over 3} \sec x\tan x\sqrt {\sec x + 2} \, dx $ using a substitution of your choice Please don't use \limits on the title of a question: it doesn't render prettily on the main page. Thanks. |
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May 10 |
revised |
Evaluate $\int_0^{\pi \over 3} \sec x\tan x\sqrt {\sec x + 2} \, dx $ using a substitution of your choice edited title |
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May 6 |
awarded | Caucus |
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Apr 29 |
awarded | Notable Question |
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Apr 21 |
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Why is the determinant equal to the index? For a self-contained proof of this theorem, see theorem 1.17 in Stewart & Tall's Algebraic Number Theory (3rd. ed). |
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Apr 17 |
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There is an orthogonal projection of a subspace onto a canonical subspace that is an isomorphism Oh, I see you wrote \not\perp but somehow I see it rendered as $\perp$. I'll reread your answer then. |
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Apr 16 |
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There is an orthogonal projection of a subspace onto a canonical subspace that is an isomorphism Ok, I finished grasping your answer, I like it, it's slick :) Thank you. |
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Apr 16 |
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There is an orthogonal projection of a subspace onto a canonical subspace that is an isomorphism I can certainly empathize with that :) But in this case $n$ will be the dimension of a manifold, and $k$ the dimension of the ambient space it sits in. |
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Apr 16 |
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There is an orthogonal projection of a subspace onto a canonical subspace that is an isomorphism I think you mean that the matrix has rank $n$, as $n\leq k$, so you select $n$ linearly independent rows... |
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Apr 16 |
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There is an orthogonal projection of a subspace onto a canonical subspace that is an isomorphism @MarcvanLeeuwen: yes, I mean isomorphism of plain vector spaces. I don't think the inner product is irrelevant, how do you make sense of the orthogonal projection if it isn't there? |
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Apr 16 |
revised |
There is an orthogonal projection of a subspace onto a canonical subspace that is an isomorphism added 77 characters in body |
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Apr 16 |
answered | Is vector subtraction commutative? |
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Apr 16 |
revised |
Is vector subtraction commutative? deleted 2 characters in body; edited title |
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Apr 16 |
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There is an orthogonal projection of a subspace onto a canonical subspace that is an isomorphism I think I don't understand your argument. Why can you choose $i_1$ such that $e_{i_1}\perp W$? The same example you gave on your comment to my answer is a counterexample. |
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Apr 16 |
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There is an orthogonal projection of a subspace onto a canonical subspace that is an isomorphism @Berci: shame on me, of course, you're right. |
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Apr 16 |
answered | Lie bracket of vector fields and differential of diffeomorphism in its definition |
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Apr 16 |
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There is an orthogonal projection of a subspace onto a canonical subspace that is an isomorphism Thank you for your answer! While riding back home on the bus today I found a similar argument, I believe, which I think is easier to justify by reductio ad absurdum. Would you care to look at it? Perhaps you can help me with the combinatorial argument :) |
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Apr 16 |
answered | There is an orthogonal projection of a subspace onto a canonical subspace that is an isomorphism |
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Apr 15 |
revised |
There is an orthogonal projection of a subspace onto a canonical subspace that is an isomorphism edited title |
