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olivier dot begassat dot cours at gmail dot com
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2h |
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Proving that some power of a number gives 999…90…0 number This is impossible when $n$'s last digit is one. |
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3h |
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which of the followings are positive definite: What can you say about the nullspace of a positive definite matrix? Use this to solve 2 and 3. As for 4, $AB$ will not necessarily be symmetric, and you can try and look at the $2\times 2$ case to find counterexamples. |
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3h |
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which of the followings are positive definite: Maybe tell us your definition of positive definitenes and where you're stuck. It's probably not the best idea for us to solve the problem for you when it is most likely an immediate application of the definition. |
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17h |
accepted | Example 1.K in A User's Guide to Spectral Sequences |
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17h |
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Function spaces and transitive group actions @johndoe I made a mistake, I thought the action would be free, but it won't be in general! What I wrote works if we assume the action is free instead of merely faithful, sorry. |
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18h |
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Function spaces and transitive group actions And in this case the above behavior is ruled out. |
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18h |
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Function spaces and transitive group actions @johndoe if the action is faithful and transitive, then upon fixing a point $\mathrm{pt}\in F$, we get an identification $\psi:G\stackrel{\approx}{\rightarrow} F,~g\mapsto g\cdot\mathrm{pt}$. This is only a continuous bijection, not necessarily a homeomorphism, but in case it is a homeomorphism (for instance if $G$ is compact and $F$ is Hausdorff), any function $f:B\to F$ defines a continuous map $\tilde{f}:B\to G$ with $\tilde{f}=\psi^{-1}\circ f$ such that $f=\tilde{f}\cdot\mathrm{pt}$ where this time $\mathrm{pt}$ stands for the constant map $B\to F$ associated to $\mathrm{pt}\in F$. |
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19h |
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Conjugacy classes for su(2) What do you mean conjugacy classes in the Lie algebra su(2)? Do you mean under the action adjoint of $SU(2)$? |
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19h |
revised |
Find integer solutions of $x^2 -px +q=0$, where $p$ and $q$ are prime added 87 characters in body |
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19h |
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Find integer solutions of $x^2 -px +q=0$, where $p$ and $q$ are prime @CameronBuie wow, that's what I get for getting an answer out in a rush. |
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19h |
awarded | Mortarboard |
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19h |
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Find integer solutions of $x^2 -px +q=0$, where $p$ and $q$ are prime It seems I was wrong after all :D |
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19h |
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Find integer solutions of $x^2 -px +q=0$, where $p$ and $q$ are prime it happens. 15characters |
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20h |
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Find integer solutions of $x^2 -px +q=0$, where $p$ and $q$ are prime We posted exactly the same solution, even with the same notation, and almost the same wording ^^. |
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20h |
answered | Find integer solutions of $x^2 -px +q=0$, where $p$ and $q$ are prime |
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20h |
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Why is Lie derivative smooth? @MuratGüngör You can certainly put it that way. As you note, it essentially boils down to the fact that when a map is smooth, so are (by definition!) its partial derivatives. |
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20h |
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Dimension of the set of self-adjoint operators There are indeed $n(n-1)/2$ coefficients to choose, but you choose them $\underline{\text{in }\Bbb C}$! That means you have to choose twice as many real numbers (the real and imaginary parts). The set of all herminitan operators (or matrices) is not a complex subspace, but merely a real subspace! |
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20h |
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Function spaces and transitive group actions Do you want me to expand on this answer? Is it clear enough? |
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21h |
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Dimension of the set of self-adjoint operators added 210 characters in body |
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21h |
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Dimension of the set of self-adjoint operators @KevinLee self adjoint operators are diagonalisable, but not in any orthonormal basis. The basis we chose at the get go will certainly not diagonalize all hermitian operators. So there are more than just the diagonal matrices. Also every operator commutes with itself, self adjoint or not ^^ |
