Recent content by rmehta
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Graduate Dimension of a multivariate polynomial space
I think the issue is that there are actually more polynomials that "vanish" when you restrict to the unit sphere (FYI, mathematicians use the word "ball" to include the interior of the sphere, where \sum x^2 \leq 1). For example, if n=2 you know that x^2 + y^2 - 1 = 0 (on the unit sphere)...- rmehta
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- Forum: Linear and Abstract Algebra
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Undergrad General Solution of the first order differential equation
Here's an idea. It shouldn't be too hard to find particular solutions to dy/dt + y = sin(nt)/n^2, where n is an unspecified constant. Do that and then sum the solutions. You'll end up with a weird infinite series, but it should be convergent because of the n^2 in the denominator. And this is...- rmehta
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- Forum: Differential Equations
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Graduate How Does the Lemma Support the Uniqueness Theorem in ODEs?
I think the difference is that the theorem says the solution exists and is unique in some "sufficiently small interval". The lemma asserts that the uniqueness (but not the existence) is more robust. The lemma could be rephrased as: "if a solution exists on any interval, then it is the unique...- rmehta
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- Forum: Differential Equations
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Undergrad Finding eigenvectors of similar matrices
No problem. To give some intuition for this, you should think of an invertible matrix P as giving a "symmetry" of Rn. Then conjugation by P is the corresponding symmetry of the space of nxn matrices. That's really why people care about similar matrices. If two matrices are similar, then there is...- rmehta
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- Forum: Linear and Abstract Algebra
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Graduate Invariance of scalar products on Lie algebras
If the group is connected, then ad-invariance will automatically imply Ad-invariance. Basically, ad-invariance implies that this function f that you've defined is locally constant (since its differential will be 0). If G is connected, then locally constant implies constant. Hope this helps!- rmehta
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- Forum: Linear and Abstract Algebra
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Undergrad Finding eigenvectors of similar matrices
Just use Pv. Then PAP-1(Pv) = PAv = P \lambda v = \lambda Pv.- rmehta
- Post #2
- Forum: Linear and Abstract Algebra