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comment Is it useful to learn math by proving a formula/theorem?
Hi @St.ClairBij, regarding self-study of mathematics, I think there are numerous questions and answers already on this website on the topic. You might be interested in click me for example. A complete list of questions with the tag self-learning is available at click me. Finally, there are numerous questions on this topic listed under "Related" on the right-hand sidebar.
1d
comment Is it useful to learn math by proving a formula/theorem?
I voted to close this question because I think it's too broad and primarily opinion based. Do you have a particular branch of mathematics in mind? I don't think most branches of mathematics admit a list of theorems which are universally considered to be fundamental, especially more advanced or recent branches of mathematics. I also don't know if it is meaningful to break up mathematics into isolated formulas/theorems; mathematics is a story and how you tell that story matters.
1d
comment real valued functions with composition
Hi Joel, I deleted my first comment because I saw you were editing, so I thought it might not be relevant anymore. Anyway, +1 from me. :) @MalJA the key point is that $G$ does not have any identity, as the only possible identity would have to be the identity function, which does not always assume nonnegative values.
2d
comment Prove that $L=K[x]/\langle m(x)\rangle$ is an algebraic extension of $K$
Hints: $x\in L$ is algebraic over $K$ and generates $L$ over $K$. $\{1,x,\dots,x^{n-1}\}$ is a basis for $L$ as a $K$-vector space. (Actually, the latter (finite dimensionality of $L$ over $K$) implies that $L$ is an algebraic extension of $K$.)
May
25
awarded  Nice Answer
May
21
answered Winding maps of spheres?
May
15
comment Which Lie groups are also symmetric spaces?
Hi @Qiaochu, I'm sure you know this already, but you can always find a bi-invariant metric on a compact Lie group: simply pick an inner product at the identity which is Ad-invariant ("Ad" denoting the adjoint representation of the Lie group), and extend it in the obvious left-invariant manner to a Riemannian metric on all of $G$.
May
5
comment Prove or disprove any continous map $f$ from $T^2$ to $RP^2$ is null-homotopic.
Hint: Can you answer this question if $\mathbb{T}^2$ is replaced by the unit circle $\mathbb{S}^1$?
May
5
comment Prove or disprove any continous map $f$ from $T^2$ to $RP^2$ is null-homotopic.
You can't necessarily lift a map from $\mathbb{T}^2\to \mathbb{RP}^2$ to map from $\mathbb{T}^2$ to the universal cover ($\mathbb{S}^2$) of $\mathbb{RP}^2$ since $\mathbb{T}^2$ isn't simply connected ...
May
5
answered Topological manifolds (dimension)
May
4
awarded  Yearling
Apr
29
comment Degree of minimum polynomial at most n without Cayley-Hamilton?
Dear @MikeF, thank you for the correction! Yes, I meant to say "characteristic polynomial". (Although it is true that you can read off the minimal polynomial too, since it will be a product of some of the $(x-a_i)$'s.)
Apr
29
comment Degree of minimum polynomial at most n without Cayley-Hamilton?
@egreg I guess this depends on the specific statement of the Cayley-Hamilton theorem to which you are referring. I guess one formulation is that the characteristic polynomial of an operator $T$ annihilates $T$, and another (weaker) statement is that there is some polynomial of degree $\leq n$ (if $T$ is an operator on $n$-space) which annihilates $T$. The first statement does not follow from the second, so this isn't necessarily reproving the Cayley-Hamilton theorem.
Apr
29
comment Degree of minimum polynomial at most n without Cayley-Hamilton?
You could put it in rational canonical form (works over any field). Or, if you are working over the complex numbers, then you can read off the minimal polynomial if you put it in triangular form: if $a_1,\dots,a_n$ are the diagonal entries, then the minimal polynomial is simply $(x-a_1)\cdot\dots\cdot (x-a_n)$.)
Apr
29
comment Is this true: $\sum_{n=1}^\infty a_n = \sum_{n=2}^\infty a_{n-1}$ ??
@picaposo You could state that the two series are equivalent in the following sense: "one series coverges if and only if the other series does, and in this case, they both converge to the same value".
Apr
21
comment determine the closures of the set k={1/n| n is a positive integer}
Hi @Vikrant, if you are referring to the cofinite topology on $\mathbb{R}$, then yes, the closure of $k$ is $\mathbb{R}$.
Apr
20
answered What intuition do we have for a subalgebra of Lie to be abelian?
Apr
18
comment The number of choices of 3 kinds of crust and up to 6 distinct toppings
Do you know how many subsets (including the empty set) there are of a set with $n$ elements? (Hint: if you are stuck, think about this for small values of $n$ and try to identify a pattern; in order to write a formal proof, use combinations.) You can finish the problem by referring to Simon's comment.
Apr
5
revised Convolution of matrix coefficients is also a matrix coefficients
added 4 characters in body
Mar
23
answered Clarification on notation of “left invariant fields” (Lie groups)