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 Oct6 comment Are there any decompositions of a symmetric matrix that allow for the inversion of any submatrix? Assuming your matrix is real matrix (all elements are real), you are guaranteed by spectral theorem that your matrix $A = QDQ^{T}$ where $D$ is diagonal and $Q$ is orthogonal. Getting this $Q$ is algorithmically Gram-Schimdt process (so rather simple). Sep21 comment A basic question on relative singular groups\homology Example $H_{1} \subset G_{1}$ and $H_{2} \subset G_{2}$ are two groups and their subgroups and $\phi : G_{1} \rightarrow G_{2}$ is any homomorphism taking $H_{1}$ into $H_{2}$ then we get a well defined homomorphism (assuming$H_{1}$ and $H_{2}$ are both normal; which is true in abelian case) of the groups $G_{1}/H_{1} \rightarrow G_{2}/H_{2}$. Sep21 comment A basic question on relative singular groups\homology Yes you are right the boundary map descends to the quotient nicely for any subspace $A$ but the associated homology theory will not be nice unless some extra conditions on $A$ are fulfilled. Sep20 comment Methods of Multilinear Algebra in Representation Theory The study of Lie algebras over $\mathbb{C}$ becomes easy once you have understood $\mathfrak{sl}_{2}$ and it might be a great idea to work out symmetric and tensor products of representation in this case by hand. Fulton Harris does that and if I remember correctly they have a beautiful (and very verbose) discussion on that topic. Sep20 comment Methods of Multilinear Algebra in Representation Theory Thanks! for the link Sep18 comment Segre embedding This map is a closed embedding into $\mathbb{P}^{rs+r+s}$ which means that the image is a closed algebraic subscheme (or rather subvariety). To show this show that in any affine patch of $\mathbb{P}^{rs+r+s}$ the points of the Segre embedding correspond to an ideal. Writing co-ordinates for $\mathbb{P}^{rs+r+s}$ as $k[X_{ij}]$ where $0\leq i\leq r$ and $0 \leq j \leq s$ we see that the image of the Segre embedding correspond to the ideal generated by all $2 \times 2$ minors of the matrix $(X_{ij})$. This shows that the image is closed. Sep17 comment Is it possible to practice mental math too often? You might be hitting the so called "ok plateau". Sep16 comment Proving something by copying down axioms and changing variable names? You need to assert that an axiom is true in your case of this set of even functions. Then check that your assertion is true. Sep16 comment (Symmetric) group acting on a graph Here is an attempt: Let $f: X \rightarrow Y$ be a surjective map of graphs (vertices surject onto vertices and so do edges; basically it is a contraction of certain edges and vertices)then we can talk about the group Aut(X/Y) which are graph automorphisms of $X$ commuting with the map $f$. In your case group $P$ is the group $Aut(\tilde{X}/(X/G))$. Well here $\tilde{X}=$ universal cover of $X$ and $X/G$ the quotient graph of $X$ by $G$. Sep15 comment homeomorphism question relating to the topological 3-sphere @MarianoSuárez-Alvarez, I recall when I was first learning this stuff, this question stumped me for some time. Where as when you look at any sphere (after learning some basic general topology) the idea of breaking it up into two hem-spheres seems much more natural. Sep15 comment Is the set of integer coefficient polynomials countable? @Thomas Andrews, Yes I meant the polynomial algebra not a single polynomial per se. Sep14 comment Is the set of integer coefficient polynomials countable? An integral polynomial gives a function from $\mathbb{Z} \rightarrow \mathbb{Z}$. A power series doesn't. A polynomial is a finitely generated algebra over the co-efficient ring a power series is not. Sep14 comment Set of Bounded linear Operators on $l_2$ is dense on the set of bounded operators on $l_2$? A hlibert space is still a vector space...what do you mean by a operator which is not linear? Sep14 comment homeomorphism question relating to the topological 3-sphere You are talking about the usual construction of $S^{3}$ the one asked is bit unusual. In general there are many ways to triangulate a surface (and here break up a solid into polyherdral parts: essentially a CW complex). If you keep track of the smaller parts and their intersections you can decompose the solid into more manageable parts. The question is showing the sphere is two solid tori. Sep13 comment The matrix exponential: Any good books? For finite dimensional matrices you are essentially talking about exponential maps on lie groups. So look up any elementary nook on lie algebras and lie groups (may be it is worthwhile to consider compact lie groups first). Off hand I can think of M.Artin's Algebra, Bump's Lie Groups. Sep13 comment What branch of mathematics is most needed in the industry or how one can make living with mathematics (apart from teaching)? I think you need to define relevant job more clearly. If you wish to maximize your chances at a financial institution studying economics or perhaps mathematical economics is more useful. Studying mathematics for that purpose is not the most efficient route. Sep13 comment Proving that there are infinitely many primes with remainder of 2 when divided by 3 Use Dirichlet's theorem on primes in Arithmetic progressions. Well it is an overkill but you don't have to do anything.en.wikipedia.org/wiki/Primes_in_arithmetic_progression Sep13 comment I've heard that some kinds of mathematics are so abstract, what's the point in it? I do not understand if you mean abstraction of programing languages in a concrete technical sense or a rather loose figurative sense. But, mathematics is no more abstract to a mathematician than playing a (difficult) piece of music to a pianist. Of course to an outside r it looks strange and esoteric. Sep13 comment Connectedness of centralizer exercise Look at any discussion on Spec of a commutative ring (like exercise to the fist chapter of Atiyah Macdonald). Sep13 comment Connectedness of centralizer exercise As you computation shows, the points lying in the centralizer of t in Sp(4) is an algebraic variety over $K$ (same in the Gl4 case but slightly more complicated). The relevant topology topology is the Zariski topology. More concretely, look at the ring $K[x_{1}. \ldots, x_{8}]$ and quotient it by the ideal $(x_{1}x_{7} -x_{5}x_{3}, x_{1}x_{8} + x_{2}x_{7} - x_{3}x_{6} -X_5}x_{4})$ and show that the resulting quotient has idempotents.