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Mathematical Physics Major at the University of Waterloo.


Oct
8
comment Triangle inequality for subtraction?
@SPRajagopal The only property we used in the proof was the triangle inequality itself, so this holds with any norm.
Sep
30
awarded  Explainer
Sep
24
awarded  Autobiographer
Sep
21
comment number of vertices a special graph
Try relating the number of vertices, the degrees and the number of edges using the Handshaking lemma.
Sep
16
comment Solving the recursion $y_n=2ny_{n-1}$ with Wolfram|Alpha
Just as a note in case you haven't noticed. Wolfram Alpha generally has a description of all special functions and notations at the bottom right corner of the section. When I tried this equation in Wolfram Alpha there was a description available saying that "$(a)_n$ is the Pochhammer symbol", complete with additional documentation.
Sep
10
comment Is it proper to write $\int \partial x$
Normally the explanation of why a notation isn't seen is simply that it's unnecessary. Notation is only introduced if it simplifies or clarifies calculations and it could be just that this notation isn't all that useful. Under what context would you consider using this notation?
Aug
25
comment Eigenspace and polynomials?
@Hobbit6094 I am not assuming that $A$ is diagonalizable. You can think of this as restricting the diagoanlization procedure solely to one of the eigenspaces, i.e. the matrix $P$ is not the full matrix of eigenvectors but rather the matrix with only the first few columns as eigenvectors (belonging to a single eigenspace). The resulting matrix $P^{-1}AP$ is therefore not fully diagonalized, but partially, with the number of diagonal entries depending on the number of eigenvectors you've included in $P$, i.e. the geometric multiplicity.
Aug
24
comment Using a corollary of Tutte's theorem to prove a statement
@Jack It's difficult to say whether finding bipartite graphs will always be a good idea, but when faced with problems involving matches and coverings it's worth a shot. This is simply because we have so many powerful theorems to deal with bipartite graphs (i.e. Hall's, König's) that if we manage to reduce successfully to a bipartite graph, that's often a big step towards the solution. For your latter question, the procedure is exactly as you described; you join all vertices incident to the component to the collapsed vertex.
Aug
23
comment Using a corollary of Tutte's theorem to prove a statement
@Jack This assumption is hidden within Tutte's theorem. Notice that this condition has to hold for any set of $W$. In particular, this must hold for when $W$ is the empty set of vertices. If the order of the graph is odd, then we have an odd component, which is larger than the size of the empty set $W$.
Aug
23
revised Using a corollary of Tutte's theorem to prove a statement
added 37 characters in body
Aug
23
comment Using a corollary of Tutte's theorem to prove a statement
@Jack Yes you are absolutely right. I should add the condition that the internal edges of $W$ should be removed. That will still preserve the condition $\deg(w_i) \le n$ which is what's required for the proof.
Aug
23
answered Using a corollary of Tutte's theorem to prove a statement
Aug
18
comment Let $J$ be a $k \times k$ jordan block, prove that any matrix which commutes with $J$ is a polynomial in $J$
Another easy way to prove the result (well depending on how much theory you know) is the fact that each Jordan block necessarily has a cyclic vector since the minimal and characteristic polynomials for a Jordan block are equal. This in turn implies (via the cyclic vector theorem) that each matrix which commutes with the $J$ is a polynomial in $J$.
Aug
18
comment Adding a constant to a matrix
The constant terms in the polynomial is normally translated into a multiple of the identity matrix. In this case, you would add $I$ to your sum. If you need to add some other constant, say $c$, then you would add the corresponding multiple of the identity, $cI$.
Aug
18
answered A question about a corollary of the Spectral Theorem.
Aug
11
comment A real matrix whith rows generating $U$ and columns generating $V$
@KoenraadvanDuin There are two immediate ways to see this. First you note that the kernel is invariant under $X\mapsto AX$. The rowspace is the orthogonal complement of the kernel, hence it is also left invariant. Alternatively, if $A$ is invertible then it can be written as a product of elementary matrices, so the overall effect of $A$ is to perform a sequence of elementary row operations on $X$, which does not change the rowspace.
Aug
11
answered A real matrix whith rows generating $U$ and columns generating $V$
Aug
11
comment Showing exitence of a path in Graph Theory
What makes you think they exist?
Aug
11
comment Showing exitence of a path in Graph Theory
What is $u_{n+1}$ and $v_{n+1}$?
Aug
9
comment Proof Checking - Hamiltonian Decomposition/Graph Theory
It's a fine proof. There's no need to expound on the definition of vertex set, although it might be clearer to say "each cycle in the decomposition contains all the vertices of $G$" instead of "each decomposition contains the same vertex set".