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Mathematical Physics Major at the University of Waterloo.


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".
Aug
8
answered Relationship between Archimedean spiral and Theorem of Pythagoras
Aug
8
asked Uniform perturbative solutions to the Mathieu equation
Aug
7
awarded  Self-Learner
Aug
7
comment Rearrangement inequality for multiple sequences.
A sum involving sequences of length $n+1$ has $n+1$ total terms in the sum. Of the $n+1$ terms, we must have a two term sum of the form $$a_1^1\cdot a_{i_2}^2 \cdot a_{i_3}^3 \cdots a_{i_{n+1}}^{n+1} + a_{j_1}^1\cdot a_1^2\cdot a_{j_3}^3 \cdots a_{j_{n+1}}^{n+1}$$ If we regard $(a_1^1,a_{j_1}^1),\ (a_1^2, a_{i_1}^2)$ and $(a_{i_k}^k,a_{j_k}^k)$ for $3 \le k \le n+1$ as our two term sequences, then our base case applies.
Aug
7
comment Rearrangement inequality for multiple sequences.
The idea is that if the maximal elements are not together, say without loss of generality that $a_1^1$ and $a_1^2$ were separated, then we can apply the base case to the two terms involving $a_1^1$ and $a_1^2$ individually.
Aug
7
comment Rearrangement inequality for multiple sequences.
@Mark Yea, I probably should have written the answer with non-decreasing sequences to be consistent. I forgot why I changed it. Probably a lapse in judgement. Sorry for the confusion.