All Questions
7 questions
22
votes
3
answers
3k
views
Splitting the determinant polynomial into linear factors - a Dedekind problem
Here's the question in a nutshell. For some $n\in\mathbb N$, we consider the polynomial
$\det\left(\left(X_{i,j}\right) _ {1\leq i\leq n,\ 1\leq j\leq n}\right)\in\mathbb Z\left[X_{i,j}\mid 1\leq i\...
11
votes
2
answers
558
views
Classification of algebras of finite global dimension via determinants of certain 0-1-matrices
I restrict to the elementary problem that is equivalent to give a classification when Morita-Nakayama algebras have finite global dimension (see the end of this post for some background).
A Morita-...
7
votes
1
answer
462
views
On a problem for determinants associated to Cartan matrices of certain algebras
This is a continuation of Classification of algebras of finite global dimension via determinants of certain 0-1-matrices but this time with a concrete conjecture and using the simplification suggested ...
6
votes
0
answers
375
views
Monomial base change and the Vandermonde
Denote the falling factorials by $(x)_k=x(x-1)\cdots(x-k+1)$.
The Vandermonde determinant is given by $\det\left[x_i^{j-1}\right]_1^n=\prod_{i<j}(x_j-x_i)$.
It is well-known that in as much as ...
5
votes
0
answers
620
views
Is there a method to simultaneously block-diagonalize a set of group matrices?
Assume that you are explicitly given the representation matrices of a group.
How does one go about finding that common basis which will find the irreducible components of all of them simultaneously?
...
4
votes
1
answer
2k
views
Determinant and symmetric power
Let $V$ be a vector space over some field $k$ and $T \in \mathrm{GL}(V)$. Then, we can view $T\in \mathrm{GL}(\mathrm{Sym}^k(V))$ where $\mathrm{Sym}^k(V)$ denotes the $k^\mathrm{th}$ symmetric power ...
4
votes
1
answer
237
views
A (bi)alternant formula for Wronskian
We know that there exists similarities between power functions and derivative of a function (in particular, Newton binomial formula and Leibniz rule for derivation of a product can be deduced from ...