44,458 reputation
23191
bio website www-math.univ-poitiers.fr/…
location Poitiers, France
age 54
visits member for 2 years, 8 months
seen 19 hours ago

professor of mathematics at the Université de Poitiers (France)


2d
revised How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$
added 413 characters in body
2d
revised How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$
deleted 1 character in body
Jul
23
answered Suppose that $V_1$ and $V_2$ are subsets of a vector space…
Jul
21
comment Making 400k random choices from 400k samples seems to always end up with 63% distinct choices, why?
It does not follow rigorously from the fact that any particular element has a probability $\exp(-1)$ of not being marked that the fraction of marked elements is $1-\exp(-1)$, since the probabilities considered are not independent (as extreme case, if no other element than $x$ is marked, then it is certain that $x$ is marked). I'm not saying the conclusion is wrong (the dependencies are probably quite weak, and the experiment seems to confirm it) but the argument is not quite complete.
Jul
14
revised When are minimal and characteristic polynomials the same?
make a bit more complete
Jul
6
revised If $x$ is a rational number, then $1/x$ is a rational number
added 3 characters in body
Jul
5
answered Proving linear independence of matrices
Jul
4
comment To minimize $x^TAx$ where $A$ is not necessarily positive semi-definite with constrains?
Does your final formula simplify to $2x^4$? If not, what should one make of the parentheses?
Jul
4
awarded  Tag Editor
Jul
4
revised combinations wiki excerpt
added 34 characters in body; added 8 characters in body
Jul
4
revised counting the number of solution
edited tags
Jul
4
comment If $K = \frac{2}{1}\times \frac{4}{3}\times \cdots \times \frac{100}{99}.$ Then value of $\lfloor K \rfloor$
You might want to add to the question whether you specifically want a proof that the calculated value is correct. This would require, if using floating point calculations or any (other) approximation method, a rigorous error estimate placing the value of $K$ between two specific integers. Currently only one of the answers does that, more or less.
Jul
3
comment Why do determinants have their particular form?
You must qualify "multilinear" to either "mutlilinear by rows" or "multilinear by columns", because these are very different notions. It happens that you can use either of them to characterise the determinant, but that is not a justification for not specifying which one you are using.
Jul
3
comment Why do determinants have their particular form?
What's an intuitive way to think about the determinant? gives many answers about intuition for determinants
Jul
3
comment Determinant of a non-square matrix
@NikolajK: Well, it is important for the answer that $\det(BA)\neq0$, so the entries do matter a bit. However the first example that came to my mind (honestly!) is taking $A$ to be the $n\times0$ matrix and $B$ the $0\times n$ matrix, for some $n>0$; then $AB$ is a $n\times n$ zero matrix so $\det(AB)=0$, while $BA$ is the $0\times0$ (identity) matrix, so $\det(BA)=1$. In this example there aren't even any entries of $A$ or $B$ to worry about.
Jul
3
comment Determinant of a non-square matrix
Also what you say in this answer is not correct, since the product of all nonzero singular values fails to have property$~$2. Being nonzero by definition, the product does not for square matrices detect the condition of being singular, which is by far the most basic application of determinants. And even if the matrix is non-singular, the product of the (nonzero) singular values only gives the absolute value of the determinant.
Jul
2
comment points with integer coordinates inside triangles in $\Bbb{R}^3$
No, both the pair $(v,w)$ has been changed and the basis in $\Bbb Z^3$ has changed, so you cannot say that $v' = [ 1,0,0]$ etc.; indeed that vector is not in the lattice spanned by $v,w$. Every row operation is a basis change, and every column operation changes the pair $v,w$. There are only few column operations, so $v'=2v+w$ and $w'=7v+3w$ are easy to see. To find the adapted basis, apply all the row operations to an initially identity $3\times3$ matrix; the result is a matrix that computes the coordinates of vectors on the new basis. For the basis itself, take the columns of its inverse.
Jul
2
comment Determinant of a non-square matrix
Without condition 3, one would not be asking for anything very spectacular. Just define the function to have any constant value for all non-square matrices, and of course the usual value for square matrices, and it satisfies 1, 2, 4.
Jul
2
awarded  Curious
Jul
2
comment Is there a 'conjugation' on every algebraically closed field?
While it was shown at mentioned other question that every automorphism of a subfield extends to an automorphism of the algebraically closed field, that extended automorphism may have infinite order while the original automorphism has finite order. I would hesitate to call an automorphism a "conjugation" if it does not have order$~2$. For instance over $\Bbb F_3$ the polynomial $x^2+1$ is irrededucible, and there is a (Frobenius) automorphism of order$~2$ of $\Bbb F_9$ that interchanges its roots, but the algebraic closure of $\Bbb F_3$ does not have any automorphism of order$~2$.