Hot answers tagged

4

Write $\omega = \alpha_1 + \dots + \alpha_n$ where $\alpha_i = dx_{2i - 1} \wedge dx_{2i}$. The important observation is that if you expand $$ \omega^n = (\alpha_1 + \dots + \alpha_n) \wedge \dots \wedge (\alpha_1 + \dots + \alpha_n) = \sum_{i_1, \dots, i_n} \alpha_{i_1} \wedge \dots \wedge \alpha_{i_n} = \sum_{I} \alpha_I$$ then if $I$ contains a repeated ...


4

For $R=\mathbb Z/4\mathbb Z$ the permanent has all the properties you list, and it differs from the determinant when $n\ge 2$. (For any matrix, the difference between its permanent and its determinant is a multiple of $2$, so in $\mathbb Z/4\mathbb Z$ the permanent is a unit iff the determinant is). In general, we can set $R=\mathbb Z/m\mathbb Z$ ...


2

So your bilinear form is non-degenerate if there does not exist a vector $v \neq 0$ such that $B(v,w) = 0$ for any vector $w \in V$. In other words, there doesn't exist a vector that sends all $V$ to zero. Consider a bilinear form $B$ and matrix $A$. If $A$ is non-invertible, then the kernel of $A$ is non-trivial. Suppose $k$ is a vector in the kernel of $...


1

I'll try to give an answer to this question that doesn't require knowing what a tensor is. Of course it will require some comfort with linear algebra, but I'll even try to keep the multilinear algebra to a minimum. Bye_World's answer in the comments is a good one, although there is slightly more that you can say: a matrix is naturally identified with a rank-...


1

Your list includes some subjects, some algebraic structures, and some objects within algebraic structures. I'll separate them to organize the list a little better. Anyone can feel free to contribute to/ edit this list as I'm certainly not an expert in all of this. Subjects: Linear Algebra: the study of vector spaces and the linear transformations ...


1

According to Wikipedia and some papers: Exterior algebra = Grassmann algebra (= differential forms, since they are a construction of the exterior algebra) (=derivations, since derivations are just one possible construction of the dual object to differential forms). EDIT: To make the claim that "exterior algebra=differential forms" precise, since as it ...


1

$\Rightarrow)$ To see that $rank(A)=n$, we have to proof that the columns of $A$ are linearly independant. So we notice that the $j$-th column is $(\alpha_{ij})_i$. If we have $\lambda_1,\dots,\lambda_n\in F$ such that $\sum_{j=1}^n\lambda_j(\alpha_{ij})_i=0$, then $(\sum_{j=1}^n\lambda_j\alpha_{ij})_i=0$, which means that for each $i=1,\dots,n$, $\sum_{j=...


1

Suppose $U$ is a vector space over a field of characteristic different from $2$. If you have any linear map $\alpha$ on $U$ such that $\alpha^{2} = I$, then either $\alpha = \pm I$, or $x^{2} - 1$ is the minimal polynomial of $\alpha$. In the latter case the minimal polynomial has two distinct roots $1$ and $-1$, thus $U$ decomposes as the direct sum of ...



Only top voted, non community-wiki answers of a minimum length are eligible