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Other that simplifying into $\bigwedge R=R\oplus R$, you are completely right. You can also go for a different approach, by defining the degree$-n$ terms as follows. Namely by $\bigwedge^nM=M^{\otimes n}/N$, where $N$ is the submodule generated by the dublicate terms. Then we have $\bigwedge M=\bigoplus_n \bigwedge^n M$. From this definition, it is ...

2

Fix a basis $\{e_1, \ldots, e_n\}$ of $V$, and consider the dual basis $\{f_1, \ldots, f_n \}$ of $V^\ast$. Then we have a basis $$\{e_1\otimes f_1,\ldots, e_i \otimes f_j, \ldots, e_n \otimes f_n\}$$ for $V \otimes V^\ast$, and the matrix $$A = (a_{ij})$$ is just a way of representing the element $$\sum_{i=1}^n \sum_{j=1}^n a_{ij} \; e_i \otimes f_j \in V \... 2 Supplementing Morgan's answer with the following. Assume that Q is a quadratic form on n\ge3 variables ranging over the field \Bbb{F}_p,p>2. By the result recalled by Morgan we know that Q is of the form$$ Q(v)=\lambda_1v_1^2+\lambda_2v_2^2+\lambda_3v_3^2+\cdots+\lambda_n v_n^2. $$I claim that for some vector (v_1,v_2,v_3)\neq(0,0,0) from \... 2 Such a map v\mapsto vAv^{T} defines a quadratic form on \mathbb{F}^{n}, and your desired property that for all v \neq 0, vAv^{T}\neq0 is that the quadratic form is anisotropic. Basically, we define map Q : \mathbb{F}^{n} \to \mathbb{F} by Q(v) = v A v^{T}. This map has the property that, if a \in \mathbb{F}, then Q(av) = a^{2}Q(v) (this is ... 1 You can characterise \vec{v} \otimes \vec{w} as the unique n \times n matrix such that$$ \forall \vec{x} \in \mathbb{R}^n, \quad (\vec{v} \otimes \vec{w})\vec{x} = \langle \vec{w}, \vec{x} \rangle \vec{v},  where $\langle \vec{w}, \vec{x} \rangle = \vec{w}^T\vec{x}$ denotes the usual inner product on $\mathbb{R}^n$. From this perspective, then, it ...

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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-...

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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 ...

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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 ...

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