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In general the tensor product of a right module with a left module is an abelian group and thus - via a forgetful functor - an abelian monoid. Please look at your other related question and my other related answer So , if you say that a tensor is an element of a tensor product, then it is an element of its related abelian monoid. However in category theory ...

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Everything is an element of a vector space, tensor or not. Let $x$ be anything, and let $F$ be any field. The simplest example of an $F$-vector space containing $x$ is the vector space in which $x$ is the only element. That is: The set vectors is $\{ x \}$ Addition is defined by $x + x = x$ Scalar multiplication is defined by $r \cdot x = x$ A more ...

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I think tensors are by definition elements of the tensor product of vector spaces. Any vector space $V$ over $K$ is naturally isomorphic to $V\otimes K$, so yes, you can say that any vector in a vector space is tautologically a tensor.

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Hint Let us abbreviate $\dot{x^k}$ for $\frac{d x^k}{ds}$. Now we must manipulate $\frac{\partial L}{\partial{\dot{x^k}}}$ as if $\dot{x^k}$ were a simple variable. So, restricting to dimension two and instead of multi-indexing we use: $$L=\sqrt{g_{11}(\dot{x^1})^2+2g_{12}\dot{x^1}\dot{x^2}+g_{22}(\dot{x^2})^2}.$$ Or better: ...

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Hints: Prove that a cofactor $C_{k}{}^{\ell}$ of an $n\times n$ matrix $A=(a^i{}_j)_{1\leq i,j\leq n}$ is given by the formula $$C_{k}{}^{\ell}~=~ \frac{1}{(n-1)!}\sum_{1\leq i_2,\ldots, i_n,j_2,\ldots, j_n\leq n}\varepsilon_{ki_2\ldots i_n}~\varepsilon^{\ell j_2\ldots j_n}\prod_{r=2}^n a^{i_r}{}_{j_r},$$ where $\varepsilon_{i_1\ldots i_n}$ denotes the ...

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This worried me at one time as well. The way I thought about it was by working at a fixed point and using the Gram-Schmidt process for inner products on the coordinate basis $\partial_1,...,\partial_n$ to produce an orthonormal basis $e_1,...,e_n$. It's a standard and easy fact that the matrices that represent these bilinear forms are related by $I=A^tgA$, ...

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Based on the problem as you have stated we have a matrix: A that can be described as $$A(x,y,z)$$ where we input x, y, z and get out a matrix A. From here it becomes clear that changing variables is independent of the matrix definition: in other words $$A(x,y,z) = A(r \sin(\phi_1) \cos(\phi_2 ) , r \sin(\phi_1)\sin(\phi_2), r \cos(\phi_1))$$ ...

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Lost 1 is correct what you wrote is totally absurd....it is utter nonsense. Recall : $$\epsilon_{ijk} = -\epsilon_{ikj} = -\epsilon_{jki} = -\epsilon_{kji}.$$ We realize that product vanishes, i.e. it becomes zero....which I must say it's wrong for we don't have enough info to justify that. So we rewrite the product as follows. $$C_x = (\vec{A} \times ... 3 There are different solutions; however, I like this. Consider the following exact sequence:$$\mathbb Z\xrightarrow{f} \mathbb Z\xrightarrow{g}\mathbb Z_n\xrightarrow{}0$$where g(a)=\overline{a} and f(a)=na. By tensor theorems, the sequence$$\mathbb Z\otimes \mathbb Z_n\xrightarrow{f\otimes id} \mathbb Z\otimes \mathbb Z_n \xrightarrow{g\otimes ...

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When you take $A^i B_j$, without any repeated indices, then, indeed, you're forming a $(1,1)$-tensor $C^i_{\;j} = A^i B_j$ with matrix $$\begin{pmatrix} 2\\3\\4\end{pmatrix}\begin{pmatrix}1&2&3\end{pmatrix} = \begin{pmatrix}2&4&6\\3&6&9\\4&8&12\end{pmatrix}.$$ When you repeat an index, so that it appears both as a ...

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At the lowest level of understanding a tensor $T$ of rank $r$ is an $r$-dimensional array (think of a spreadsheet) whose "side-lengths" are all equal to a given $n\geq1$. Therefore $T$ has $n^r$ entries, which we assume to be real numbers in the following. When we are setting up such a tensor we have some application in mind, say in geometry or physics. ...

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