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3

In fact $a\otimes b\mapsto a\otimes b$ is an isomorphism $A\otimes_DB\leftrightarrow A\otimes_FB$ for any spaces (or algebras) $A$ and $B$ over a field $F$ which is the fraction field of a domain $D$. The reason is that $$\begin{array}{ll} \displaystyle \color{Blue}{\frac{1}{y}}a\otimes b & \displaystyle =\frac{1}{y}a\otimes \color{Green}{y}\frac{1}{y}b ...


2

I don't have a complete answer, but I think I can add some information. In the spirit of you last comment, the way toweards characterizing the cone of simple tensors is looking at maximal linear subspaces contained in it. Each line in $V$ determins a linear subspace of dimension $dim(W)$ contained in the cone and vice versa. These two families are exactly ...


2

It is $${\rm Alt}(\phi_1\otimes\phi_2\otimes\phi_3)= \phi_1\otimes\phi_2\otimes\phi_3-\phi_1\otimes\phi_3\otimes\phi_2+\phi_2\otimes\phi_3\otimes\phi_1-\phi_2\otimes\phi_1\otimes\phi_3+\phi_3\otimes\phi_1\otimes\phi_2-\phi_3\otimes\phi_2\otimes\phi_1$$ but some other will define $${\rm Alt}(\phi_1\otimes\phi_2\otimes\phi_3)=\frac{1}{6}( ...


1

You'd write it as $$\begin{align} \operatorname{Alt}(\phi_1\otimes\phi_2\otimes\phi_3) = \frac{1}{6}(&\phi_1\otimes\phi_2\otimes\phi_3 + \phi_2\otimes\phi_3\otimes\phi_1 + \phi_3\otimes\phi_1\otimes\phi_2\\ &- \phi_1\otimes\phi_3\otimes\phi_2 - \phi_2\otimes\phi_1\otimes\phi_3 - \phi_3\otimes\phi_2\otimes\phi_1) \end{align}$$ More generally, for ...


1

For symmetric or skew-symmetric forms $(u,v)=0 \implies (v,u) = 0$. But you are right, the statement does not need any of them, just decide what $\perp$ means. The nice thing is that if non-degeneracy holds for one map, it holds for the flip one ( since, if a matrix is nonsingular, then its transpose also is). Here is a quick proof to have in your toolbox ( ...


1

Note that $(6,9)=3(2,3)=3T(1,1)\not=T(3,3)=(1,4)$


1

You have been asked to find $a$, $b$ and $c$ such that (in a least square sense) $$\left[ {\begin{array}{*{20}{c}}5\\1\\1\\1\end{array}} \right] = a\left[ {\begin{array}{*{20}{c}}1\\1\\1\\1\end{array}} \right] + b\left[ {\begin{array}{*{20}{c}}1\\{ - 1}\\{ - 1}\\1\end{array}} \right] + c\left[ {\begin{array}{*{20}{c}}1\\1\\{ - 1}\\{ - 1}\end{array}} ...


1

Clifford algebra can make this manipulation simpler. To use it here, first define the geometric product of vectors. This is traditionally denoted by juxtaposition: i.e. the geometric product of a vector $a$ and $b$ is denoted $ab$. Given vectors $a, b, c$, the geometric product has the following properties: $$aa \equiv \langle a,a\rangle \implies ab = ...



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