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 Oct 7 awarded Nice Answer Sep 5 revised Golden Ratio Approximation Correct typo in title. Sep 5 suggested approved edit on Golden Ratio Approximation Sep 4 comment Why do we use a Least Squares fit? Was this really four years ago? And apparently I'm still checking math.stackexchange with some regularity :) Aug 20 awarded Popular Question Aug 9 answered Solve an equation with complex numbers Jul 27 awarded Popular Question Jun 16 answered Differential Equations Lectures or books from a theoretical perspective? May 23 awarded Critic May 21 comment Uniqueness of Rectifying Coordinates: Question for Arnold's ODE Book @Evgeny Yes, that was my initial intuition also, that perhaps the basis vectors were stretched. But I can't get the maths to work that way :) May 19 asked Uniqueness of Rectifying Coordinates: Question for Arnold's ODE Book May 6 awarded Yearling May 5 awarded Necromancer Apr 13 awarded Nice Question Apr 9 answered analytical ability and logical reasoning Apr 7 comment Tensor product in multilinear algebra This looks something like $(U\times V)^* \simeq U^* \otimes V^*$. Is this some alternative formulation of the tensor product? Or maybe it should be $\text{Hom}(U,V^*) \simeq U^* \otimes V^*$, because $Functions(X\times Y)$ is referring to the space of bilinear forms... Apr 7 comment Tensor product in multilinear algebra @paulgarrett Hi, I have one more quick question :) It makes sense to me that the tensor product is a space whose dual is isomorphic to the space of bilinear forms (using some previous notation $(U\otimes V)^* \simeq \text{Hom}(U,V^*)$), and the link between elements of $U\times V$ and $U\otimes V$ is defined by a natural bilinear form $b$. But I've also seen that we should want our tensor product to satisfy a different property, something like $Functions(X\times Y) = Functions(X)\otimes Functions(Y)$. See the 2nd paragraph here: math.harvard.edu/archive/25b_spring_05/tensor.pdf Apr 6 comment Tensor product in multilinear algebra @paulgarrett OK, thanks, this makes so much more sense now! I really like your explanation because it makes the link between the space of all bilinear forms over $U$ and $V$ and the space of all linear functionals on $U\otimes V$ extremely explicit. In other places I've heard that these two spaces should be "isomorphic", but at least in the finite-dimensional case (I don't know beyond that :) ) to be isomorphic they just need to be the same dimension, so this isn't saying much. But by your definition you must also exhibit the 'link' $b$ between the two spaces. You've made my day :) Apr 6 comment Tensor product in multilinear algebra @paulgarrett Sorry about the $\oplus$ - I meant $\otimes$ actually - $B$ is a linear functional on $V\otimes W$, or more generally a linear map $V\otimes W \rightarrow X$. I'm still a little unsure now about how that composition works. If $\beta$ maps from $V\times W \rightarrow X$, how can we compose it with $B$ which maps from $V\otimes W \rightarrow X$? Wouldn't we need $B: X \rightarrow V\otimes W$ so that $b(x,y) = B(\beta(x,y)) \in V\otimes W$? Apr 6 comment Tensor product in multilinear algebra @paulgarret However, you then go on to say that the link between $B$ and $\beta$ is that $B$ is the unique linear functional on $V\oplus W$ with the property that '$b = B \otimes \beta$'. I'm totally lost here. You've used $\otimes$ to signify a tensor product between two vector spaces, but $B$ and $\beta$ are linear and bilinear functionals respectively. I just don't know how to interpret what you mean by $B\otimes \beta$. Thanks