Can a rank three tensor act as a trilinear, bilinear, and linear map? Similarly, a matrix (a representation of a rank two tensor) can be bilinear, taking in two vectors and spitting out a scalar for instance, but it can also be a linear map on just one vector. What's going on here?
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asked Aug 20, 2018 at 19:13
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3 Answers
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Let's simplify this a bit and take a $(1,1)$ tensor over a real vector space $V$. We might take this, by definition, to be a bilinear map $V\times V^* \to \mathbb R.$ So we can think of it as taking a vector and a covector argument and looking like $T(v,v^*) = a.$
However, we can also think of it as a linear map $V\to V,$ or as a linear map $V^*\to V^*.$ To see how this works, take $v^*\in V^*$ fixed, and consider $T(v,v^*)$ as a function of $v.$ This is a linear map that takes a vector $v$ and produces a real number. In other words, it is a covector. So we can also define the tensor as the map which takes a covector $v^*$ and returns the covector $T(\_,v^*)$ we just just discussed. This is a way of viewing it as a linear map $V^*\to V^*.$ Similarly we can view it as $v\mapsto T(v,\_)$ a map from $V\to V.$ (Recall that a linear map that takes a covector and produces a vector is just a vector).
This reasoning works out on higher ranks as well. An $(n,m)$ tensor can be looked at a multilinear map $V^m\times (V^*)^n \to \mathbb R,$ or we can think of it as a map that takes a vector (or another tensor) and contracts it with the tensor to produce a lower-rank tensor. For instance if you give an $(n,m)$ tensor a vector, it will act linearly on that vector and return an $(n,m-1)$ tensor. In other words, it's equally well viewed as a linear map $$ V\to V^{\otimes (m-1)}\otimes V^{\otimes n}$$
answered Aug 20, 2018 at 19:39
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$\begingroup$ So just to check, all n-linear maps are also m-linear to m variables, when m is between 0 and n. Multilinear maps are a type of linear map on a variable. $\endgroup$ Aug 20, 2018 at 20:50
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$\begingroup$ @BenjaminThoburn Yes. I think the other two answers addressed this explicitly. If you fix some of the arguments in a multilinear map, the map defined this way is multilinear in the remaining arguments $\endgroup$ Aug 20, 2018 at 20:53
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Let $v,w,$ be fixed vectors, and $x$ be arbitrary.
Then $T_1(x)=x\otimes v\otimes w,$ $T_2(x)=v\otimes x\otimes w$ and $T_3=v\otimes w\otimes x$ are all linear maps.
To see why, let $x_1$ and $x_2$ be arbitrary vectors, with $\alpha$ any scalar, then
$$T_3(\alpha x_1 +x_2)=v\otimes w\otimes (\alpha x_1+x_2)=v\otimes w\otimes (\alpha x_1)+v\otimes w\otimes x_2=\underbrace{\alpha(v\otimes w\otimes x_1)+v\otimes w\otimes x_2}_{=\alpha T_3(x_1)+T_3(x_2)}.$$
The other two maps have identical arguments to show linearity.
Similarly, fixing one input, say $v\otimes x_1\otimes x_2$ defines a bilinear map.
All of this is a result of the how tensors are defined. That is, they must satisfy $x\otimes (v+w)=x\otimes v +x\otimes w$, and they're balanced over the scalar field
$$\alpha(v\otimes w)= (\alpha v)\otimes w=v\otimes (\alpha w).$$
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A multilinear map is linear in all its arguments. If you make some of them constant, you obtain a multilinear map of lower order, which can be described by a tensor of lower order.