1
$\begingroup$

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?

$\endgroup$

3 Answers 3

2
$\begingroup$

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 $$ V\to V^{\otimes (m-1)}\otimes V^{\otimes n}$$

$\endgroup$
2
  • $\begingroup$ 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
  • $\begingroup$ @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
2
$\begingroup$

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

$$\alpha(v\otimes w)= (\alpha v)\otimes w=v\otimes (\alpha w).$$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.