2
$\begingroup$

Consider that we know the intuitive notion about tensor products: a tool to bring multilinear algebra to linear algebra and also some notation about maps. Now, consider the following:

By Universal property we have that:

$$B(v,w) = L(v\otimes w)$$

So we can write:

$$B(v,w) = L(v\otimes w) \iff$$ $$\iff B(\sum v^i\vec{e}_i,\sum w^j\vec{e}_j) = L(\sum v^i\vec{e}_i\otimes \sum w^j\vec{e}_j )\iff $$

$$\iff \boxed{\sum_{i}\sum_{j} v^iw^jB(\vec{e}_i,\vec{e}_j) =\sum_{i}\sum_{j} v^iw^j L(\vec{e}_i\otimes \vec{e}_j )}$$

Well, cleary $B(\vec{e}_i,\vec{e}_j)$ is a bilinear function and I KNOW that the core of the intuitive notion lies on expression up above. I mean: we "change" $B(\vec{e}_i,\vec{e}_j)$ multilinear study for $L(\vec{e}_i\otimes \vec{e}_j )$ linear study. But,first of all, I don't "see" the linearity of $L(\vec{e}_i\otimes \vec{e}_j )$ (I mean, how can I write the definition of a linear map?) and I can't even think about how to start an "pratice-the-concept" exercise. I need easy examples to start to grasp. I'm not a math student, but I want to know all the rigor of Tensors.

$\endgroup$
10
  • $\begingroup$ As you have said, tensor product is a tool to bring multilinear algebra to linear algebra. With that in mind, I think it is clear $L$ must be supposed to be linear, don't you? (a map is linear if $L(a\vec v+b\vec{w})=aL(\vec{v})+bL(\vec w)$). $\endgroup$
    – Dog_69
    Feb 26, 2018 at 10:48
  • $\begingroup$ @Dog_69 I know that is ask too much, but could you please show me how can I study the bilinear map defined as $B(v,w) = x \cdot _{\mathbb{R}} y$ in this framework? $\endgroup$
    – M.N.Raia
    Feb 26, 2018 at 11:34
  • 1
    $\begingroup$ Sorry. What map?? $\endgroup$
    – Dog_69
    Feb 26, 2018 at 11:41
  • $\begingroup$ @Dog_69 this map: $$B: \mathbb{R} \times \mathbb{R} \to \mathbb{R} ; (v,w) \to B(v,w) := xy$$ $\endgroup$
    – M.N.Raia
    Feb 26, 2018 at 11:45
  • $\begingroup$ I think the $L$ associated to the scalar should be the trace operator. But I must confess I'm not very familiarized with this formalism. $\endgroup$
    – Dog_69
    Feb 26, 2018 at 11:51

1 Answer 1

1
$\begingroup$

Seeking to get intuition is very good. Slowly and calmly understand where you are coming from and where you're going to.

Goal: to show $L$ is a linear map.

The right statement is actually to show that there is always a linear map L that works. What you are asking for is a conceptual clarification of the proof of the universal property, so my answer will be mostly conceptual and verbose (actually, sometimes the universal property can be taken as a definition of tensor product, in which case you want a clarification of its statement and to show a construction to help with intuition). This verbose style is not the way one should write a final proof, but helps you can understand and come up with one (one usually cleans up all the mess after understanding, sadly also leaving insight out)

Solution:

1) Understand where the map $L$ is defined

$L$ is defined on $V \otimes W$. Conceptually, this is a vector space of pairs of vectors with certain rules for adding pairs and multiplying them by scalars. As you know we take $\alpha(v,w) = (\alpha v,w) = (v,\alpha w)$ etc, in some sense, and if these rules hold for pairs of vectors, then we mark pairs with this algebra as $v\otimes w$. If you use quotient vector spaces or whatever, this is just one possible construction (or "implementation", as a computer scientist would say) of $V\otimes W$. What matters is that such algebraic rules over pairs of vectors reflect the fact that $V\otimes W$ is some vector space, the codomain of a generic bilinear map $\otimes: V\times W \to V\otimes W$ (pairs without linear algebra “exterior” to them, to pairs endowed with some linear algebra)

2) If $L$ is judiciously construted from $B$, verify that it is linear from the properties of $B$ (which is what you are seeking for grasping the intuitive notion)

To do this, you have to find an $L$ such that $L(v\otimes w) \doteq B(v,w)$ and that $$L(\alpha t) = \alpha L(t),\text{ for all $t$ in $V \otimes W$, and all scalars $\alpha$}$$ $$L(t_1 + t_2) = L(t_1) + L(t_2),\text{ for all $t_1, t_2$ in $V \otimes W$}$$

But any $t$ in $V \otimes W$ looks like a linear combination of simple tensor products. So we can try first to prove rules with the simple tensors; for instance, we can try to show that we can always get $L$ such that:

$$L(\alpha v\otimes w) = \alpha L(v \otimes w)$$ and $$L(v_1\otimes w_1 + v_2\otimes w_2) = L(v_1\otimes w_1) + L(v_2\otimes w_2)$$ for all such vecors $v$, $w$, $v_i$, $w_i$, and scalar $\alpha$.

The scalar property you get for free from what you already know about $L$ in terms of $B$:

$$L(\alpha\,\,v\otimes w) = L((\alpha v) \otimes w) = B(\alpha v, w) = \alpha B(v,w) = \alpha L(v\otimes w).$$ (You could have put $\alpha$ inside with the $w$ vector just as well)

To get the sum property to work, you have to figure out how you could construct more rules on $L$ from $B$. You want

$$L(v_1\otimes w_1 + v_2\otimes w_2) = L(v_1\otimes w_1) + L(v_2\otimes w_2) = B(v_1,w_1) + B(v_2,w_2).$$

Then simply set (yes, you set it because you want it, then see it works):

$$L(v_1\otimes w_1 + v_2\otimes w_2) \doteq B(v_1,w_1) + B(v_2,w_2).$$

Now, we can sum it all up to show that such $L$ indeed a linear map like you wanted all along:

Let us show that the map $L$ given by the rules $$L(v\otimes w) \doteq B(v,w),$$ and $$L(v_1\otimes w_1 + v_2\otimes w_2) \doteq B(v_1,w_1) + B(v_2,w_2)$$ indeed exists / is well defined and is linear. In fact, since any element from $V\otimes W$, the domain of $L$, is a linear combination of simple vectors $v\otimes w$, we can use these rules to take $L$ of any element of $V\otimes W$, and it will satisfy the linearity conditions for all of $V\otimes W$.

$\endgroup$

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.