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.

• 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 $L(a\vec v+b\vec{w})=aL(\vec{v})+bL(\vec w)$). Feb 26, 2018 at 10:48
• @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? Feb 26, 2018 at 11:34
• Sorry. What map?? Feb 26, 2018 at 11:41
• @Dog_69 this map: $$B: \mathbb{R} \times \mathbb{R} \to \mathbb{R} ; (v,w) \to B(v,w) := xy$$ Feb 26, 2018 at 11:45
• 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. Feb 26, 2018 at 11:51

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 $V\otimes W$.