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I'm trying to get my head round tensor products of vector spaces (I'm happy to see arguments in a more general setting, though).

I am concerned principally with two statements:

i) If $U,V,W$ are vector spaces then there is a one-to-one correspondence $\{ \mathrm{linear \ maps} \ V \otimes W \to U \} \longleftrightarrow \{ \mathrm{bilinear \ maps} \ V\times W \to U \} $.

ii) There is a natural (basis-independent) isomorphism $ (U \oplus V) \otimes W \to (U \otimes W) \oplus (V \otimes W)$

For the first of these statements, I can see map from left to right; any linear map $\phi : V \otimes W \to U$ gives rise to a bilinear map $ V \times W \to V \otimes W \to U$, where the first of these maps is the canonical map $p: (v,w) \mapsto v \otimes w$ and the second is $\phi$. I can't see, however, why any bilinear map $V \times W \to U$ necessarily factors into $\phi \circ p$ for some suitable linear map $\phi$.

I haven't got much experience with commutative diagrams. I think I've convinced myself that ii) is true with a commutative diagram, but I don't know if it's correct (and I also don't know how to LaTeX it easily...)

Any help would be appreciated. Thanks!

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i) is usually part of the definition of a tensor product. What definition are you working with? –  Qiaochu Yuan Mar 22 '12 at 19:00
@QIaochu: it sounds like Matt has a construction of tensor products anbd is trying to prove it has the correct properties. –  Mariano Suárez-Alvarez Mar 22 '12 at 19:05
@QiaochuYuan The definition I have is that if $V,W$ are vector spaces (with bases $v_1, \ldots , v_m $ and $w_1, \ldots w_n $), then the tensor product of $V$ and $W$ is the vector space with basis $ \{ v_i \otimes w_j \ | \ 1 \leq i \leq m, 1 \leq j \leq n \} $. My notes then go on to define the tensor product of $v \in V$, $w \in W $ to be $v \otimes w = \sum \lambda_i v_i \otimes \sum \mu_j w_j = \sum_{i,j} \lambda_i \mu_j (v_i \otimes w_j)$ –  Matt Mar 22 '12 at 19:33
This definition does seem strange to me, since it doesn't come with any intuition (it just seems like formal sums of symbols, which I don't like) –  Matt Mar 22 '12 at 19:35
Matt: your definition of tensor products of vector spaces sucks, because it depends on a choice of basis and to prove a basis-independent feature of $V \otimes W$ (like its relation to direct sums) you'd have to work through what happens if you change bases. See up through Theorem 5.3. –  KCd Mar 22 '12 at 20:46

2 Answers 2

up vote 1 down vote accepted

In your definition (you should really look up the universal property of the tensorproduct!) you can argue as follows:

If $b$ is a bilinear map $V\times W \rightarrow U$, you can simply define a linear map $$ v_i \otimes w_j \mapsto b(v_i,w_j) $$ since you know the $v_i\otimes w_j$ are a basis and a linear map can be defined by choosing arbitrary values for the elements of a basis.

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The approach to tensor products of vector spaces is this: think of them abstractly but you should have concrete examples in mind.

The first problem arises from the fact that if $V$ and $W$ are vector spaces (abstract or concrete) then $V \otimes W$ is another vector space that can be constructed in more than one ways and, depending on the situation, one way may better than another.

I suggest you focus on $V=W =$ the set of real-valued functions on a set $S$. Suppose that $S$ is finite even, in which case $V$ is isomorphic to the $|S|$-dimensional Euclidean space. But think of $V$ as a space of functions. If $x \in V$, let $x_i$ be the value of the function $x$ at the point $i\in S$. Now think of functions of two variables, i.e., of functions on $S \times S$. What is the simplest way to create a function of two variables? Well, take two functions $x, y$ of one variable (i.e., $x, y \in V$) and form the function $$ (i,j) \mapsto x_i y_j. $$ Give a name to this function: $x \otimes y$. So, $$ (x \otimes y)(i,j) := x_i y_j. $$ The set of these functions (call them "elementary") is not a linear space. Taking linear combinations of them you create new functions of two variables. Define then $$ V \otimes V := \text{ all finite linear combinations of elementary functions.} $$

Theorem (you can prove it): If $S$ is finite then $V \otimes V$ is the set of ALL functions of $2$ variables.

This is what the tensor product in a concrete case is. You can repeat the argument with $n$ copies of $V$, tit for tat.

The abstract case is just an embodiment of this concrete idea.

If we could think of an abstract vector space as a space of functions then we can repeat the concrete process and construct tensor products in the same way. Can we think of an abstract vector space as a space of functions? Yes. By thinking of an element $v \in V$ as a function taking a linear functional $f: V \to \mathbb R$ into the number $f(v)$. Using this idea you can concretely construct $V \otimes V$ and $V \otimes V \otimes V$, etc, and even $V \otimes U \otimes W$, etc.

But it's not advisable to always think of this concrete way of constructing tensor products for abstract spaces. Why not? To understand, you need experience.

So, then, we have an abstract way of constructing the tensor product space $T= V \otimes V$, and it is this:

Definition. $T$ is SOME vector space with the property that there is some bilinear map $\otimes : V \times V \to T$ such that, for any vector space $W$ and for any bilinear map $\phi: V \times V \to W$ there is a LINEAR map $\overline \phi: V \otimes V \to W$ with $$ \phi(v_1, v_2) = \overline\phi(v_1\otimes v_2). $$


1) Show that such a $T$ exists. (We've done it.)

2) Show that if $T, T'$ are two tensor product spaces then there is a linear bijection $A: T \to T'$ and a linear bijection $B: T' \to T$. Hence the two spaces are ``the same''. Hence, no matter how we define $T$, we essentially get the same thing.

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