If V and W are vector spaces over a common field with bases $V_B = ${$v_i : i \in I$} and $W_B = ${$w_j : j \in J$}, then is {$v_i \otimes w_j: i \in I, j \in J$} a basis for $V \otimes W$ ?

I have several references for the finite dimensional case but they seem to involve either a dimensional argument or reference to dual spaces, neither of which would seem to be appropriate in the infinite dimensional case.

The references that I can find for infinite dimensional spaces focus on "orthonormal bases" (usually in the context of Hilbert Spaces).

So, I'd appreciate any help answering

  1. is this actually true if {$v_i : i \in I$} and {$w_j : j \in J$} are the Hamel (algebraic) bases of infinite dimensional spaces and what is the proof ?
  2. Does it make any difference if only one space is infinite dimensional ?
  3. Is there any difference between countable and larger infinities ?

Addendum 15 March 2015:

Why not simply define the tensor product as $F(V_B \times W_B)$, i.e. the free vector space generated by the set of ordered pairs of basis vectors ? There seems to be a bilinear mapping from $V \times W$ and it appears to satisfy the universal property.

Have I missed something here ?

  • $\begingroup$ As a side remark:In infinite dimensions algebraic bases do not span the vector space (there are elements that are not finite linear combinations of the basis elements). So one can consider algebraic tensor products (which is what you probably have in mind) and also completed tensor product (denoted by $\widehat{\otimes}$ in both cases and algebraic basis is given by $v_{i} \otimes w_{j}$. $\endgroup$
    – DBS
    Commented Mar 12, 2015 at 11:33
  • 1
    $\begingroup$ @DBS. Thanks for your interest, but I beg to differ. I think that in an infinite dimensional inner-product space, the orthonormal basis aligns with what you are saying. But in any space, the algebraic basis does by definition span the space. $\endgroup$ Commented Mar 12, 2015 at 12:18
  • $\begingroup$ The example I had in mind is take $V= L^{2}(S^{1})$, then the functions $e^{2 \pi inx}$ forms an orthogonal basis for $V$ but there are $L^{2}$ functions which do not have infinitely many non zero fourier coeff. So the algebraic span (which can only have finitely many non-zero basis vectors) is dense (in the natural topology) of $V$ but it doesn't span $V$. $\endgroup$
    – DBS
    Commented Mar 13, 2015 at 13:50
  • 2
    $\begingroup$ @DBS So, we agree. An orthonormal basis may not span an infinite dimensional space. But an orthonormal basis is not the same thing as an algebraic basis, and by definition an algebraic basis does span its space. Now, in some infinite vector spaces the algebraic basis may not be easy to find, or of much use if found, but it always exists (proof from Zorn's lemma). $\endgroup$ Commented Mar 13, 2015 at 14:20
  • $\begingroup$ Yes we agree. i don't know if there is an explicit description of an algebraic basis in the above example though. I would be interested if there is one. $\endgroup$
    – DBS
    Commented Mar 13, 2015 at 14:51

2 Answers 2


Yes it is true independent of the cardinality of the bases of the vector spaces. Use the universal property: Let $T$ be any vector space over the given field.

A bilinear map $V \times W \to T$ is uniquely determined by the images of the pairs $(v_i,w_j)_{i \in I,j \in J}$, so we get

$$Bil(V \times W, T) = Abb(\{(v_i,w_j), i \in I, j \in J\},T)= Abb(\{v_i \otimes w_j, i \in I, j \in J\},T)$$

By the universal property $V \otimes W$ is the vector-space with the property

$$Hom(V \otimes W,T) = Bil(V \times W, T)$$

, so we obtain

$$Hom(V \otimes W,T) = Abb(\{v_i \otimes w_j, i \in I, j \in J\},T)$$

, which precisely states that $\{v_i \otimes w_j, i \in I, j \in J\}$ is a basis of $V \otimes W$.

Let us precise this:

Given a map in $f \in Abb(\{v_i \otimes w_j, i \in I, j \in J\},T)$, which means that we are given $f(v_i \otimes w_j)$ for all $i \in I, j \in J$, we have to show that this extends uniquely to a map $F:V \otimes W \to T$.

We define the bilinear map $\hat f: V \times W \to T$ by $\hat f(v_i,w_j) := f(v_i \otimes w_j)$. By the universal property, this gives us a unique map $F:V \otimes W \to T, F(v \otimes w)=\hat f(v_i,w_j)=f(v_i \otimes w_j)$. Hence this is the desired unique map.

The uniqueness of $F$ is trivial anyway, because $\{v_i \otimes w_j, i \in I, j \in J\}$ is clearly a set of generators. We only need to show the existence of $F$ here (And this corresponds to the linear independence).

  • 2
    $\begingroup$ Thanks for the response. Unfortunately I don't know what Abb() is and I don't get anything relevant from Google - could you please enlighten me ? $\endgroup$ Commented Mar 6, 2015 at 12:27
  • 1
    $\begingroup$ This is just the set of all set maps between two sets. $\endgroup$
    – MooS
    Commented Mar 6, 2015 at 12:36
  • $\begingroup$ Thanks again. But now when I get to $Bil(V \times W, T) = Abb(\{(v_i,w_j), i \in I, j \in J\},T)= Abb(\{v_i \otimes w_j, i \in I, j \in J\},T)$ I can only see the stated equalities as bijections. So that gives me a bijection between $Hom(V \otimes W,T)$ and $Abb(\{v_i \otimes w_j, i \in I, j \in J\},T)$, is that sufficient to make $ \{v_i \otimes w_j, i \in I, j \in J\}$ a basis ? $\endgroup$ Commented Mar 6, 2015 at 14:01
  • $\begingroup$ The bijections are natural. Think about this while having the commutative diagram from the universal property in mind. $\endgroup$
    – MooS
    Commented Mar 9, 2015 at 8:23
  • $\begingroup$ To complete my understanding of this can you comment on "Why not simply define the tensor product as $F(V_B \times W_B)$ ?" This would seem to be a lot simpler than involving quotient spaces to show the existence of a tensor product. $\endgroup$ Commented Mar 15, 2015 at 13:59

I give a "conceptual" answer.

Let $k$ be a field, and $V, W$ vector spaces over $k$. Recall that $k$-vector space $V\otimes_k W$ has the following universal property : if we note $\pi : V\times W\to V\otimes_k W$ the canonical $k$-bilinear map $(v,w)\mapsto v\otimes w$ and $\textrm{Bil}_k (V,W;P)$ the set of $k$-bilinear applications $V\times W\to P$ for a $k$-vector space $P$, then we have : for each $k$-vector space $P$ the map $\textrm{Hom}_k (V\otimes_k W,P)\to \textrm{Bil}_k (V,W;P)$ definied by $f\mapsto f\circ\pi$ is a bijection.

As I think that any answer to your question should emphasize what is true in general, what is proper to the situation (vector spaces = modules over a field) and what things are independent. I will prove the following three general lemmas, with the following hypothesis : from now on, $k$ is not a field but a commutative ring with unit, and $V,W$ are modules over $k$.

Lemma 0. The elements of the form $v\otimes w$ for $v\in V$ and $w\in W$, called elementary tensors (or also simple tensors), generate $V\otimes_k W$ over $k$.

Proof. I give a proof of this lemma using only the universal property of $V\otimes_k W$ and not any of its constructions (as quotient of $A^{(W\times V)}$ for instance). Note $Q$ the sub-$k$-module of $V\otimes_k W$ generated by elementary tensors. Note $\pi' : V\times W\to Q$ the map defined by $(v,w)\mapsto v\otimes w \in Q$. I pretend that $Q$ and $\pi'$ satisfies the same universal property that $V\otimes_k W$ and $\pi$ do. Let $P$ be a $k$-module. The map $\textrm{Hom}_k (Q,P)\to \textrm{Bil}_k (V,W;P)$ defined by $f\mapsto f\circ\pi'$ is a injective. Indeed : if $f\circ \pi' = g\circ \pi'$ for $f,g\in \textrm{Hom}_k (Q,P)$ then it means that $f$ and $g$ are equal on elementary tensors, and as elementary tensor generate $Q$ by definition of $Q$, we have $f=g$. Now, let's show that this map is surjective, so take $b\in \textrm{Bil}_k (V,W;P)$. By the universal property of $V\otimes_k W$ we can find $f\in \textrm{Hom}_k (V\otimes_k W,P)$ such that $b = f\circ\pi$, which obviously imply that $b = f' \circ \pi'$, where the $k$-linear map $f' : Q\to P$ is the restriction of $f$ to the submodule $Q$ of $V\otimes_k W$, that is, the $k$-linear map $f' : Q\to P$ defined by $q\mapsto f(q)$ for $q\in Q$. This shows the surjectivity of the map $\textrm{Hom}_k (Q,P)\to \textrm{Bil}_k (V,W;P)$, which is finally injective and surjective, that is, bijective. This shows that $Q$ and $\pi'$ satisfies the same universal property that $V\otimes_k W$ and $\pi$ do. Thanks to this, we will now show that $Q = V\otimes_k W$. Note $i : Q \to V\otimes_k W$ the canonical injection : it is the map sending an element of $Q$ to the same element but viewed as element of $V\times_k W$. Rephrased, what we want to show is that $i$ is surjective. Now $\pi' : V\times W \to Q$ is a $k$-bilinear map, so that by the universal property of $V\otimes_k W$ and $\pi$, there exists a unique $k$-linear map $\varphi : V\otimes_k W \to $ such that $\varphi\circ \pi = \pi'$. But then $(i\circ\varphi)\circ\pi = i\circ(\varphi\circ\pi) = i\circ\pi' = \pi = \textrm{Id}_{V\otimes_k W}\circ\pi$, and by the universal property $i\circ\varphi = \textrm{Id}_{V\otimes_k W}$, showing that $i$ has a right inverse, and is therefore surjective. $\square$

Lemma 1. If $(v_i)_{i\in I}$ (resp. $(w_j)_{j\in J}$) is a familly of generating elements of $V$ (resp. of $W$) over $k$, then the familly $(v_i\otimes w_j)_{(i,j)\in I\times J}$ is a familly of of generating elements of $V\otimes_k W$ over $k$.

Proof. As $\pi$ is $k$-bilinear, it is clear that any elementary tensor will be linear combination of $v_i\otimes w_j$'s over $k$, and by the previous lemma, the elementary tensors generate $V\otimes_k W$, showing thereby that the elements $v_i\otimes w_j$'s generate V\otimes_k W$

From now on we suppose that $k$ is a field.

Lemma 2. If $(v_i)_{i\in I}$ (resp. $(w_j)_{j\in J}$) is a familly of independant elements of $V$ (resp. of $W$) over $k$, then the familly $(v_i\otimes w_j)_{(i,j)\in I\times J}$ is a familly of independant vector of $V\otimes_k W$ over $k$. $\square$

Proof. Suppose we have a dependance relation $$\sum_{(i,j)\in I\times J} c_{i,j} v_i\otimes w_j = 0$$ where $(c_{i,j})_{(i,j)\in I\times J}$ is a familly a elements in $k$ such that $c_{i,j}\not=0$ only for finitely many indexes $(i,j)$'s. Complete $(v_i)_{i\in I}$ (resp. $(w_j)_{j\in J}$) in a basis $(v_i)_{i\in \overline{I}}$ (resp. $(w_j)_{j\in \overline{J}}$) of $V$ resp. (of $W$) over $k$. Take $(i_0,j_0)\in I\times J$ and let's show that $c_{i_0,j_0} = 0$. For this, consider the bilinear function $V\times W\to k$ sending $(v,w)=\left( \sum_{i\in \overline{I}} \lambda_i v_i, \sum_{j\in \overline{J}} \mu_j w_j \right)$ to $\lambda_{i_0} \mu_{j_0}$. Take by the universal property $f : V\otimes_k W \to k$ to be the linear map associated to this bilinear form. Applying the linear function to the above dependance equation gives $c_{i_0,j_0} = 0$. $\square$

Lemma 2 and 3 applied to the case of a field do answer your question.

Remark. No cardinality hypothesis were made.

  • $\begingroup$ Thanks for the response: I haven't had chance to digest it all yet. $\endgroup$ Commented Mar 13, 2015 at 14:22
  • $\begingroup$ If you need details, don't hesitate $\endgroup$
    – Olórin
    Commented Mar 13, 2015 at 14:25
  • $\begingroup$ I just corrected some typos in my answer. $\endgroup$
    – Olórin
    Commented Mar 20, 2018 at 12:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .