# Linear independence of tensor product basis $\{ v_i \otimes w_j\}$ for $\{v_i\}$ and $\{w_j\}$ linearly independent.

Show that the set $$\{v_i \otimes w_j\}$$ is a linear independent subset of $$V\otimes W$$ when $$\{v_i\}$$ and $$\{w_j\}$$ are independent subsets of V and W respectively.

I want to find an error in a proof of this statement.

I've found great proofs of this in the literature, for example these notes and this book.

They all have the same core of the proof: take a linear form $$\varphi_i: V\to \mathbb{F}$$ such that $$\varphi_i(v_j)=\delta_{ij}$$, which exists thanks to independence of $$\{v_i\}$$. Then we have a linear function $$\phi_i : V\otimes W \to W$$ given by $$\phi_i(v\otimes w)= \varphi_i(v)w$$ which is as close to projection $$V\otimes W \to W$$ as one can hope to get. We can transport any linear relation from $$V\otimes W$$ to $$W$$ and use linear independence of $$W$$.

Details are in the links above.

Okay, so here is the argument I am skeptical of:

$$\DeclareMathOperator{span}{span}$$We can assume that $$\{v_i\}$$ and $$\{w_j\}$$ are bases, instead we can consider $$V'=\operatorname{span}\{v_i\}$$ and $$W'=\operatorname{span}\{w_j\}$$. Let $$F=F(\{v_i\}\times\{w_j\})$$ be a free vector space with a basis $$\{(v_i,w_j)\}$$. We define a bilinear function $$\Phi:V\times W \to F$$ by $$\Phi(v,w) = \Phi\left(\sum_i a_iv_i,\sum_j b_jw_j\right) = \sum_{i,j} a_ib_j(v_i,w_j)$$ As $$\{v_i\}$$ and $$\{w_j\}$$ are bases, therefore the coefficients $$a_i,b_j$$ are unique, so this function is well defined. By the universal property, there is a linear function $$\varphi: V\otimes W \to F$$ such that $$\varphi(v_i\otimes w_j)=\Phi(v_i,w_j)=(v_i,w_j)$$. We see that it is an isomorphism, because we can define $$\phi^{-1}$$ on a basis: $$\phi^{-1}((v_i,w_j))=v_i\otimes w_j$$ which is clearly a linear inverse. Therefore $$F$$ is isomorphic to $$V\otimes W$$, so if $$\{(v_i,w_j)\}$$ is a basis for $$F$$, $$\phi[\{(v_i,w_j)\}] = \{v_i\otimes w_j\}$$ must also be a basis.

It uses no projection or a variant thereof, and proves the statement almost tautologically. I've tried to find the error, but I couldn't. I have a really hard time accepting this proof, because I haven't seen it anywhere (and there are plenty of places) and it really feels logically less demanding as there are no linear functionals used, the isomorphism $$F\simeq V\otimes W$$ is constructed directly.

Actually, there are a few faulty proofs similar to the above that were already corrected on this site, namely here and here (one of the answers).

If the proof above is correct, can you direct me to the the sources or, ideally, explain why this proof don't appear in the sources I linked to?

• His claim of $(v_i, w_j)$ being basis is not true. The free vector space shall be the infinite dimensional space with every single pair of $(v,w)$ being its basis. Aug 27, 2023 at 9:14
• It's the free vector space spanned on the $(v_i,w_i)$ so it is true, it's not $F(V)$.
– mz71
Aug 29, 2023 at 8:23
• Hmm I mean $V \otimes W$ is usually defined via the quotient of $F(V \times W)$ and also the universal property referred were usually this one. Maybe this proof (i don't know its validity) skips the essential bit of verifying the universal property for $V\times W$, which is different from $F(V\times W)$ and thus makes it appear to be much more easier? Aug 30, 2023 at 2:49