# Equal tensor product implies equal basis element

Let $\beta$ and $\gamma$ be bases for finite dimensional vector spaces U and V respectively. Prove that if $u_1, u_2 \in \beta$ and $v_1, v_2 \in \gamma$ and $u_1 \otimes v_1 = u_2 \otimes v_2$, then $u_1 = v_1$ and $v_1 = v_2$. Is it true if $u_1, u_2 \in U$ and $v_1, v_2 \in V$

Proof

Because $u_1$ and $v_1$ are bases element of U and V it follows that $u_1 \otimes v_1$ is a basis element of $U \otimes V$. Similarly $u_2 \otimes v_2$ is a basis element of $U \otimes V$. Since $u_1 \otimes v_1 = u_2 \otimes v_2$, they are the same basis element, therefore $u_1 = v_1$ and $v_1 = v_2$

If $u_1, u_2 \in U$ and $v_1, v_2 \in V$, then this need not be true. Given any nonzero scalar k, $(ku_1) \otimes v_1 = u_1 \otimes (kv_2)$ but obviously $(ku_1) \neq u_1$ and $v_1 \neq kv_1$.

Is my argument correct? It looks so simple that I feel like I'm overlooking something.

• Your argument is not correct. In the first line, what do you mean that "it follows that $u_1\otimes v_1$ is a basis element of $U\otimes V$"? Any nonzero vector could be a basis element of $U\otimes V$". – ChocolateAndCheese Dec 7 '15 at 3:43
• What I have in my note is that if $\beta = \{u_1,...,u_m\}$ and $\gamma = \{v_1,...,v_n\}$ are bases for U and V respectively, then $\{u_i \otimes v_j | i = 1,...,m$ and $j = 1,...,n\}$ is a basis for $U \otimes V$ – user295188 Dec 7 '15 at 3:49
• Sure, but that theorem is saying that there exists a specific basis of $U\otimes V$. In general, any nonzero vector in a vector space can be a "basis vector". This is a subtle but important distinction. The first line of your proof should state that you have chosen bases $\{u_i\}$ and $\{v_j\}$ of $U$ and $V$, respectively. Then by a theorem $\{u_i\otimes v_j\}$ is a basis of $U\otimes V$. – ChocolateAndCheese Dec 7 '15 at 3:55

## 1 Answer

A good way to show that a tensor $t$ is nonzero in $U \otimes V$ is to show that there is a bilinear map $U \times V \to W$ that is nonzero on $t$. In this case, you'd like to show that $u_1 \otimes v_1 - u_2 \otimes v_2 \neq 0$. Try building a bilinear map (I suggest a map $U \times V \to F$, the underlying field) that is nonzero on this, assuming for example that $u_1 \neq u_2$. Use that it is friendly to build a bilinear form on a finite dimensional vector space when you're given a basis.