Necessary and sufficient condition for equality of two tensor products

Question

Let $v_1, v_2 \in V; w_1, w_2 \in W$. What is necessary and sufficient condition for equality $v_1 \otimes w_1 = v_2 \otimes w_2$ in $V \otimes W$?

Answer 1

You have the following cases:

1. If $w_1 = 0$ then $v_1 \otimes w_1 = v_2 \otimes w_2$ iff $v_2 = 0$ or $w_2 = 0$.
2. If $w_1 \neq 0$ and $w_2 = c w_1$ for some $c \in \mathbb{F}$ then $v_1 \otimes w_1 = v_2 \otimes w_2$ iff $cv_2 = v_1$.
3. If $w_1,w_2$ are linearly independent then $v_1 \otimes w_1 = v_2 \otimes w_2$ iff $v_1 = v_2 = 0$

To see why, remember that if $(e_i)$ is a basis of $V$ and $(f_j)$ is a basis of $W$ then $(e_i \otimes f_j)$ is a basis of $V \otimes W$ and use this to analyze each case.

Answer 2

The other answer has given the standard "linear-algebra" perspective. Another way to answer this question (more common in functional analysis and abstract algebra) is via the universal property that defines a tensor product.

The following are equivalent:

1. $v_1 \otimes w_1 = v_2 \otimes w_2$
2. For every $f \in (V \otimes W)^*$, $f(v_1 \otimes w_1) = f(v_2 \otimes w_2)$
3. For every bilinear form $f: V \times W \to \Bbb F$, $f(v_1,w_1) = f(v_2,w_2)$

From the above, "3" tends to be particularly useful.

For instance: take $v_1 = w_2 = e_1$ and $v_2 = w_1 = e_2$, where $e_1,e_2$ denote the standard basis vectors in $\Bbb R^2 = V = W$. We can define the bilinear form $f:V \times W \to \Bbb R$ by $$ f(x_1e_1 + x_2e_2,y_1e_1 + y_2e_2) = x_1y_2 $$ We see that $f(v_1,w_1) = 1$, whereas $f(v_2,w_2) = 0$. Thus, $v_1 \otimes w_1$ and $v_2 \otimes w_2$ are distinct elements in $V \otimes W$.