张量积的交集：如何用集合等价条件证明子空间张量积等式

Proving $(U_{1}\otimes V) \cap (U_{2}\otimes V) = (U_{1} \cap U_{2}) \otimes V$ via Mutual Inclusion #

To prove two sets are equal, we just need to show each is a subset of the other—this is exactly the criterion you mentioned: $A = B$ iff $A \subseteq B$ and $B \subseteq A$. Let's break this down step by step.

1. Left-to-Right Inclusion: $(U_{1} \cap U_{2}) \otimes V \subseteq (U_{1}\otimes V) \cap (U_{2}\otimes V)$ #

Take any element $x \in (U_{1} \cap U_{2}) \otimes V$. By definition of the tensor product, $x$ can be written as a finite sum:
$$x = \sum_{i=1}^k u_i \otimes v_i$$
where each $u_i \in U_1 \cap U_2$ and $v_i \in V$.

• Since every $u_i$ is in $U_1$, the sum $\sum u_i \otimes v_i$ clearly belongs to $U_1 \otimes V$.
• Similarly, every $u_i$ is also in $U_2$, so the same sum is an element of $U_2 \otimes V$.

By definition of set intersection, this means $x$ is in the overlap of $U_1 \otimes V$ and $U_2 \otimes V$. So the left-hand side is a subset of the right-hand side.

2. Right-to-Left Inclusion: $(U_{1}\otimes V) \cap (U_{2}\otimes V) \subseteq (U_{1} \cap U_{2}) \otimes V$ #

Now take any element $x \in (U_1 \otimes V) \cap (U_2 \otimes V)$. This means $x$ lives in both tensor products at once.

Let's use a basis argument here—tensor products have unique representations with respect to a basis, which makes this straightforward. Let ${v_j}_{j \in J}$ be a basis for $V$.

• Since $x \in U_1 \otimes V$, we can write $x$ uniquely as:
  $$x = \sum_{j \in J} u_{1j} \otimes v_j$$
  where $u_{1j} \in U_1$, and only finitely many $u_{1j}$ are non-zero (since tensor products are finite sums).
• Since $x \in U_2 \otimes V$, we can also write $x$ uniquely as:
  $$x = \sum_{j \in J} u_{2j} \otimes v_j$$
  where $u_{2j} \in U_2$, again with only finitely many non-zero terms.

By the uniqueness of tensor product representations relative to a basis of $V$, the coefficients for each $v_j$ must be identical: $u_{1j} = u_{2j}$ for all $j$. Let's call this common element $u_j = u_{1j} = u_{2j}$.

Since $u_j$ is in both $U_1$ and $U_2$, it's in their intersection $U_1 \cap U_2$. So we can rewrite $x$ as:
$$x = \sum_{j \in J} u_j \otimes v_j$$
which is exactly an element of $(U_1 \cap U_2) \otimes V$. This proves the right-hand side is a subset of the left-hand side.

Final Conclusion #

Since we've shown both directions of inclusion, we can apply the set equality criterion to confirm:
$$(U_{1}\otimes V) \cap (U_{2}\otimes V) = (U_{1} \cap U_{2}) \otimes V$$

内容的提问来源于stack exchange，提问作者Humberto Gimenes Macedo