tensor rank of an element in a tensor product

Question

Let $V$ and $W$ be finite-dimensional vector spaces over $k$ with $\text{dim}(V)=n$ and $\text{dim}(W)=m$.

How can I see that every element $t \in V \otimes_k W$ has tensor rank at most $\text{min}\{m,n\}$.

Answer 1

Suppose rank $t=r$ and write $$ t = \sum_{i=1}^r v_i \otimes w_i $$ where $w_i \in W$ and $v_i \in V$. I claim that $\{v_i\}_{i=1}^r$ and $\{w_i\}_{i=1}^r$ must be linearly independent sets.

Assume toward I contradiction that $\{v_i\}$ is not a linearly independent set. WLOG we can write $v_r=\sum_{i=1}^{r-1} \lambda_i v_i$. Observe that $$ v_r \otimes w_r = \left(\sum_{i=1}^{r-1} \lambda_i v_i\right) \otimes w_i=\sum_{i=1}^{r-1} \left( v_i \otimes \lambda_i w_r \right) $$ Then we have $$ t = v_r \otimes w_r +\sum_{i=1}^{r-1} v_i \otimes w_i= \sum_{i=1}^{r-1} v_i \otimes \lambda_i w_r+\sum_{i=1}^{r-1} v_i \otimes w_i = \sum_{i=1}^{r-1} v_i \otimes (w_i +\lambda_i w_r), $$ contradicting the fact that $t$ has rank $r$. So we must have $\{v_i\}_{i=1}^r$ is a linearly independent set. A near identical argument works for $\{w_i\}_{i=1}^r$. It follows that $\dim V \geq r$ and $\dim W \geq r$.

Answer 2

Fix a basis $v_1, \ldots, v_n$ of $V$. Then any tensor $t \in V \otimes W$ can be written $t = v_1 \otimes w_1 + \cdots + v_n \otimes w_n$ for some vectors $w_i$ depending on $t$. This shows the tensor rank is at most $n$, and the argument to show that it is at most $m$ is identical.