Ytukyg

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 $text{min}{m,n}$.

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
$$
v_r otimes w_r = left(sum_{i=1}^{r-1} lambda_i v_iright) otimes w_i=sum_{i=1}^{r-1} left( v_i otimes lambda_i w_r right)
$$
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 $dim W geq r$.

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.