Tensor Product with Trivial Vector Space

Notice that the space of bilinear maps $f:V\times \{0\}\to k$ consists of exactly the zero map, therefore the constant map $w:V\times \{0\}\to\{0\}$ satisfies the universal property of the tensor product: $w$ is bilinear and any bilinear map $f:V\times\{0\}\to k$ is the constant zero map, therefore the linear map $0:\{0\}\to k$ satisfies $f=0\circ w$. $0$ is also the only linear map $\{0\}\to k$, so it's a fortiori the only linear map $g$ such that $f=g\circ w$.

Recall that the tensor product $V\otimes W$ of two finite-dimensional vector spaces $V$ and $W$ satisfy the dimension formula $\dim(V\otimes W)=\dim(V)\cdot\dim(W)$.

So, tensoring a finite-dimensional vector space $V$ with the trivial vector space $0$ yields a vector space $V\otimes 0$ with dimension $$ \dim(V\otimes 0)=\dim(V)\cdot\dim(0)=\dim(V)\cdot 0 = 0 $$ This implies that $V\otimes0$ is itself the trivial vector space $V\otimes 0=0$.