# Tensors The outer product of two $(1,0)$ [[Tensor|tensors]] is the $(2,0)$ [[Tensor]] given by: $(\pmb{V} \otimes \pmb{W})(\tilde{p}, \tilde{q}) \equiv \pmb{V}(\tilde{q})\pmb{W}(\tilde{p})$ This generalizes to two $(N, N')$ and $(M, M')$ tensors giving an $(N + M, N'+ M')$ [[Tensor]]. The outer product of tensors is also known as the direct product or the [[Tensor]] product. Note that a general $(2,0)$ [[Tensor]] cannot be written as an outer product of two $(1,0)$ tensors since it the outer product has only $(n+n = 2n)$ [[components#^8b9d84|components]] while a general $(2,0)$ [[Tensor]] has $n^2$ [[components]], where $n$ is the [[dimension]] of the space. Moreover, observe that we can get the [[components]] of $\pmb{V} \otimes \tilde{\omega}$ as: $ (\pmb{V} \otimes \tilde{\omega})(\tilde{\omega}^i; \pmb{e}_j) = \pmb{V}(\tilde{p}) \tilde{\omega}(\pmb{e}_j)= V^i \omega_j $ Thus the [[components]] of $\pmb{V} \otimes \tilde{\omega}$ are given by the set $\{V^i \omega_j\}$, $i = 1, \ldots N$ and $j = 1, \ldots N'$. Moreover, note that the set of all $(N, N')$ [[Tensor|tensors]] at a point $P$ forms a [[vector space]] as seen by analogy to the [[One-Form#^918ab6|linearity of one-forms]]. The [[vector space]] is called $\bigotimes_{i=1}^{N} T^*_P \otimes \bigotimes_{j=1}^{N'} T_P$ and has a [[basis]] $\{\bigotimes_{k=1}^N \tilde{\omega}^{i_k} \otimes \bigotimes_{k=1}^{N'} \pmb{e}_{j_k}\}$ for $i = 1 \ldots N$ and $j = 1 \ldots N'$