> [!definition] > > Let $V$ be a [[Vector Space|vector space]] over a [[Field|field]] $F$. A [[Multilinear Map|multilinear map]] $T: V^k \to F$ is called a $k$**-tensor** on $V$. > [!definition] > > Let $V$ be a vector space, then $\mathcal{J}^k(V)$ is the space of all $k$-tensors over $V$. $\mathcal{J}^k(V)$ is also a vector space over the same field. > [!definition] > > Let $V$ be a vector space, then the map > $ > \mathcal{J}^l(V) \times \mathcal{J}^k(V) \to \mathcal{J}^{k + l}(V) \quad (S, T) \mapsto S \otimes T > $ > where > $ > (S \otimes T)(v, w) = S(v) \cdot T(w) > $ > for any $v \in V^l$, $w \in V^k$, is the [[Tensor Product|tensor product]]. > [!theorem] > > Let $\alpha = \seqf{v_i}$ be a basis for $V$, and let $\seqf{\phi_i}$ be the [[Topological Dual|dual]] basis, then the set of all $k$-fold tensor products > $ > \bigotimes_{j = 1}^{k}\phi_{i_j} \quad \bracs{i_j}_1^k \text{ strictly increasing} > $ > is a basis for $\mathcal J^k(V)$, which therefore has dimension $n^k$. > > *Proof*. Let $x \in V^k$ and $A \in V^{n \times k}$ such that $x = A\alpha$ and $T \in \mathcal J^k(V)$, then by multilinearity > $ > \begin{align*} > Tx &= TA\alpha \\ > &= \sum_{j \in [n]^k} \prod_{i = 1}^{k}j_i \cdot T(v_{j_1}, \cdots, v_{j_k}) \\ > &= \sum_{j \in [n]^k} \bigotimes_{i = 1}^{k}\phi_{j_i}(x) \cdot T(v_{j_1}, \cdots, v_{j_k}) > \end{align*} > $ > we can write any $k$-tensor as a sum of the tensor products. > > Suppose that there exists numbers $a_{i_1, \cdots, i_k}$ such that > $ > \sum_{i \in [n]^k}a_i\bigotimes_{j = 1}^k\phi_{i_j} = 0 > $ > Applying both sides to $(v_{j_1}, \cdots, v_{j_k})$ yields $a_i = 0$ for all $i \in [n]^k$.