# Tensor product of linearly independent vector fields is also linearly independent?

For simplicity, let us think of $$3-$$dimensional torus $$\mathbb{T}^3$$, which is a compact $$3-$$dimensional smooth manifold without boundary.

Then, we can regard the space of smooth vector fields on $$\mathbb{T}^3$$ simply as $$$$C^\infty(\mathbb{T}^3, \mathbb{R}^3):= \{ X : \mathbb{T}^3 \to \mathbb{R}^3 \mid \text{each component of }X \text{ is a smooth real-valued function} \}$$$$

Now, consider $$C^\infty(\mathbb{T}^3, \mathbb{R}^3)$$ as a vector space over $$\mathbb{R}$$and let $$\{ X_1, \cdots, X_n\}$$ be a linearly independent subset of $$C^\infty(\mathbb{T}^3, \mathbb{R}^3)$$.

Then, I know that the tensor product of smooth vector fields $$X,Y : M \to TM$$ is defined by $$$$(X \otimes Y)(p) := X_p \otimes Y_p \in T_pM \otimes T_pM$$$$ for each $$p \in M$$, where $$M$$ is any smooth manifold.

In the case above, then the tensor products of any ordered pair from $$X_1, \cdots, X_n$$ live in $$C^\infty(\mathbb{T}^3, \mathbb{R}^9)$$. In case of $$X_1$$ and $$X_2$$, we have the smooth tensor field $$$$X^i_1(x)X^j_2(x)$$$$ where $$i,j=1,2,3$$ and $$x \in \mathbb{T}^3$$.

My question is that, are these $$n^2$$ tensor products linearly independent over $$\mathbb{R}$$ in $$C^\infty(\mathbb{T}^3, \mathbb{R}^9)$$?

I am not sure in the context of differential geometry, so I would like to ask to be sure..

I don't think you have to consider vector fields. Ordinary vectors in $$\mathbb R^d$$ are enough. For simplicity I assume $$d=2\,.$$ When $$e_1,e_2$$ is the canonical basis of $$\mathbb R^2$$ then the four matrices $$e_1\otimes e_1=\begin{pmatrix}1&0\\0&0\end{pmatrix}\,,\quad e_1\otimes e_2=\begin{pmatrix}0&1\\0&0\end{pmatrix}\,,\quad e_2\otimes e_1=\begin{pmatrix}0&0\\1&0\end{pmatrix}\,,\quad e_2\otimes e_2=\begin{pmatrix}0&0\\0&1\end{pmatrix}\,,\quad$$ are obviously a basis of $$\mathbb R^2\otimes\mathbb R^2\simeq\mathbb R^4\,.$$
When $$X_1,X_2$$ are linearly independent then the matrix $$S=\begin{pmatrix}X_{11}&X_{21}\\X_{12}&X_{22}\end{pmatrix}$$ is invertible and maps $$e_i$$ to $$X_i\,.$$
Therefore, \begin{align} 0&= \alpha_{11}\,X_1\otimes X_1+\alpha_{12}\,X_1\otimes X_2+\alpha_{21}\,X_2\otimes X_1+\alpha_{22}\,X_2\otimes X_2\\ &=S\Big(\alpha_{11}\,e_1\otimes e_1+\alpha_{12}\,e_1\otimes e_2+\alpha_{21}\,e_2\otimes e_1+\alpha_{22}\,e_2\otimes e_2)\,S^\top\,. \end{align} This implies linear independence, i.e., that $$\alpha_{11}=\alpha_{12}=\alpha_{21}=\alpha_{22}=0\,.$$