In $\mathbb{R}^2$ we have the linear transformation $R_{\pi/2}$ which rotates everything by $90^\circ$. A nice thing about this is that a non-zero vector field $X$ on $\mathbb{R}^2$ can be given a smooth orthogonal complement by just applying $R_{\pi/2}$ at each point. I thought about the possibility of such a phenomenon in dimensions besides $2$.
In odd dimensions, a real linear transformation must have at least one real eigenvalue, and if the transformation is to be invertible, there is no possibility of $\langle Ax, x\rangle$ being zero for all $x \in \mathbb{R}^{2k +1}$. In even dimension $2k$, we can have a block matrix of $k$ copies of $R_{\pi/2}$ which will have the property that $\langle Ax, x\rangle = 0$ for all $x \in \mathbb{R}^{2k}$. But can we find linear transformations $A_{0}=I$, $A_{1}, \dotsc, A_{2k-1}$ such that $\langle A_{i}(x), A_{j}(x)\rangle = 0$ for all $i \neq j$? Or, weaker, so that they are all linearly independent?
There exist linear transformations $A_1,\dots,A_n\in M_n(\mathbb{R})$ such that $(A_1v,\dots,A_nv)$ is linearly independent for all nonzero $v\in\mathbb{R}^n$ iff $n=1,2,4,$ or $8$.
First, let's reformulate the problem a bit. Note that $(A_1v,\dots,A_nv)$ is linearly independent for all nonzero $v$ iff every nontrivial linear combination of the $A_i$ is invertible. Identifying the linear span of the $A_i$ with $\mathbb{R}^n$, this gives a bilinear binary operation $\cdot:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}^n$ with the property that $x\cdot y=0$ implies $x=0$ or $y=0$. Explicitly, this multiplication is defined by the formula $$(x_1,\dots,x_n)\cdot y=\sum x_i A_iy.$$ We can think of this as a (possibly nonassociative) multiplication on $\mathbb{R}^n$ for which division is possible.
By a famous theorem that uses hard results from algebraic topology, such a possibly nonassociative division algebra structure on $\mathbb{R}^n$ exists iff $n=1,2,4,$ or $8$. In those dimensions, examples of such division algebra structures are given by the real numbers, the complex numbers, the quaternions, and the octonions, respectively.
If you require additionally that $\langle A_i v,A_jv\rangle=0$ for $i\neq j$ and that each $A_i$ is an orthogonal transformation, then this is equivalent to asking that the multiplication on $\mathbb{R}^n$ satisfy $|xy|=|x||y|$, where $|\cdot|$ is the Euclidean norm. In this case, it is much easier to prove that such a multiplication exists only for $n=1,2,4,$ or $8$, and in particular there is a purely algebraic proof. See here for one such proof. Since the multiplications on the real numbers, the complex numbers, the quaternions, and the octonions all satisfy $|xy|=|x||y|$, they give examples of $A_1,\dots, A_n$ such that $(A_1v,\dots,A_nv)$ is not just linearly independent but orthogonal for any nonzero $v$.