 Solution

In an oriented orthonormal coordinate system, the dot product and cross product areThe scalar triple product is $a\cdot (b\times c)$; equivalently it is the determinant of the matrix whose columns are $a,b,c$. The three vectors have linear independence when

Let $U=(a_1{a}_2{a}_3)$ and $V=(b_1{b}_2{b}_3)$ be column matrices. Since $(U^{T}V{)}_{ij}=a_i\cdot b_j$, we have $S=U^{T}V$. Multiplicativity of the determinant and its invariance under matrix transpose yieldA square matrix has maximal matrix rank exactly when its determinant is nonzero, equivalently when its columns have linear independence. A product of two real numbers is nonzero exactly when both factors are nonzero. Hence $\boxed{\mathrm{rank}S=3⟺(a_i{)}_{i=1}^3\text{ and }(b_j{)}_{j=1}^3\text{ are both independent}}$.

The same argument works in every dimension. Set $U_n=(c_1\cdots c_n)$ and $V_n=(d_1\cdots d_n)$. Thenso $\boxed{\mathrm{rank}T=n⟺(c_i{)}_{i=1}^n\text{ and }(d_j{)}_{j=1}^n\text{ are both independent}}$.

For the last construction, choose two standard basis vectors $e_1,e_2$ in ${\mathbb{R}}^n$ and take $c_j=e_1+j{e}_2$ for $j=1,\dots ,n$. If $\alpha c_i+\beta c_j=0$ with $i\ne j$, its first two coordinates give $\alpha +\beta =0$ and $i\alpha +j\beta =0$, hence $\alpha =\beta =0$. Every pair therefore has linear independence. Every triple lies in the two-dimensional linear span of $e_1,e_2$ and has linear dependence. More explicitly, for distinct $i,j,k$,a nontrivial dependence. Thus $\boxed{c_j=e_1+j{e}_2(1\le j\le n)\text{ works for every }n>2}$.