Past exam of the mathematical methods I course of the University of Oxford / 2014 resit / 12

In components the scalar triple product is $A_i{ϵ}_{ijk}{B}_j{C}_k$. A cyclic permutation has positive sign of a permutation, so ${ϵ}_{ijk}={ϵ}_{kij}$. Relabelling the summed indices therefore giveswhich proves $A\cdot (B\times C)=(A\times B)\cdot C$. Equivalently both expressions are the same three-column determinant.

Apply the contraction of two Levi-Civita symbols to the cross product:Thus the vector triple product identity is

Three vectors form a basis of ${\mathbb{R}}^3$ exactly when they are linearly independent. The determinant with those vectors as columns is their scalar triple product, so the condition isGeometrically their parallelepiped must have nonzero volume.

If two indices of the n-dimensional Levi-Civita symbol are equal, exchanging them leaves the component unchanged. Antisymmetry of a multilinear map simultaneously negates it, so $ϵ=-ϵ$. Over the real numbers this implies $\boxed{ϵ=0}$.

If all $n$ indices are distinct and each lies in $\{1,\dots ,n\}$, their list is a permutation of $(1,\dots ,n)$. Sorting it takes some number $N$ of transpositions. The n-dimensional Levi-Civita symbol equals $(-1{)}^N$ because its sorted component is $1$. ThereforeThe parity of a permutation is independent of the chosen sequence of transpositions.

Reversal has one inversion of a permutation for every pair of indices, hence $\left(\genfrac{}{}{0ex}{}{n}{2}\right)=n(n-1)/2$ inversions of a permutation. The sign of a permutation givesIt is $+1$ for $n\equiv 0,1(\mathrm{mod}4)$ and $-1$ for $n\equiv 2,3(\mathrm{mod}4)$.

Every nonzero component of the n-dimensional Levi-Civita symbol is $\pm 1$, so its square is one. There are exactly $n!$ permutations of the $n$ distinct indices; all remaining index lists have a repeat and contribute zero. Hence

Let $M$ be the square matrix whose $j$th column is $A_j$. The Leibniz formula for determinants is precisely the stated contraction:A square matrix has zero determinant exactly when its columns are linearly dependent, proving the required equivalence. One direction also follows directly from multilinearity and antisymmetry of a multilinear map: replacing one column by a linear combination of the others leaves only terms with repeated columns, each zero. Conversely, nonzero determinant gives an inverse matrix and hence independent columns.