Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 1 / 8C (Algebra and Geometry)

Yes. Suppose ${\alpha }_1{a}_1+{\alpha }_2{a}_2+{\alpha }_3{a}_3=0$. Taking its dot product with $a_2\times a_3$ eliminates the last two terms and givesThe nonzero scalar triple product forces ${\alpha }_1=0$. Cyclically using $a_3\times a_1$ and $a_1\times a_2$ forces ${\alpha }_2={\alpha }_3=0$ as well. This is the definition of linear independence. Equivalently the three-column matrix has nonzero determinant.

No. For every $n\ge 2$, takein ${\mathbb{R}}^n$. These form a minimally dependent vector family: the full family is dependent because $a_n-{\sum }_{j=1}^{n-1}{a}_j=0$.

If the omitted vector is $a_n$, the remaining coordinate vectors are independent. If the omitted vector is $e_k$, a relation among the retained vectors has an $e_k$ coordinate coming only from $a_n$, so its coefficient must be zero; the coefficients of all retained coordinate vectors must then be zero. Thus every $(n-1)$-element subfamily is linearly independent, as required for a counterexample. When $n=1$, the empty subfamily is independent while the one-vector family consisting of the zero vector is not, so the answer remains negative.

A direct construction, valid for every $0\le t<1$, isThe first coordinates give $a_1\cdot a_2=a_1\cdot a_3=t$, andThe first two norms are one. For the third,Thus the required Gram matrix is $(1-t)I+t11^T$. The column matrix is upper triangular and has determinantso these vectors are linearly independent. This is an explicit instance of unit vectors with a common inner product; at $t=0$ it reduces to the standard orthonormal basis.