Paper 1, Section II, 6C6 \mathbf{C}

Vectors and Matrices
Part IA, 2010

Let a1,a2\mathbf{a}_{1}, \mathbf{a}_{2} and a3\mathbf{a}_{3} be vectors in R3\mathbb{R}^{3}. Give a definition of the dot product, a1a2\mathbf{a}_{1} \cdot \mathbf{a}_{2}, the cross product, a1×a2\mathbf{a}_{1} \times \mathbf{a}_{2}, and the triple product, a1a2×a3\mathbf{a}_{1} \cdot \mathbf{a}_{2} \times \mathbf{a}_{3}. Explain what it means to say that the three vectors are linearly independent.

Let b1,b2\mathbf{b}_{1}, \mathbf{b}_{2} and b3\mathbf{b}_{3} be vectors in R3\mathbb{R}^{3}. Let SS be a 3×33 \times 3 matrix with entries Sij=aibjS_{i j}=\mathbf{a}_{i} \cdot \mathbf{b}_{j}. Show that

(a1a2×a3)(b1b2×b3)=det(S)\left(\mathbf{a}_{1} \cdot \mathbf{a}_{2} \times \mathbf{a}_{3}\right)\left(\mathbf{b}_{1} \cdot \mathbf{b}_{2} \times \mathbf{b}_{3}\right)=\operatorname{det}(S)

Hence show that SS is of maximal rank if and only if the sets of vectors {a1,a2\left\{\mathbf{a}_{1}, \mathbf{a}_{2}\right., a3}\left.\mathbf{a}_{3}\right\} and {b1,b2,b3}\left\{\mathbf{b}_{1}, \mathbf{b}_{2}, \mathbf{b}_{3}\right\} are both linearly independent.

Now let {c1,c2,,cn}\left\{\mathbf{c}_{1}, \mathbf{c}_{2}, \ldots, \mathbf{c}_{n}\right\} and {d1,d2,,dn}\left\{\mathbf{d}_{1}, \mathbf{d}_{2}, \ldots, \mathbf{d}_{n}\right\} be sets of vectors in Rn\mathbb{R}^{n}, and let TT be an n×nn \times n matrix with entries Tij=cidjT_{i j}=\mathbf{c}_{i} \cdot \mathbf{d}_{j}. Is it the case that TT is of maximal rank if and only if the sets of vectors {c1,c2,,cn}\left\{\mathbf{c}_{1}, \mathbf{c}_{2}, \ldots, \mathbf{c}_{n}\right\} and {d1,d2,,dn}\left\{\mathbf{d}_{1}, \mathbf{d}_{2}, \ldots, \mathbf{d}_{n}\right\} are both linearly independent? Justify your answer with a proof or a counterexample.

Given an integer n>2n>2, is it always possible to find a set of vectors {c1,c2,,cn}\left\{\mathbf{c}_{1}, \mathbf{c}_{2}, \ldots, \mathbf{c}_{n}\right\} in Rn\mathbb{R}^{n} with the property that every pair is linearly independent and that every triple is linearly dependent? Justify your answer.