Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 1 / 8C / iii

Table of contents
- Solution iii

Solution

A direct construction, valid for every

0 \leq t < 1

, is

a_{1} a_{2} a_{3} = (1, 0, 0), = (t, 1 - t^{2}, 0), = (t, t \frac{1 - t}{1 + t}, \frac{( 1 - t ) ( 1 + 2 t )}{1 + t}) .

The first coordinates give

a_{1} \cdot a_{2} = a_{1} \cdot a_{3} = t

, and

a_{2} \cdot a_{3} = t^{2} + t (1 - t) = t .

The first two norms are one. For the third,

∣ a_{3} ∣^{2} = t^{2} + \frac{t ^{2} ( 1 - t ) + ( 1 - t ) ( 1 + 2 t )}{1 + t} = 1.

Thus the required Gram matrix is

(1 - t) I + t 1 1^{T}

. The column matrix is upper triangular and has determinant

det (a_{1} a_{2} a_{3}) = (1 - t) 1 + 2 t > 0,

so 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.

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