 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/digital-systems-linear-algebra-introduction-to-formal-proof/4/e/solution

u_{1} = a_{1} = (1, 1, 1, 1)^{T}

, of norm two. Next

a_{1}^{T} a_{2} / (a_{1}^{T} a_{1}) = 2

, so

u_{2} = a_{2} - 2 u_{1} = (1, - 1, - 1, 1)^{T} .

For the third column, both coefficients onto

u_{1}, u_{2}

equal one, giving

u_{3} = a_{3} - u_{1} - u_{2} = (1, - 1, 1, - 1)^{T} .

These three nonzero vectors are mutually orthogonal and each has norm two. The resulting orthonormal basis is

q_{1} = \frac{1}{2} 1111, q_{2} = \frac{1}{2} 1 - 1 - 1 1, q_{3} = \frac{1}{2} 1 - 1 1 - 1 .

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