 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-i/1/a/iii/solution

Put

w_{1} = v_{1}

and

w_{j} = v_{j} - v_{1}

for

j \geq 2

. Since

v_{1} = w_{1}

and

v_{j} = w_{j} + w_{1}

, the two families have the same linear span. Moreover,

\sum_{j = 1}^{n} a_{j} w_{j} = (a_{1} - \sum_{j = 2}^{n} a_{j}) v_{1} + \sum_{j = 2}^{n} a_{j} v_{j} .

If the

v_{j}

are linearly independent, a vanishing expression forces

a_{j} = 0

for

j \geq 2

and then

a_{1} = 0

. Conversely,

\sum_{j = 1}^{n} b_{j} v_{j} = (\sum_{j = 1}^{n} b_{j}) w_{1} + \sum_{j = 2}^{n} b_{j} w_{j},

w_{j}

b_{j} = 0

. Thus both the spanning set condition and linear independence are equivalent for the two families, proving the desired equivalence of bases.

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