Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 2 / 15E / c

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- Solution c

Solution

Assume for contradiction that

n > m

. Apply part b to the matrix of the given relation coefficients, whose columns are nonzero. For its first dependent prefix choose

b_{1}, \dots, b_{k}

with

\sum_{j = 1}^{k} a_{ij} b_{j} = 0 for all i, z_{i} := \sum_{j = 1}^{k} a_{ij} λ_{j} b_{j} not all zero .

Multiply the

j

th given vector relation by

b_{j}

and sum. Interchanging the finite sums gives

0 = \sum_{i = 1}^{m} (\sum_{j = 1}^{k} a_{ij} b_{j}) v_{i} + \sum_{i = 1}^{m} (\sum_{j = 1}^{k} a_{ij} λ_{j} b_{j}) w_{i} = \sum_{i = 1}^{m} z_{i} w_{i} .

The

w_{i}

are a basis, so all

z_{i}

must be zero, contradicting part b. Therefore $n \leq m$ .

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