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 Incoming links: Sum of mutually orthogonal projectors

q_{i}^{T} q_{j} = 0

for

i \neq = j

and one for

i = j

. Therefore

P_{i} P_{j} = q_{i} (q_{i}^{T} q_{j}) q_{j}^{T}

equals zero for different indices and

P_{i}

for equal indices. Expanding the finite sum yields

\overline{P}_{k}^{2} = \sum_{i = 1}^{k} \sum_{j = 1}^{k} P_{i} P_{j} = \sum_{i = 1}^{k} P_{i} = \overline{P}_{k} .

Thus every stated partial sum is a projector. Its image of a linear map is the span of the first

k