Proving that every vector space has a norm.

linear-algebra functional-analysis vector-spaces

  1. No, you do not necessarily have $F_1 = F_2$. But as you have already shown the coefficiencts of the indices in $(F_1 \setminus F_2) \cup (F_2 \setminus F_1)$ are zero.

  2. This combination is unique, up to zero coefficients, cf. 1.

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