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  1. 9.4.5

9.4.5 #

Theorem 9.4.5 (Maximality and minimality of bases)Let 𝑉 be a vector space over some field 𝑘 and take 𝑒1,…,𝑒𝑛∈𝑉. The following are equivalent:(a)The 𝑒𝑖 form a basis.(b)The 𝑒𝑖 are spanning, but no proper subset is spanning.(c)The 𝑒𝑖 are linearly independent, but adding any other element of 𝑉 makes them not linearly independent.Solution by kiwiyou(a) ⟹ (b)By definition of a basis, for all 𝑣∈𝑉, 𝑣=∑𝑛𝑖=1𝑎𝑖⋅𝑒𝑖 for unique 𝑎𝑖∈𝑘. It is clear that {𝑒𝑖} is spanning.Suppose that 𝑆={𝑒𝑖}𝑚𝑖=1 is spanning for some 𝑚<𝑛.Since 𝑆 is spanning, there exists 𝑏𝑖 such that 𝑒𝑛=∑𝑚𝑖=1𝑏𝑖⋅𝑒𝑖.𝑒𝑛 has two different representations, which is a contradiction.Therefore, {𝑒𝑖} is spanning, but no proper subset is spanning.∎(a) ⟹ (c)It is clear that {𝑒𝑖} is linearly independent. Also, every nonzero vector 𝑣∈𝑉 has a unique representationwith at least one nonzero coefficient.Let 𝑣=∑𝑛𝑖=1𝑎𝑖⋅𝑒𝑖 be the representation of 𝑣 in {𝑒𝑖}.Since 0=(−1)𝑣𝑖+∑𝑛𝑖=1𝑎𝑖⋅𝑒𝑖, {𝑒𝑖}∪{𝑣} is not linearly independent.∎(b) ⟹ (a)Suppose that there exists a vector 𝑣0∈𝑉 with two different representations. In other words, there exists𝑎𝑖,𝑏𝑖 such that 𝑣0=∑𝑛𝑖=1𝑎𝑖⋅𝑒𝑖=∑𝑛𝑖=1𝑏𝑖⋅𝑒𝑖 for some 𝑎𝑖,𝑏𝑖∈𝑘 and, without loss of generality, 𝑎𝑛≠𝑏𝑛.For all 𝑣∈𝑉, let 𝑣=∑𝑛𝑖=1𝑐𝑖⋅𝑒𝑖 be the representation of 𝑣 in 𝑆. Then we have:𝑣=𝑣+0=∑𝑛𝑖=1𝑐𝑖⋅𝑒𝑖+𝑐𝑛𝑎𝑛−𝑏𝑛∑𝑛𝑖=1(𝑎𝑖−𝑏𝑖)⋅𝑒𝑖=∑𝑛𝑖=1(𝑐𝑖+𝑐𝑛𝑎𝑛−𝑏𝑛(𝑎𝑖−𝑏𝑖))⋅𝑒𝑖Since the coefficient of 𝑒𝑛 is always 0, {𝑒𝑖}𝑛−1𝑖=1 is spanning. This is a contradiction.Therefore all vectors have unique representations in 𝑆.∎Bonus: (c) ⟹ (a)For all 𝑣∈𝑉∖{𝑒𝑖}, {𝑒𝑖}∪{𝑣} is not linear independent, which means there exists 𝑐≠0 and 𝑎𝑖 such that0=𝑐⋅𝑣+∑𝑛𝑖=1𝑎𝑖⋅𝑒𝑖.Then every vector has its representation:𝑣=∑𝑛𝑖=1−𝑎𝑖𝑐⋅𝑒𝑖If 𝑣 has two different representations, then 0=𝑣−𝑣 has two different representations, which is a contradiction.Therefore, every vector has a unique representation.∎Notice the use of division in the proofs (b) ⟹ (a) and (c) ⟹ (a). This shows why we needthe field instead of a commutative ring.