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4.2: Elementary properties of vector spaces

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    We are going to prove several important, yet simple, properties of vector spaces. From now on, \(V\) will denote a vector space over \(\mathbb{F}\).

    Proposition 4.2.1. Every vector space has a unique additive identity.

    Proof. Suppose there are two additive identities \(0\) and \(0'\) Then

    \[ 0'=0+0'=0, \]

    where the first equality holds since \(0\) is an identity and the second equality holds since \(0'\) is an identity. Hence \(0=0'\), proving that the additive identity is unique.

    Proposition 4.2.2. Every \(v \in V\) has a unique additive inverse.

    Proof. Suppose \(w\) and \(w'\) are additive inverses of \(v\) so that \(v+w=0\) and \(v+w'=0\) Then

    \[ w = w+0 = w+(v+w') = (w+v)+w' = 0+w' =w'. \]

    Hence \(w=w'\), as desired.

    Since the additive inverse of \(v\) is unique, as we have just shown, it will from now on be denoted by \(-v\) We also define \(w-v\) to mean \(w+(-v)\) We will, in fact, show in Proposition 4.2.5 below that \(-v=-1 v\)

    Proposition 4.2.3. \(0v = 0\) for all \(v \in V\).

    Note that the \(0\) on the left-hand side in Proposition 4.2.3 is a scalar, whereas the \(0\) on the right-hand side is a vector.

    Proof. For \(v\in V\), we have by distributivity that

    \[ 0v=(0+0)v=0v+0v. \]

    Adding the additive inverse of \(0v\) to both sides, we obtain

    \[ 0 = 0v-0v = (0v+0v)-0v=0v.\]

    Proposition 4.2.4. \(a0=0\) for every \( a \in \mathbb{F}\).

    Proof. As in the proof of Proposition 4.2.3, if \(a\in \mathbb{F}\), then

    \[ a0=a(0+0)=a0+a0. \]

    Adding the additive inverse of \(a0\) to both sides, we obtain \(0=a0\), as desired.

    Proposition 4.2.5. \((-1)v=-v\) for every \(v \in V\).

    Proof. For \(v\in V\), we have

    \[ v+(-1)v= 1v+(-1)v = (1+(-1))v=0v=0, \]

    which shows that \((-1)v\) is the additive inverse \(-v\) of \(v\).