# Example 97 infinite sequences let k be a field a

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Example 9.7: Infinite sequencesLetKbe a field. Asequenceof elements ofKis an infinite list(a0,a1,a2,a3...),whereaiKfor alli.We also use the notation(ai)iN, or occasionally(ai), to denote such asequence. LetSeqKbe the set of sequences of elements ofK. We add two sequences by addingtheirithelements:(ai)iN+(bi)iN=(ai+bi)iN.We scale a sequence by scaling each of its elements:k(ai)iN=(kai)iN.ThenSeqKis a vector space.Example 9.8: Vector space of polynomials of unbounded degreeLetKbe a field, and letPbe the set of all polynomials (of any degree) with coefficients fromK,i.e., expressions of the formp(x)=anxn+an1xn1+...+a1x+a0,wheren0anda0,...,anK. Addition and scalar multiplication of polynomials are defined inthe usual way . ThenPis a vector space.
9.1. Definition of vector spaces331We conclude this section by deriving some initial consequences of the vector space axioms.Proposition 9.9: Elementary consequences of the vector space axiomsIn any vector space, the following are true:(a) The additive unit is unique. In other words, wheneveru+v=u, thenv=0.(b) Additive inverses are unique. In other words, wheneveru+v=0, thenv=u.(c)0u=0for all vectorsu.(d) The followingcancellation lawholds: ifu+w=v+w, thenu=v.Proof.We prove the first three properties, and leave the last one as an exercise. AssumeVis any vectorspace over a fieldK.(a) Consider arbitrary vectorsu,vVand assumeu+v=u.Applying the law (A1) (commutative law) to the left-hand side, we havev+u=u.Addinguto both sides of the equation, we have(v+u)+(u)=u+(u).Applying the law (A2) (associative law) to the left-hand side, we havev+(u+(u))=u+(u).Applying the law (A4) (additive inverse law) to both sides of the equation, we havev+0=0.Applying the law (A3) (additive unit law) to the left-hand side, we havev=0.This proves that wheneveru+v=u, thenv=0, or in other words,v=0is the only element actingas an additive unit.(b) Consider arbitrary vectorsu,vVand assumeu+v=0.Applying the law (A1) (commutative law) to the left-hand side, we havev+u=0.
332Vector spacesAddinguto both sides of the equation, we have(v+u)+(u)=0+(u).Applying the law (A2) (associative law) to the left-hand side, we havev+(u+(u))=0+(u).Applying the law (A4) (additive inverse law) to the left-hand side, we havev+0=0+(u).Applying the law (A1) (commutative law) to the right-hand side, we havev+0=u+0.Applying the law (A3) (additive unit law) to both sides of the equation, we havev=u.This proves that wheneveru+v=0, thenv=u, or in other words,v=uis the only elementacting as an additive inverse ofu.(c) First, note that the scalar 0Ksatisfies the property 0+0=0, by property (A3) of the definition ofa field. Now letuVbe any vector. Using the vector space law (SM2) (distributive law over scalaraddition) and 0+0=0, we have0u+0u=(0+0)u=0u.Next, we use a small trick: add(0u)to both sides of the equation. This gives(0u+0u)+((0u)) =0u+((0u)).

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Matrix Theory and Linear Algebra