5I. Preliminaries: Linear Algebra** linear space **Functional Analysis plays in linear spaces of functions. For all practical purposes, alinear space (=: ls) X is a (nonempty) collection of functions f, all on the same domainT , and this collection is closed under pointwise addition and scalar multiplication.This means that, with f,g ∈ X, their sum f + g, i.e., the mapf + g : t 7→ f(t)+g(t),is also in X, as is the product αf of such f with any scalar α, i.e., the mapαf : t 7→ αf(t).For this to make sense, the maps in X must all have a common target, and it must bepossible to add elements in the target and to multiply them with scalars.The prime example is the collectionIRT:= { f : T → IR }of all real-valued functions on some set T , with addition and scalar multiplication definedpointwise, i.e., in the above way. In this case, the underlying scalar field is IR := the realnumber field. The most important special case occurs when T = { 1, 2,...,n},inwhichcase we get the n-dimensional coordinate spaceIRn:= IR{1,2,...,n}whose elements I will never write as 1-column matrices but, rather, as n-sequences x =(x(i):i =1,...,n)=(x(1),...,x(n)). In the discussion of inner product spaces and ofeigenvalues, we will also consider complex linear spaces, i.e., linear spaces for which thescalar field is C:= the complex number field. Agreement: if nothing is said, then thescalars are real. If I don’t care, I’ll write IF for the scalar field. Note that the collectionIF0of all empty sequences in IF consists of exactly one element, (). Even this coordinate spaceis useful at times.If X is a linear space, then the collectionXTof all functions f on the same domain T into X is also a linear space (under pointwiseaddition and scalar multiplication).Another way to get linear spaces from linear spaces is by considering linear subspaces(=: lss’s). These are nonempty subsets Y that are closed under addition and scalarmultiplication, i.e.,Y + Y := {y + y0: y ∈ Y,y0∈ Y }⊂Y, αY := {αy : y ∈ Y }⊂Y.E.g., for T ⊂ IRn,C(T ):={f ∈ IRT: f is continuous}is a lss of IRT.linear spacec2002 Carl de Boor6 I. Preliminaries: Linear AlgebraH.P.(1) Verify that any sum and any intersection of lss’s is again a lss.Here, for the record, is the formal definition of a ls:(1) Definition. To say that X is a linear space (=:ls)(ofvectors) over the (commuta-tive) field IF (of scalars) means that there are two maps, (i) X × X → X :(x, y) 7→ x + ycalled (vector) addition; and (ii) IF × X → X :(α, x) 7→ αx =: xα called scalarmultiplication, that satisfy the following rules.(a) X is a commutative group with respect to addition; i.e., addition(a.1) is associative: f +(g + h)=(f + g)+h;(a.2) is commutative: f + g = g + f;(a.3) has neutral element: ∃{0}∀{f} f +0=f;(a.4) has inverse: ∀{f}∃{g} f + g =0.(s) scalar multiplication is(s.1) associative: α(βf)=(αβ)f;(s.2) field addition distributive: (α + β)f = αf + βf;(s.3) vector addition distributive: α(f + g)=αf + αg;(s.4) unitary: 1f = f .It is standard to denote the element g ∈ X for which f + g =0by−f since such g isuniquely determined by the requirement that f + g = 0. I will denote the neutral elementin X by the same symbol, 0, used for the zero scalar. In particular, as the sole element ofIF0, the empty sequence () is denoted by 0.H.P.(2) Prove: For f in the ls X, (−1)f = −f and 0f =0.Also,αf =0with f 6=0implies α =0.H.P.(3) Prove: For any set T and any field IF , IFTis a ls with respect to pointwise addition and scalarmultiplication.H.P.(4) Prove: Any lss is a ls (with respect to the addition and scalar multiplication as restricted tothe lss).lss’s are often given as the range or the kernel of a linear map.** linear map **We deal extensively with linear maps(=: lm’s) (or operators, transformations,mappings, functions, etc. all much longer than ‘map’), i.e., with A : X → U satisfyingA(f + g)=Af + Ag, all f, g ∈ X (additivity)A(αf)=α(Af), all α ∈ IF ,f ∈ X (homogeneity)where X and U are ls’s. U could be X. The simplest examples are:0:X → U : x 7→ 0,α: X → X : x 7→ αx.We denote byL(X, U)the collection of all lm’s from the ls X to the ls U, and writeL(X):=L(X, X).linear mapc2002 Carl de BoorLinear spaces and linear maps 7L(X, U) is a ls under pointwise addition and scalar multiplication. In addition, the col-lection of linear maps is closed under composition: If A ∈ L(X, U) and C ∈ L(U, W ),thenCA : X → W : f 7→ C(Af)is in L(X,W ). Also, if A ∈ L(X, U) is invertible (as a map), then A−1∈ L(U, X). Com-position (like all map composition) is associative and combines with addition and scalarmultiplication of linear maps in the expected way. But, composition is not commutative.H.P.(5) Prove that an additive map A : X → U is homogeneous for all rational scalars.H.P.(6) Prove thatthe inverse of a lm is linear.H.P.(7) Prove:If X is a ls and A : X → U is a lm with respect to some addition and scalar multiplicationonU,thenran A is a ls (even if U fails to be a ls). (See the definition of quotient space below for an instructiveexample.)** special case: column maps, especially matrices **An important special case is L(IFn,X) which provides us with a first opportunity topractice a basic step in fa, namely representation. Here, we show that L(IFn,X) is nicelyrepresentable by Xn, the set of n-sequences in the ls X.Indeed, each sequence (v1,...,vn) ∈ Xngives rise to the corresponding map[v1,...,vn]:IFn→ X : a 7→nXj=1vja(j)which is evidently linear. As a special example, withek:= (0,...,0|{z }k−1 zeros, 1, 0,...,0|{z }n−k zeros) ∈ IFnthe kth unit vector in IFn,[e1,...,en]:IFn→ IFn: a 7→Xjeja(j)=ais the identity, 1 = 1n,onIFn.If also A ∈ L(X, U), then A(Pjvja(j)) =Pj(Avj)a(j), henceA[v1,...,vn]=[Av1,...,Avn] ∈ L(IFn,U).In particular,∀{A ∈ L(IFn,X)} A = A1n=[Ae1,...,Aen]and this shows that every A ∈ L(IFn,X) is uniquely representable as [v1,...,vn] (withvj= Aej,allj). This sets up the invertible linear mapXn→ L(IFn,X):(vj: j =1,...,n) 7→ [v1,...,vn].special case: column maps, especially matricesc2002 Carl de Boor8 I. Preliminaries: Linear AlgebraThis is quite familiar for the special case that also X is a coordinate space, X =IFmsay. In that case, each vjis an m-vector, and one associates [v1,...,vn] with the m × nmatrix whose jth column contains the entries of vj. This sets up the invertible linear mapIFm×n=(IFm)n→ L(IFn, IFm):M → [M(:,