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Linear Transformations

4 Definition. Let V and be vector spaces. A ◦ Let : M22 be the linear transformation : V W is a from V to W such that a → ab T( + v)=T(u)+T(v) →and T(cu)=cT(u) L cd = ⎡⎤ . !" #$ for all u, v ∈ V and for all c ∈ R. ⎢ ⎥ ⎣ ⎦ Examples: The maps defined below are all linear. ◦ L : Mm, Mn, defined by The verifications are easy in each case. L(A)=AT . ◦ An n × m A defines a → n m T LA : R R defined by Here A denotes of A. R LA(⃗v)=A⃗v. ◦ Let C ( ) be the of all → infinitely differentiable functions on R. ∞ All linear maps from Rn to Rm has this form. Define D : C (R) C (R) by

R ∞ ∞ ◦ Let Tr : Mn be the function D()=f′ (the of f) → Tr(( aij)) = a11 + a22 + ···+ ann. → ◦ Let C([0, 2]) be the space of all continuous R In other words, Tr(A) is the sum of the functions on [0, 2].Defineev1 : C([0, 2]) diagonal entries of A.WesayTr(A) is the by ev (f)=f(1). of A. 1 → and range

Let T : V W be a linear transformation (also ◦ If T : V W is a linear map, then verify that called a linear map). ker(T) is a subspace of V and range(T) is a subspace of W. ◦ V is→ called the domain of T. → ◦ Let A be an n × m matrix. Recall that A ◦ W is called the of T. n m defines a linear map LA : R R defined ◦ The range of T,denotedrange(T),meansthe by LA(⃗v)=A⃗v.Verifythat of all vectors of W that can be written in → the form T(v) for some v ∈ V. ker(LA)=nul(A) and range(LA)=col(A). ◦ The kernel of T,denotedbyker(T),means the set of all v ∈ V such that T(v)=0. Examples Definitions.

◦ Let V be a vector space. Let {v1, ··· , vp} be an indexed set of vectors in V.

◦ A of v1, ··· , vp} is an expression of the form

c1v1 + c2v2 + ···+ cpvp.

where c1, ··· , cp are scalars.

◦ We say that {v1, ··· , vp} is linearly dependent,ifthevectorequation

c1v1 + c2v2 ···+ cpvp = 0 only have the trivial solution c1 = c2 = ···= cp = 0. ◦ Linearly independent means not linearly dependent. and Definition. Let V be a vector space. An indexed Examples. subset B of V is called a basis of V if B is linearly independent and span(B)=V. We say that a vector space is finite dimensional,if it has a basis consisting of finitely many elements. We shall shoon see that: ◦ every finite dimensional vector space V has afinitebasisB,and ◦ any two basis of V have the same number of elements. This number is called the dimension of V and is denoted by dim(V).

If a vector space V is not finite dimensional, we Exercise Find a basis for the in R3 given by say that V is infinite dimensional and we write − 3y + 2z = 0. dim(V)= .

∞ Some results on linear independence, span, basis and dimension

Lemma 1. Let be a subset of a vector space . Lemma 2. If a vector space H can be spanned by p vectors, then H cannot contain (p + 1) or more (a) Suppose S spans H and S is linearly dependent. linearly independent vectors. Then some proper subset of S also spans H. sketch of proof. Suppose v1, v2, ··· , vp spans of H. (b) Suppose S is linearly independent but does not Let u1, u2, ··· , up+1 any p + 1 vectors in H.Since span H.ThenS is properly contained in a subset of H v1,....,vp spans H,wecanwrite that is also linearly independent. p sketch of proof. (a). Write a nontrivial dependence uj = cijvi for = 1, 2, ··· , p + 1. S v S i=1 relation among elements of and choose a in # that has nonzero coefficient in this linear combination. Then that v can be written as a The matrix C =((cij)) is of size p × (p + 1).SoC has more columns than rows. So reduced row linear combination of S − {v},andoneverifiesthat ⃗ ⃗ S − {v} still spans H. echelon form has a pivot free column. So Cx = 0 a1 . (b) Choose v in H − span(S) and verify that has a nontrivial solution . .Notethat S ∪ {v} is still indep. +ap+1,

p+1 p p+1

ajuj = vi cijaj = 0. = = = #j 1 #i 1 #j 1 So u1, ··· , up+1 are linearly dependent. Lemma 3. Any finite spanning set of a vector space Lemma 5. Suppose a vector space H has a basis of p H contains a basis of H.Inparticular,afinite elements. Then a linearly independenet subset S of H dimensional vector space has a finite basis. of size p is already a basis of H. sketch of proof. Let T be a finite spanning subset of sketch of proof. Let v in H − S.SinceH can be H.AmongallsubsetsofT that spans H,choose spanned by p vectors, lemma 3 implies that one of smallest possible size. Call it S.IfS is S ∪ {v} must be linearly dependent. Write a linearly dependent, then a proper subset of S nontrivial dependence relation among S ∪ {v}.The would still span H (by Lemma 1(a)), contradicting coefficient of v must be nonzero because the minimality of S.SoS must be linearly otherwise we get a nontrivial dependence relation independent, hence a basis of H. among S.Rearranging,itfollowsv can be written as a combination of elements of S.SoS must Lemma 4. Let V be a vector space that can be spanned span H. by n elements. Then any subspace H of V has a basis of at most n elements. Lemma 6. Any two basis of a finite dimensional vector space H must have the same number of sketch of proof. Since V can be spanned by n elements. elements, by lemma 2, V cannot contain more than n linearly independent elements. So nor can sketch of proof. Let S and T be two basis of H. H.Inotherwords,anylinearlyindependent Since S spans H and T is linearly independent, subset of H has size at most n.Soamongallthe lemma 2 implies that size of T is at most size of S. linearly independent subsets of H we can choose Now interchange the role of S and T. one of largest size. Call it S.IfS does not span H, then it can be enlarged to a bigger linearly independent subset of H (by lemma 1(b)) contradicting the maximality of S.SoS must span H,hencemustbeabasisofH. Lemma 7. Suppose a vector space H has a basis of p Theorem. Let T : V W be a linear map. Assume V elements. Then a spanning subset S of H of size p is is finite dimensional. Then ker(T) and range(T) are already a basis of H. also finite dimensional and → sketch of proof. If S is not a basis, then by lemma 3, dim(ker(T)) + dim((T)) = dim(V). apropersubsetofS would give a basis of H of size smaller than p,contradictinglemma6. Lemma 8. Let V be a vector space of dimension n.Let H be a subspace of V.Thendim(H) ! n.If dim(H)=n,thenH = V. sketch of proof. The first statement follows from lemma 4. Now suppose dim(H)=n.LetS be a basis of H.ThenS is a linearly independent subset of V of size n = dim(V),henceS must be a basis of V by lemma 5. Coordinate systems