Page 1 of 20 Maths Revision Notes Daniel Guetta, 2007 Matrices & Linear Algebra Vector Spaces Scalarsaretheelementsofanumberfield(for example, R and C), which oIs a set of elements on which the operations of additionandmultiplicationaredefined,and satisfytheusuallawsofarithmetic (commutative, associative and distributive). oIs closed under addition and multiplication oIncludesidentityelementsforadditionand multiplication (0 and 1). oIncludes inverses (negatives and reciprocals) for addition and multiplication, except 0. Vectorsareelementsofalinearvectorspacedefined over a number field F. A vector space V oIs a set of elements on which the operations of vectoradditionandscalarmultiplicationare defined and satisfy certain axioms. oIs closed under these operations. oIncludesanidentityelement(0)forvector addition. IfthenumberfieldFoverwhichthelinearvector space is defined is real, then the vector space is real. Notes: oVector multiplication is not, in general, defined for a vector space. oThe basic example of a vector space is a list of nscalars,Rn.Vectoradditionandscalar multiplication are defined component-wise. Page 2 of 20 Maths Revision Notes Daniel Guetta, 2007 oR2 is not exactly the same as C, because C has a rule for multiplication. oSimilarly,R3isnotquitethesameasphysical space,becausephysicalspacehasarule (Pythagoras)forthedistancebetweentwo points. The Inner Product Theinnerproductisusedtogiveameaningto lengths and angles in a vector space. It is a scalar function,, x yof two vectors x and y. An inner product must oBe linear in the second argument , ,, , ,a a =+ = +x y x yx y z x y x z oHave Hermitian symmetry *, , = y x x yoBe positive definite , 0 x xwith equality if and only if x = 0. Notes: oTheinnerproducthasanexistencewithout reference to any basis. oThehermitiansymmetryisrequiredsothat , x xis real. It is not required in a real vector space. oItfollowsfromtheabovethattheinner product is antilinear in the first argument: *, ,, , ,a a =+ = +x y x yx y z x z y z oInCn,thestandard(Euclidean)innerproduct, the dot product, is: *,i ix y = x yPage 3 of 20 Maths Revision Notes Daniel Guetta, 2007 The complex conjugation is needed to maintain Hermitiansymmetry,andtoensurethatthe product is linear in the second argument. The Cauchy-Schwarz Inequality states that 2, , , x y x x y yOr, equivalently: , x y x yWith equality if and only ifa = x y . Toprove,assumethat, x y 0 (inwhichcasethe inequality is trivial). We first say: ( )* *** *22 2| | || | | || | | |2Re |a a a a aa a aaa a aaa a- - = - - -= - - += - - += + -x y x y x yx x yyx x y x x y yyx x x y x y yyx y x yNow,thisquantitymustbepositive,becauseofthe positivedefinitepropertyoftheinnerproduct.Ifwe choosethephaseofa(whichisarbitrary)suchthat a | x y isrealandnon-negative,sothat | a a | = x y x y , we then have that: ( )22 222 | 02 2 | 0a aa a a+ - - + - x y x yx y x y x y And now, if we choose/ a= x y , we get: | x y x yAs required. WecanusetheCauchy-Schwarzinequalitytoprove thetriangleinequality( + + x y x y ),bywriting 2+ x y intermsoftheinnerproduct,expanding, using the inequality, and factorising. The Cauchy-Schwarz Inequality allows us to define, in realvectorspace,theangleq betweentwovectors, though , cos q = x y x yPage 4 of 20 Maths Revision Notes Daniel Guetta, 2007 [Thisispossiblebecause,bytheCauchySchwartz, ,1 x yx y.If, 0 = x y ,inanyvectorspace,xandy are said to be orthogonal. A knowledge of the inner product of the basis vectors is sufficient to determine the inner product of any two vectors x and y. Let: |ijG =i je eThen *,ij i jG x y = x yWhere the Gij are the metric coefficients of the basis. The Hermetian Symmetry of the inner product implies that *ij jiG G =The matrix G is hermitian. Foranorthonormalbasis,inwhich|ijd =i je e ,we have that *|i ix y = x y Bases Let{ }1 2, , ,mS= e e e be a subset of the vectors in V. AlinearcombinationofSisanyvectoroftheform 1 1 2 2 m mx x x + + + e e e , where the x are scalars. The span of S is the set of all vectors that are linear combinations of S. If the span of S is the entire vector space V, then S is said to span V. The vectors of S are said to be linearly independent if 1 1 2 2 1 2, , , 0m m mx x x x x x + + + = = e e e 0 If, on the other hand, such an equation holds for non-trivialvaluesofthecoefficients,oneofthevectorsis redundant, and can be written as a linear combination of the other vectors. Ifanadditionalvectorisaddedtoaspanningset,it remains a spanning set. If a vector is removed from a Page 5 of 20 Maths Revision Notes Daniel Guetta, 2007 linearlyindependentset,itremainsalinearly independent set. A basis for a vector space V is the subset of vectors S thatspansVandisalsolinearlyindependent.The properties of basis sets are: oAllbasesofVhavethesamenumberof elements, n, which is called the dimension of V. oAny n linearly independent vectors in V form a basis of V. oAnyvectorV x canbewritten[proveby considering{ } V xas a linearly dependent set] inauniqueway[provebycontradiction]asa linearcombinationofthevectorsinabasis. Therelevantscalarsarecalledthecomponents of x with respect to that particular basis. The same vector (a geometrical identity) has different componentswithrespecttodifferentbases.Tosee howwecanchangefromonetotheother,consider two bases of V { }iS= eand{ }iS = e . oBecausebotharebases,theelementsofone basis can be written in terms of the other: j i ijR= e eWhere Rij is the transformation matrix between the two bases oNow,consideravectorV x .The representation of the vector in each basis is: j j i ix x = = x e eHowever,usingourresultfromabove,wecan write this as: i i j j i ij jx x R x = = = x e e eoFromthis,wecandeducethetransformation law for vector components: i ij jx R x =Page 6 of 20 Maths Revision Notes Daniel Guetta, 2007 Note that: The law is the reverse of that for basis vector transformation. This is to ensure that, overall, the vector x stays unchanged by transformation. The first suffix of R corresponds to the same basis in both relations. We defined Rij, above, by j i ijR= e eThe condition of the basis {ej} to be orthonormal is ( )* i j ijk ki l lj ijki lj k l ijR RR Rddd= = =e ee ee e If the second basis is also orthonormal, this becomes: *ki kj ijR R d == R R I Inotherwords,transformationsbetweenorthonormal basesisdescribedbyunitarymatrices.Inrealvector space, an orthogonal matrix does this in R2 and R3, this corresponds to a rotation and/or reflection. Givenanymvectors m 1u u thatspanann-dimensionalvectorspace(m>n),itispossibleto construct an orthogonal basis 1 ne e using the Gram-Schmidt procedure: 11rrss s-=== -1 1sr rse ue ue u ee e What we are effectively doing is taking each vector u andremovinganybitsofvectorswevealready addedtothebasisfromit,toleaveuswithafinal vector that is orthogonal to all others already added We can prove, by induction, that this works: Page 7 of 20 Maths Revision Notes Daniel Guetta, 2007 Inductive stepAssumethatvectors 1 te e havealreadybeenadded to the orthogonal basis (such that0 i j = " i je e ), andnowconsiderthevectoret+1thatwereaboutto add: 1111rtt ss s-++ +== -st 1se ue u ee e Andnow,considerdottingitwithev(v