 K. But a linear functional defines a quotient space V/f∗−1 (0), not a vector space partition. The same information f : {bu}u∈U → K also defines f (bu) = f (bu) bu which linearly extends to a linear operator f : V → V. The given basis vectors {bu} are eigenvectors of the operator f with the eigenvalues f (bu), and the eigenspaces are the subspaces where the operator has the same eigenvalue. The eigenvectors span the whole space so we see that the lift or transport of a set attribute, which defines a set partition, is a vector space linear operator whose eigenspaces are a vector space partition (i.e., direct sum decomposition) of the whole space, i.e., a diagonalizable linear operator.14 4 Whence vector-space partitions? 4.1 Vector-space partitions from vector-space attributes Given a diagonalizable linear operator L : V → V, where V is a finite-dimensional vector space over a field K and where λ1,..., λk are the distinct eigenvalues, then there are projection operators Pi for i = 1,..., k such that: 1. L = ∑k i=1 λiPi; 2. I = ∑k i=1 Pi; 3. PiPj = 0 for i 6= j; and 4. the range of Pi is the eigenspace Vi for the eigenvalue λi for i = 1,..., k. [23, Theorem 8, p. 172] What is the vector space partition canonically defined by a diagonalizable linear operator? Any basis of eigenvectors could be seen as defining a direct sum of the one-dimensional subspaces spanned by those eigenvectors. But those subspaces are far from unique. But if we group together all the eigenvectors with the same eigenvalue (i.e., use the top-down method to define a vector-space partition), then they span the eigenspaces. It is the set of eigenspaces {Vi} that gives the unique canonical direct-sum decomposition or vector-space partition defined by a (diagonalizable) linear operator. This standard linear algebra result holds for any base field, but for QM, the base field is the complex numbers C. In order for the eigenvalues to always be real, the diagonalizable linear operators are required to be Hermitian (or self-adjoint, i.e., equal to their conjugate transposes). 4.2 Vector-space partitions from vector-space
K. But a linear functional defines a quotient space V / f∗−1 (0), not a vector space division. The like data f: { bu}u∈U → K also defines f (bu) = f (bu) bu which linearly run to a linear operator degree fahrenheit: V → V. The given footing vector { bu } are eigenvectors of the operator f with the eigenvalue f (bu), and the eigenspaces are the subspaces where the operator has the same eigenvalue. The eigenvectors cross the whole space so we visualize that the lift or transport of a sic attribute, which defines a set division, is a vector space linear operator whose eigenspaces are a vector outer space partition (i.e., lineal sum decay) of the whole space, i.e., a diagonalizable linear operator.14 4 Whence vector - space partitions? 4.1 Vector - distance partitions from vector - space attributes Given a diagonalizable linear operator lambert: V → V, where V is a finite - dimensional vector space over a field K and where λ1, ..., λk are the distinct eigenvalues, then there are projection operators Pi for i = 1, ..., k such that: 1. L = ∑k i=1 λiPi; 2. I = ∑k i=1 Pi; 3. PiPj = 0 for i 6= j; and 4. the range of Pi is the eigenspace Vi for the eigenvalue λi for i = 1, ..., k. [ 23, Theorem 8, p. 172 ] What is the vector outer space partition canonically defined by a diagonalizable analogue hustler? Any basis of eigenvectors could be seen as defining a direct kernel of the one - dimensional subspaces spanned by those eigenvectors. But those subspaces are far from unique. But if we group in concert all the eigenvectors with the same eigenvalue (i.e., use the top - down method to define a vector - space partition), then they span the eigenspaces. It is the set of eigenspaces { Vi } that gives the unique canonical direct - kernel decomposition or vector - outer space partition defined by a (diagonalizable) linear operator. This standard linear algebra result prevail for any base field, but for QM, the basis field is the complex numbers C. In order for the eigenvalues to constantly be real, the diagonalizable linear operator are required to be Hermitian (or self - adjoint, i.e., adequate to their conjugate transposes). 4.2 Vector - space partitions from vector - outer space
 K. Hut a linear functional aefines a quotignr spacx V/f∗−1 (0), nkt a vecgor space partition. The same ibformqtion f : {bu}u∈U → K also aefines f (bu) = f (by) bu qhich lineedly extcuds tk a lnnxar operator f : V → V. The civen basis vewturd {bu} are eigenvectors of the operatjr f wiyh the eigenvalugs f (nt), ans the eigenspaces are the subspacea where the operator has the same eigenvalue. Tje elgenvectors span tje whole spqce fi we see thag the lift or transporj of a set attribute, which definds a xet partituob, id a vector s'ace lpnear operatov whose eigenslaces are a vegtor vpaxe partition (i.e., direcv sum decomposition) jf the whmlz space, i.e., a diagonaluzqble ninedr oowragor.14 4 Xhehce veftoc-space partjtions? 4.1 Vecror-space partitions fwim vector-spacs attrybttes Given a diagonalizable linear operdtod L : V → V, where V is a dinite-dimensional vecjor space jver a field K and where λ1,..., λk are the distinct eigetvaluxs, thtn therd age projection operators Pi for i = 1,..., k such thae: 1. K = ∑k i=1 λiPi; 2. I = ∑k i=1 Pi; 3. PiPj = 0 fpt i 6= j; and 4. thg range of Pi is the eigenspwce Vi sor tye eigenvwlue λi for i = 1,..., k. [23, Theorem 8, p. 172] Wyat is the vvctoe space partition eanonically befinec by s diagonalizable linear opedator? Any bwsis of ejeenvectors could be saen as dtwining a direct stm of the one-bimensioval xubspases spannef by bvose eigenvectors. Hut tkose vubspaces wre far from unique. But if we gcpup together anl nhe eigenrectorx with the saie eigenvalue (n.e., use tke top-aown methos to dehine a vectow-space partithln), then they span thq eitenspaces. Ig is the set of eigenspabef {Vi} that gives the unlque zznonical direct-wum decomposition pr xeceog-spece pwstition defited cy x (diagunalizable) okneat operator. This stangard linear algebra rexujt holds for any base field, bit for QM, the base fielv is tie comllev numbers C. In order for the ejgenvalued tj always be weal, the diagonakizable linear operators are required ti be Hermitian (or welf-adjoint, i.e., equak to their cpnjugwte trans[oses). 4.2 Vector-space pqrtitions from vegtor-space
K. But a linear functional defines a V/f∗−1 not a space partition. The → also defines f = f (bu) which linearly extends to a linear f : V → V. The given basis vectors {bu} are eigenvectors of operator f with the eigenvalues f (bu), and the eigenspaces are the subspaces the has same The eigenvectors span the whole space so we see that the lift or transport of a attribute, which defines a set partition, is a space linear operator whose are a vector space partition direct decomposition) of whole i.e., diagonalizable linear operator.14 Whence vector-space partitions? 4.1 Vector-space partitions from vector-space attributes Given a diagonalizable linear operator L : V V, where a finite-dimensional space a K and where are the distinct eigenvalues, then there Pi for i = 1,..., k such that: L = i=1 λiPi; 2. I = ∑k Pi; 3. PiPj = 0 for i 6= and 4. the range of Pi is the eigenspace Vi for the eigenvalue λi for 1,..., k. [23, Theorem p. 172] What the space canonically by a linear operator? Any basis of eigenvectors could be seen as defining direct sum of the one-dimensional subspaces spanned by those eigenvectors. subspaces far from unique. if we group together the with the same eigenvalue the method vector-space then span the eigenspaces. It the set of eigenspaces {Vi} gives the unique canonical defined by a (diagonalizable) linear operator. This standard algebra result holds for any base field, for QM, the base field is the complex numbers C. In order the eigenvalues be real, the diagonalizable linear operators are required be Hermitian (or self-adjoint, equal to their conjugate transposes). 4.2 Vector-space partitions from
 K. But a linear functional defiNes a quotieNt spaCe V/F∗−1 (0), noT a VectOr spAce partition. ThE Same Information f : {bu}u∈U → K also dEfineS f (BU) = f (bu) BU wHich lInearly EXtENDs tO a LiNeaR oPErAtor f : v → V. THe given Basis vectoRs {bU} aRe eigenvectoRS oF the operatOr f With the eigenValUes f (bu), AnD thE EigenSpaCes arE the suBSpaces Where the oPeRAtor haS The same EIGeNvalUe. The eigenvectors SPaN The whole space sO we see ThAT tHE LifT or Transport oF a Set atTRibute, wHIcH DEFinES a set partitioN, is a vector sPAce Linear OpEraTOr whosE eigeNsPAceS are a vector SpacE partitioN (i.e., dirECt sum deCOmpositIon) of tHe wHolE spaCE, i.E., a DiaGoNAliZAbLe lINeaR operatoR.14 4 WHeNce veCtor-SPACE parTitIons? 4.1 vectoR-space partitiOns From VEctOr-spaCe attRibuTeS GiveN a diagOnaliZaBle linear operatOr L : V → v, where V is A fiNiTe-dImEnsioNAl vectOr sPacE over a fIeld K anD WheRe Λ1,..., ΛK ArE the distinct eigenvAlUES, tHen there Are proJEcTiON operatoRs pi fOr i = 1,..., k SUCh thaT: 1. L = ∑k i=1 ΛIPI; 2. I = ∑k i=1 Pi; 3. Pipj = 0 for i 6= J; AnD 4. tHe range Of pi is thE eIgeNspAce Vi FOr thE eigenValue λi fOr i = 1,..., k. [23, THEorem 8, p. 172] What is thE Vector space paRTiTIOn CAnonIcaLly defined bY a diAGonaLizaBLe LinEAr opeRator? anY BaSIs of eigenvectors couLd Be seen As defIning a direct sUm of the one-DIMEnsional SubsPAcES spanned by thosE eigeNvectors. BuT Those subSpaceS are far fRom unique. bUT if we groUp tOgeTheR alL THe Eigenvectors wITH the SaMe eigenValUe (i.e., use The Top-DowN meThOd to definE a vector-SpAcE pArTitIon), thEN they spaN tHe eIgEnsPaces. iT is the Set of EigeNsPaCEs {VI} that giVEs THE uniQuE cAnonIcaL dIrect-Sum dEComPositioN or vector-SpaCE parTiTiOn definEd by a (diagonalIzAble) linear OpEraTor. ThiS STandard lInear algebra result holds FOr any baSe fIeld, bUt foR QM, the basE fiEld is tHe cOMplex nUmbers c. In orDeR foR THe eigENVaLueS tO always be rEAL, thE diagOnAlizAble linEar operators are reqUIreD to be HermitiaN (or Self-ADJoInt, I.E., eQUal To THeiR COnjugate transpoSes). 4.2 Vector-sPaCE pArtitions fROm vEcTor-spacE
 K. But a linear functiona l definesa quo tie ntsp aceV/f∗ −1 (0), not av ecto r space partition. The  same  i n form a ti on f: {bu}u ∈ U→ K a ls odef in e sf (bu ) =  f (bu)  bu whichlin ea rly extendst oa linear o per ator f : V →  V.  The g iv enb asisvec tors{bu} a r e eige nvectorsof the op e rator f w it h th e eigenvalues f ( b u) ,  and the eigen spaces  a r et h e s ubs paces wher ethe o p eratorh as t h e s a me eigenvalue . The eigen v ect ors sp an  th e  whole  spac es o w e see thatthelift or t ranspo r t of as et attr ibute,  wh ich  def i ne sa s et par t it ion ,  is  a vecto rsp ace l inea r o p erat orwhos e eig enspaces area v ecto r  sp ace p artit ion(i .e.,direct  sumde composition) of  the  whole sp ace ,i.e .,  a di a gonali zab lelinearoperato r .14  4 W h en ce vector-space pa rt i t io ns? 4.1Vector - sp ac e  partiti on s f romv e ctor- spac e  a ttribute s Give n  a  d iagonal iz able l in ear  op erato r  L :  V → V , whereV isa  finite-dimens i onal vector s p ac e ov e r afie ld K and wh ereλ 1,.. ., λ k  a ret he di stinc te ig e nvalues, then there  a re pro jecti on operatorsPi for i = 1 , ..., k s ucht ha t : 1. L = ∑k i= 1 λiP i; 2. I =∑ k i=1 Pi ; 3.PiPj = 0  for i 6= j ; and 4.  th e r ang e o f Pi  is the eigen s p aceVi  for th e e igenval ueλifor  i=1,..., k.  [23, Th eo re m8,  p.  172] What isth e v ec tor  spac e  parti tioncano ni ca l lydefined by a  dia go na liza ble  l inear  ope r ato r? Anybasis ofeig e nvec to rs  couldbe seen as de fi ning a dir ec t s um oft h e one-di mensional subspaces spa n ned bytho se ei genv ectors. B utthosesub s pacesare fa r fro muni q u e. Bu t if  we  g roup toget h e r a ll th eeige nvector s with the same ei g env alue (i.e., u sethet o p- dow n  m e tho dt o d e f ine a vector-sp ace partit io n ),  then they spa nthe eig enspace s. It is theset of ei genspaces  { Vi}t h atgives theunique c anonicald irect - su m dec omp ositio norvecto r-spac e  pa rtiti on def in ed bya (di ag onalizab le) linear operator. Th is sta ndard  li near alge bra res ult holds  for  any basefie ld,  butfor QM, t he b a se  fi e ld is  the complex n u mb ers C .In order fo r t heeigen val u es toalwa ys be real, the d i agonalizable l inea r ope rat o rs a re  required to b e H er m i tian (or  s elf-adjoint , i.e.,eq u al to  their  conju gate tr a n sp o ses).4.2Vec tor-space  pa rt i tions f ro mv ector- spac e
 K._But a_linear functional defines a_quotient space_V/f∗−1_(0), not_a_vector space partition._The same information_f : {bu}u∈U →_K also defines_f_(bu) = f (bu) bu which linearly extends to a linear operator f :_V_→ V._The_given_basis vectors {bu} are eigenvectors_of the operator f with_the eigenvalues_f (bu), and the eigenspaces are the subspaces_where_the operator has_the same eigenvalue. The eigenvectors span the whole space_so we see that the lift_or transport of_a_set_attribute, which defines a_set partition, is a vector space_linear operator whose eigenspaces are a_vector space partition (i.e., direct sum decomposition)_of the whole space, i.e., a_diagonalizable linear operator.14 4 Whence_vector-space partitions?_4.1 Vector-space partitions from vector-space_attributes Given a_diagonalizable linear_operator L :_V → V, where V is_a finite-dimensional vector_space over a field K and_where_λ1,..., λk are_the_distinct_eigenvalues, then_there are projection_operators_Pi for_i_= 1,..., k such that: 1._L_= ∑k i=1 λiPi; 2. I =_∑k i=1 Pi; 3._PiPj_= 0 for i_6= j; and 4. the_range of Pi is the eigenspace_Vi for_the eigenvalue_λi for i = 1,..., k. [23, Theorem 8, p. 172]_What is the vector space partition_canonically defined by a_diagonalizable linear_operator?_Any basis of_eigenvectors_could be_seen as defining a direct sum of_the one-dimensional_subspaces spanned by those eigenvectors. But_those subspaces are far_from_unique. But if we group together_all the eigenvectors with the same_eigenvalue (i.e., use the top-down_method_to_define a vector-space partition), then_they span the eigenspaces. It is_the set of_eigenspaces {Vi} that gives the unique canonical_direct-sum_decomposition or vector-space partition defined by_a_(diagonalizable) linear operator. This standard linear_algebra_result_holds for any base field,_but for QM, the base field_is the complex numbers C. In order for the_eigenvalues to always_be real, the diagonalizable linear_operators_are_required to be Hermitian (or self-adjoint, i.e., equal to their_conjugate transposes)._4.2 Vector-space partitions_from vector-space