Why is the determinant Mn(R)→R continuous? - Math Stack ...

It's continuous because it's computable as a function from R ... MathematicsStackExchangeisaquestionandanswersiteforpeoplestudyingmathatanylevelandprofessionalsinrelatedfields.Itonlytakesaminutetosignup. Signuptojointhiscommunity Anybodycanaskaquestion Anybodycananswer Thebestanswersarevotedupandrisetothetop Home Public Questions Tags Users Unanswered Teams StackOverflowforTeams –Startcollaboratingandsharingorganizationalknowledge. CreateafreeTeam WhyTeams? Teams CreatefreeTeam Teams Q&Aforwork Connectandshareknowledgewithinasinglelocationthatisstructuredandeasytosearch. Learnmore Whyisthedeterminant$M_n(\mathbbR)\to\mathbbR$continuous? AskQuestion Asked 10years,2monthsago Modified 9years,9monthsago Viewed 16ktimes 46 21 $\begingroup$ Whyisthedeterminantasafunctionfrom$M_n(\mathbb{R})$to$\mathbb{R}$continuous,pleasecananyoneexplainpreciselyandrigorously?SofarIknowtheexplanationwhichcomesfromthefacts:polynomialsarecontinuous,sumandproductofcontinuousfunctionsarecontinuous.AlsoIhavetheconfusionregardingthemetricon$M_n(\mathbb{R})$ real-analysis Share Cite Follow editedMar18,2012at21:23 DylanMoreland 19k22goldbadges4747silverbadges7676bronzebadges askedMar18,2012at20:44 MarsoMarso 31.2k1818goldbadges106106silverbadges241241bronzebadges $\endgroup$ 2 8 $\begingroup$ Isn'titclearthatpolynomialsarecontinuousfunctions?$M_n(\mathbbR)$isthesameas$\mathbbR^{n^2}$underadifferentdisguise. $\endgroup$ – azarel Mar18,2012at20:50 1 $\begingroup$ Non-rigorous,butperhapshelpful....Determinantscanbeshowtobeequivalenttothehyper-volumeofahyper-parallelepipedwithalltheedgesextendingfromonevertexdefinedbythecolumns(orrows)ofamatrix.Giventhatthisvolumechangescontinuouslywithachangeinanycomponentvector,thedeterminantmaybeseentobecontinuous. $\endgroup$ – Tpofofn May17,2012at2:08 Addacomment  |  4Answers 4 Sortedby: Resettodefault Highestscore(default) Datemodified(newestfirst) Datecreated(oldestfirst) 50 $\begingroup$ $M_n(\mathbbR)$isjust$\mathbbR^{n^2}$withtheeuclidianmetric. detiscountinous,becauseitisapolynomialinthecoordinates $$ \text{det}(x_{i,j})=\sum_\sigma\text{sgn}(\sigma)\prod_{i=1}^{n}x_{\sigma(i),i} $$ Share Cite Follow editedAug11,2012at13:48 Marso 31.2k1818goldbadges106106silverbadges241241bronzebadges answeredMar18,2012at20:51 BlahBlah 5,0241717silverbadges2020bronzebadges $\endgroup$ 4 7 $\begingroup$ Itshouldbenotedthat$M_n(\mathbb{R})$isoftenalsogiventhemetricinducedbytheoperatornorm.Ofcoursethepointis,itdoesn'tmatterbecauseallnormsonafinitedimensionalspaceinducethesametopology. $\endgroup$ – user12014 May17,2012at1:04 1 $\begingroup$ Idonotunderstandyourexpressionattherightsideoftheinequality.alsowhatdoyoumeanby$\det(x_{i,j})$? $\endgroup$ – Marso Dec16,2013at9:37 $\begingroup$ Isthisargumentindependentofthemetricyoudefineon$M_n(\mathbb{R})$?(haven'ttakentopologyyet) $\endgroup$ – ChrisJing Apr12,2020at2:50 $\begingroup$ @ChrisJingallnormsinducethesametopology $\endgroup$ – zaira Dec7,2020at8:10 Addacomment  |  37 $\begingroup$ Recallthatthedeterminantcanbecomputedbyasumofdeterminantsofminors,thatis"sub"-matricesofsmallerdimension. Nowwecanprovebyinductionthat$\det$iscontinuous: For$n=1$,$A\inM_1(\mathbbR)$issimplyascalarwehavethat$\detA=A$,andsurelytheidentityfunctioniscontinuous. Supposethatfor$n$wehavethat$\det$iscontinuouson$M_n(\mathbbR)$,let$A\inM_{n+1}(\mathbbR)$.Weknowthat$\detA$canbecalculatedasthealternatingsumoveroneoffirstrow,whencalculatingthe$\det$oftheappropriateminor. So$\detA$iswrittenasasumandscalarmultiplicationof$\det$onasmallerdimension.Fromtheinductionhypothesisthesearecontinuousandtherefore$\det$iscontinuouson$n+1\timesn+1$matrices. Share Cite Follow answeredMay16,2012at23:42 AsafKaragila♦AsafKaragila 370k4141goldbadges552552silverbadges948948bronzebadges $\endgroup$ Addacomment  |  13 $\begingroup$ Thecoefficientmaps$A\longmapstoa_{i,j}$arecontinuousbecausetheyarelinearonthefinite-dimensionalvectorspace$M_n(\mathbb{R})$.Hereyouwanttorefertothetopologyofthelatterasanormedspace,whichdoesnotdependonthenormsincetheyareallequivalentinfinitedimension.Thenthedeterminantisapolynomialinthecoefficients,soitiscontinuousbycompositionofcontinuousmaps. Share Cite Follow editedMar18,2012at21:13 answeredMar18,2012at20:50 user26770user26770 $\endgroup$ Addacomment  |  5 $\begingroup$ It'scontinuousbecauseit'scomputableasafunctionfrom$\mathbb{R}^{n^2}$to$\mathbb{R}$. Share Cite Follow answeredMay17,2012at1:10 QuinnCulverQuinnCulver 4,21711goldbadge2525silverbadges4545bronzebadges $\endgroup$ 7 $\begingroup$ Ididn'tdownvoteyou,butI'mcuriousifyoucanelaborateonwhycomputablefunctions(intherelevantsense)arecontinuous? $\endgroup$ – ElchananSolomon May26,2012at0:10 $\begingroup$ @IsaacSolomonIt'sessentiallythefiniteuseprincipal.Seehereforexample. $\endgroup$ – QuinnCulver May26,2012at18:23 1 $\begingroup$ @QuinnCulverIfsomeonedownvoted,itwasprobablybecausetheythoughttheterminologywasalittlevague.Icanseeyouridea(andthisishowIthinkofit,too)isthatthedeterminantisacompositionoffinitelymanyadditionandmultiplicationoperations,allofwhicharejointlycontinuous,andsothecompositioniscontinuous. $\endgroup$ – rschwieb Jun8,2012at14:12 $\begingroup$ @rschwiebActuallythenotionofacomputablefunctionon$\mathbb{R}^{n}$isapreciselydefinedone.Seeforexample,ComputableAnalysisbyWeihrauch. $\endgroup$ – QuinnCulver Jun9,2012at15:12 3 $\begingroup$ @QuinnCulverNoobjectionhere,sincetheveracityofyourclaimhasneverbeenindoubt:onlytheaccessibilityofit.It'sgoodpolicytoassumeothershavenotreadeverythingyouhave. $\endgroup$ – rschwieb Jun9,2012at17:49  |  Show2morecomments YourAnswer ThanksforcontributingananswertoMathematicsStackExchange!Pleasebesuretoanswerthequestion.Providedetailsandshareyourresearch!Butavoid…Askingforhelp,clarification,orrespondingtootheranswers.Makingstatementsbasedonopinion;backthemupwithreferencesorpersonalexperience.UseMathJaxtoformatequations.MathJaxreference.Tolearnmore,seeourtipsonwritinggreatanswers. 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