prove that the matrix A is diagonalizableBlock Diagonal Matrix DiagonalizableNew proof about normal matrix is diagonalizable.Show that matrix $A$ is NOT diagonalizable.Prove a matrix is not diagonalizableHow to use inner products in C(n) to prove normal matrix is unitarily diagonalizable after knowing that normal matrix is diagonalizable?Is the Matrix Diagonalizable if $A^2=4I$Prove that $A$ is diagonalizable.Prove that a general matrix is diagonalizableDetermine $a$ to make matrix $A$ diagonalizableDiagonalizable block-diagonal matrix

Twin primes whose sum is a cube

Why "Having chlorophyll without photosynthesis is actually very dangerous" and "like living with a bomb"?

What mechanic is there to disable a threat instead of killing it?

Took a trip to a parallel universe, need help deciphering

What is going on with Captain Marvel's blood colour?

I Accidentally Deleted a Stock Terminal Theme

If human space travel is limited by the G force vulnerability, is there a way to counter G forces?

Why is consensus so controversial in Britain?

How do I write bicross product symbols in latex?

How to model explosives?

Withdrawals from HSA

What does it mean to describe someone as a butt steak?

intersection of two sorted vectors in C++

Should I tell management that I intend to leave due to bad software development practices?

Blender 2.8 I can't see vertices, edges or faces in edit mode

Cronab fails because shell path not found

What's the point of deactivating Num Lock on login screens?

How much of data wrangling is a data scientist's job?

What's the difference between 'rename' and 'mv'?

What exploit are these user agents trying to use?

Diode datasheet reading

Alternative to sending password over mail?

Did Shadowfax go to Valinor?

Is there a hemisphere-neutral way of specifying a season?



prove that the matrix A is diagonalizable


Block Diagonal Matrix DiagonalizableNew proof about normal matrix is diagonalizable.Show that matrix $A$ is NOT diagonalizable.Prove a matrix is not diagonalizableHow to use inner products in C(n) to prove normal matrix is unitarily diagonalizable after knowing that normal matrix is diagonalizable?Is the Matrix Diagonalizable if $A^2=4I$Prove that $A$ is diagonalizable.Prove that a general matrix is diagonalizableDetermine $a$ to make matrix $A$ diagonalizableDiagonalizable block-diagonal matrix













2












$begingroup$


We have :



$A^3-3A^2-A+3I_n = 0 $



how can i prove that A is diagonalizable .



I don't know how to do when A is written this way






linear-algebra matrices eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
    $endgroup$
    – José Carlos Santos
    1 hour ago










  • $begingroup$
    yes , you're right i edit my mistake
    $endgroup$
    – JoshuaK
    1 hour ago















2












$begingroup$


We have :



$A^3-3A^2-A+3I_n = 0 $



how can i prove that A is diagonalizable .



I don't know how to do when A is written this way






linear-algebra matrices eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
    $endgroup$
    – José Carlos Santos
    1 hour ago










  • $begingroup$
    yes , you're right i edit my mistake
    $endgroup$
    – JoshuaK
    1 hour ago













2












2








2





$begingroup$


We have :



$A^3-3A^2-A+3I_n = 0 $



how can i prove that A is diagonalizable .



I don't know how to do when A is written this way






linear-algebra matrices eigenvalues-eigenvectors diagonalization







share|cite|improve this question











$endgroup$




We have :



$A^3-3A^2-A+3I_n = 0 $



how can i prove that A is diagonalizable .



I don't know how to do when A is written this way






linear-algebra matrices eigenvalues-eigenvectors diagonalization




linear-algebra matrices eigenvalues-eigenvectors diagonalization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago







JoshuaK

















asked 1 hour ago









JoshuaKJoshuaK

254




254







  • 1




    $begingroup$
    Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
    $endgroup$
    – José Carlos Santos
    1 hour ago










  • $begingroup$
    yes , you're right i edit my mistake
    $endgroup$
    – JoshuaK
    1 hour ago












  • 1




    $begingroup$
    Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
    $endgroup$
    – José Carlos Santos
    1 hour ago










  • $begingroup$
    yes , you're right i edit my mistake
    $endgroup$
    – JoshuaK
    1 hour ago







1




1




$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
1 hour ago




$begingroup$
Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
$endgroup$
– José Carlos Santos
1 hour ago












$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
1 hour ago




$begingroup$
yes , you're right i edit my mistake
$endgroup$
– JoshuaK
1 hour ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.






    share|cite|improve this answer









    $endgroup$




















      1












      $begingroup$

      We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
        $endgroup$
        – JoshuaK
        43 mins ago











      Your Answer





      StackExchange.ifUsing("editor", function ()
      return StackExchange.using("mathjaxEditing", function ()
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      );
      );
      , "mathjax-editing");

      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "69"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader:
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      ,
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );













      draft saved

      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3175144%2fprove-that-the-matrix-a-is-diagonalizable%23new-answer', 'question_page');

      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.






      share|cite|improve this answer









      $endgroup$

















        3












        $begingroup$

        The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.






        share|cite|improve this answer









        $endgroup$















          3












          3








          3





          $begingroup$

          The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.






          share|cite|improve this answer









          $endgroup$



          The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          TheSilverDoeTheSilverDoe

          5,324215




          5,324215





















              2












              $begingroup$

              Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.






                  share|cite|improve this answer









                  $endgroup$



                  Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 59 mins ago









                  EricEric

                  513




                  513





















                      1












                      $begingroup$

                      We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.






                      share|cite|improve this answer









                      $endgroup$












                      • $begingroup$
                        Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
                        $endgroup$
                        – JoshuaK
                        43 mins ago















                      1












                      $begingroup$

                      We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.






                      share|cite|improve this answer









                      $endgroup$












                      • $begingroup$
                        Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
                        $endgroup$
                        – JoshuaK
                        43 mins ago













                      1












                      1








                      1





                      $begingroup$

                      We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.






                      share|cite|improve this answer









                      $endgroup$



                      We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 1 hour ago









                      GSoferGSofer

                      8631313




                      8631313











                      • $begingroup$
                        Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
                        $endgroup$
                        – JoshuaK
                        43 mins ago
















                      • $begingroup$
                        Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
                        $endgroup$
                        – JoshuaK
                        43 mins ago















                      $begingroup$
                      Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
                      $endgroup$
                      – JoshuaK
                      43 mins ago




                      $begingroup$
                      Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
                      $endgroup$
                      – JoshuaK
                      43 mins ago

















                      draft saved

                      draft discarded
















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …


                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.

                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3175144%2fprove-that-the-matrix-a-is-diagonalizable%23new-answer', 'question_page');

                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Are there any AGPL-style licences that require source code modifications to be public? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Force derivative works to be publicAre there any GPL like licenses for Apple App Store?Do you violate the GPL if you provide source code that cannot be compiled?GPL - is it distribution to use libraries in an appliance loaned to customers?Distributing App for free which uses GPL'ed codeModifications of server software under GPL, with web/CLI interfaceDoes using an AGPLv3-licensed library prevent me from dual-licensing my own source code?Can I publish only select code under GPLv3 from a private project?Is there published precedent regarding the scope of covered work that uses AGPL software?If MIT licensed code links to GPL licensed code what should be the license of the resulting binary program?If I use a public API endpoint that has its source code licensed under AGPL in my app, do I need to disclose my source?

                      Image
                      Flight departed from the gate 5 min before scheduled departure time. Refund options How to show a density matrix is in a pure/mixed state? Can we cancel the equality mark here? How to ask rejected full-time candidates to apply to teach individual courses? Are there any AGPL-style licences that require source code modifications to be public? Why did Israel vote against lifting the American embargo on Cuba? What should one know about term logic before studying propositional and predicate logic? Why is Rajasthan pro BJP in the LS elections but not in the state elections? New Order #6: Easter Egg Question on Gÿongy' lemma proof How to resize main filesystem Is it OK if I do not take the receipt in Germany? My mentor says to set image to Fine instead of RAW — how is this different from JPG? Can I feed enough spin up electron to a black hole to affect it's angular momentum? How do I find my Spellcasting Ability for my D&D character? Is my guitar...
                      Read more

                      2013 GY136 Descoberta | Órbita | Referências Menu de navegação«List Of Centaurs and Scattered-Disk Objects»«List of Known Trans-Neptunian Objects»

                      Image
                      Objetos do disco dispersoObjetos transnetunianos objeto transnetunianodisco dispersoSistema Solarcorpo celesteressonância orbitalplanetaplanetamagnitude absolutadiâmetro9 de abril2013Outer Solar System Origins Surveyórbitaexcentricidadesemieixo maiorUAperiélioSolafélio 2013 GY136 Data da descoberta 9 de abril de 2013 Descoberto por Outer Solar System Origins Survey Categoria Transnetuniano Objeto do disco disperso Elementos orbitais Semieixo maior 55,961 UA Perélio 32,595 UA Afélio 79,327 UA Excentricidade 0,418 Inclinação 10,9° Características físicas Dimensões 133 km Magnitude absoluta 7,6 ver 2013 GY 136 , também escrito como 2013 GY136 , é um objeto transnetuniano que está localizado no disco disperso, uma região do Sistema Solar. Este corpo celeste está em uma ressonância orbital de 2:5 com o planeta planeta. Ele possui uma magnitude absoluta de 7,6 [ 1 ] e tem um diâmetro estimad...
                      Read more

                      Metrô de Los Teques Índice Linhas | Estações | Ver também | Referências Ligações externas | Menu de navegação«INSTITUCIÓN»«Mapa de rutas»originalMetrô de Los TequesC.A. Metro Los Teques |Alcaldía de Guaicaipuro – Sitio OficialGobernacion de Mirandaeeeeeee

                      Image
                      Metrô de Los TequesFundações na Venezuela em 2006 metropolitanoLos TequesGuaicaipuroMirandaGoverno da VenezuelaLinha 1Linha 2estaçõesLinha 1Linha 2IndependenciaSan AntonioLinha 3GuaicaipuroLibertadorEstação Las AdjuntasMetrô de Caracasalgarismocorestaçõesnomecorestaçõesestações Metrô de Los Teques Uma das composições que operam no sistema. Informações Proprietário Governo da Venezuela Local [[ Los Teques, Miranda]] País variant = Tipo de transporte Metropolitano Número de linhas 2 Número de estações 5 Website [1] Funcionamento Início de funcionamento 3 de novembro de 2006 (12 anos) [ 1 ] Operadora(s) C. A. Metro Los Teques Dados técnicos Extensão do sistema 11,2 km Bitola Bitola padrão (1435 mm) Velocidade máxima 80 km/h ...
                      Read more