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

1960's book about a plague that kills all white people

Were any external disk drives stacked vertically?

How to draw the figure with four pentagons?

Why does Arabsat 6A need a Falcon Heavy to launch

Facing a paradox: Earnshaw's theorem in one dimension

AES: Why is it a good practice to use only the first 16bytes of a hash for encryption?

Etiquette around loan refinance - decision is going to cost first broker a lot of money

Can I use a neutral wire from another outlet to repair a broken neutral?

Is it unprofessional to ask if a job posting on GlassDoor is real?

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

How do I write bicross product symbols in latex?

Is it legal for company to use my work email to pretend I still work there?

If a Gelatinous Cube takes up the entire space of a Pit Trap, what happens when a creature falls into the trap but succeeds on the saving throw?

Western buddy movie with a supernatural twist where a woman turns into an eagle at the end

A reference to a well-known characterization of scattered compact spaces

Fully-Firstable Anagram Sets

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

What is the intuition behind short exact sequences of groups; in particular, what is the intuition behind group extensions?

Is "remove commented out code" correct English?

Can I ask the recruiters in my resume to put the reason why I am rejected?

How can I make my BBEG immortal short of making them a Lich or Vampire?

Combinations of multiple lists

prove that the matrix A is diagonalizable

How can I tell someone that I want to be his or her friend?



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










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    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
    5 hours ago










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










  • $begingroup$
    What does "when $A$ is written this way" mean?
    $endgroup$
    – anomaly
    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










share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    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
    5 hours ago










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










  • $begingroup$
    What does "when $A$ is written this way" mean?
    $endgroup$
    – anomaly
    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










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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago







JoshuaK

















asked 5 hours ago









JoshuaKJoshuaK

264




264







  • 2




    $begingroup$
    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
    5 hours ago










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










  • $begingroup$
    What does "when $A$ is written this way" mean?
    $endgroup$
    – anomaly
    1 hour ago












  • 2




    $begingroup$
    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
    5 hours ago










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










  • $begingroup$
    What does "when $A$ is written this way" mean?
    $endgroup$
    – anomaly
    1 hour ago







2




2




$begingroup$
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
5 hours 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
5 hours ago












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




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












$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
1 hour ago




$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
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




    $begingroup$
    I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
    $endgroup$
    – Acccumulation
    2 hours ago


















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
      4 hours 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









    $endgroup$








    • 2




      $begingroup$
      I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
      $endgroup$
      – Acccumulation
      2 hours ago















    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




      $begingroup$
      I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
      $endgroup$
      – Acccumulation
      2 hours ago













    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 5 hours ago









    TheSilverDoeTheSilverDoe

    5,324215




    5,324215







    • 2




      $begingroup$
      I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
      $endgroup$
      – Acccumulation
      2 hours ago












    • 2




      $begingroup$
      I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
      $endgroup$
      – Acccumulation
      2 hours ago







    2




    2




    $begingroup$
    I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
    $endgroup$
    – Acccumulation
    2 hours ago




    $begingroup$
    I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
    $endgroup$
    – Acccumulation
    2 hours ago











    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 5 hours 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
              4 hours 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
              4 hours 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 5 hours 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
              4 hours ago
















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















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




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

            Disable / Remove link to Product Items in Cart 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?How can I limit products that can be bought / added to cart?Remove item from cartHide “Add to Cart” button if specific products are already in cart“Prettifying” the custom options in cart pageCreate link in cart sidebar to view all added items After limit reachedLink products together in checkout/cartHow to Get product from cart and add it againHide action-edit on cart page if simple productRemoving Cart items - ObserverRemove wishlist items when added to cart

            Helsingin valtaus Sisällysluettelo Taustaa | Yleistä sotatoimista | Osapuolet | Taistelut Helsingin ympäristössä | Punaisten antautumissuunnitelma | Taistelujen kulku Helsingissä | Valtauksen jälkeen | Tappiot | Muistaminen | Kirjallisuutta | Lähteet | Aiheesta muualla | NavigointivalikkoTeoksen verkkoversioTeoksen verkkoversioGoogle BooksSisällissota Helsingissä päättyi tasan 95 vuotta sittenSaksalaisten ylivoima jyräsi punaisen HelsinginSuomalaiset kuvaavat sotien jälkiä kaupungeissa – katso kuvat ja tarinat tutuilta kulmiltaHelsingin valtaus 90 vuotta sittenSaksalaiset valtasivat HelsinginHyökkäys HelsinkiinHelsingin valtaus 12.–13.4. 1918Saksalaiset käyttivät ihmiskilpiä Helsingin valtauksessa 1918Teoksen verkkoversioTeoksen verkkoversioSaksalaiset hyökkäävät Etelä-SuomeenTaistelut LeppävaarassaSotilaat ja taistelutLeppävaara 1918 huhtikuussa. KapinatarinaHelsingin taistelut 1918Saksalaisten voitonparaati HelsingissäHelsingin valtausta juhlittiinSaksalaisten Helsinki vuonna 1918Helsingin taistelussa kaatuneet valkokaartilaisetHelsinkiin haudatut taisteluissa kaatuneet punaiset12.4.1918 Helsingin valtauksessa saksalaiset apujoukot vapauttavat kaupunginVapaussodan muistomerkkejä Helsingissä ja pääkaupunkiseudullaCrescendo / Vuoden 1918 Kansalaissodan uhrien muistomerkkim

            Adjektiivitarina Tarinan tekeminen | Esimerkki: ennen | Esimerkki: jälkeen | Navigointivalikko