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

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

Is it possible to download Internet Explorer on my Mac running OS X El Capitan?

How to take photos in burst mode, without vibration?

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

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

Twin primes whose sum is a cube

prove that the matrix A is diagonalizable

Why are electrically insulating heatsinks so rare? Is it just cost?

Has there ever been an airliner design involving reducing generator load by installing solar panels?

Python: return float 1.0 as int 1 but float 1.5 as float 1.5

I Accidentally Deleted a Stock Terminal Theme

Doing something right before you need it - expression for this?

Why is Collection not simply treated as Collection<?>

Why doesn't H₄O²⁺ exist?

Is the Joker left-handed?

Why is it a bad idea to hire a hitman to eliminate most corrupt politicians?

Is "remove commented out code" correct English?

Combinations of multiple lists

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

Forgetting the musical notes while performing in concert

Infinite Abelian subgroup of infinite non Abelian group example

I'm flying to France today and my passport expires in less than 2 months

Why do I get two different answers for this counting problem?

Can a virus destroy the BIOS of a modern computer?



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











$endgroup$







  • 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
    4 hours ago










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










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











$endgroup$







  • 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
    4 hours ago










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










  • $begingroup$
    What does "when $A$ is written this way" mean?
    $endgroup$
    – anomaly
    12 mins 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 4 hours ago







JoshuaK

















asked 4 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
    4 hours ago










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










  • $begingroup$
    What does "when $A$ is written this way" mean?
    $endgroup$
    – anomaly
    12 mins 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
    4 hours ago










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










  • $begingroup$
    What does "when $A$ is written this way" mean?
    $endgroup$
    – anomaly
    12 mins 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
4 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
4 hours ago












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




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












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




$begingroup$
What does "when $A$ is written this way" mean?
$endgroup$
– anomaly
12 mins 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
    1 hour 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
      3 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
      1 hour 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
      1 hour 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 4 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
      1 hour ago












    • 2




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







    2




    2




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




    $begingroup$
    I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
    $endgroup$
    – Acccumulation
    1 hour 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 4 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
              3 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
              3 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 4 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
              3 hours ago
















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















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




            $begingroup$
            Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
            $endgroup$
            – JoshuaK
            3 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

            How should I use the fbox command correctly to avoid producing a Bad Box message?How to put a long piece of text in a box?How to specify height and width of fboxIs there an arrayrulecolor-like command to change the rule color of fbox?What is the command to highlight bad boxes in pdf?Why does fbox sometimes place the box *over* the graphic image?how to put the text in the boxHow to create command for a box where text inside the box can automatically adjust?how can I make an fbox like command with certain color, shape and width of border?how to use fbox in align modeFbox increase the spacing between the box and it content (inner margin)how to change the box height of an equationWhat is the use of the hbox in a newcommand command?

            Image
            Is there a problem with hiding "forgot password" until it's needed? Time travel short story where a man arrives in the late 19th century in a time machine and then sends the machine back into the past Go Pregnant or Go Home Why Were Madagascar and New Zealand Discovered So Late? How can my private key be revealed if I use the same nonce while generating the signature? Modify casing of marked letters Trouble understanding overseas colleagues Best way to store options for panels Was Spock the First Vulcan in Starfleet? How do I keep an essay about "feeling flat" from feeling flat? How does a character multiclassing into warlock get a focus? Why does John Bercow say “unlock” after reading out the results of a vote? The Riley Riddle Mine Print name if parameter passed to function How can I replace every global instance of "x[2]" with "x_2" Hide Select Output from T-SQL Can I convert a rim brake wheel to a disc brak...
            Read more

            Doxepinum Nexus interni Notae | Tabula navigationis3158DB01142WHOa682390"Structural Analysis of the Histamine H1 Receptor""Transdermal and Topical Drug Administration in the Treatment of Pain""Antidepressants as antipruritic agents: A review"

            Image
            Antidepressiva depressionuminsomniareceptorium H 1 agonista inversapruritusdermatitis atopicaelichenis simplicis chronici (neurodermatitis) Cave: notitiae huius paginae nec praescriptiones nec consilia medica sunt. Doxepinum Natura chemica Formula chemica C 19 H 21 NO Massa molaris 279.376 g/mol PubChem 3158 DrugBank DB01142 Natura pharmacologica Codex ATC N06AA12 ( WHO ) Tempus semivitae biologicum 17 (8-24) h Nor-Doxepinum: 31 h Metabolismus iecore (hepaticus): CYP2D6 CYP2C19 Metabolitus Nor-Doxepinum Excretio renibus (ca. 50%), faecibus: minus Ad usum therapeuticum Applicatio per os, i.v., i.m. MedlinePlus a682390 .mw-parser-output .existinglinksgray a,.mw-parser-output .existinglinksgray a:visitedcolor:gray.mw-parser-output .existinglinksgray a.newcolor:#ba0000.mw-parser-output .existinglinksgray a.new:visitedcolor:#a55858 (Anglice) Doxepinum est substantia psychoanaleptica et antidepressiva, ergo medicamentum tractationis depressionum, praesertim cum insomnia praesente, effectu...
            Read more

            inputenc: Unicode character … not set up for use with LaTeX The Next CEO of Stack OverflowEntering Unicode characters in LaTeXHow to solve the `Package inputenc Error: Unicode char not set up for use with LaTeX` problem?solve “Unicode char is not set up for use with LaTeX” without special handling of every new interesting UTF-8 characterPackage inputenc Error: Unicode character ² (U+B2)(inputenc) not set up for use with LaTeX. acroI2C[I²C]package inputenc error unicode char (u + 190) not set up for use with latexPackage inputenc Error: Unicode char u8:′ not set up for use with LaTeX. 3′inputenc Error: Unicode char u8: not set up for use with LaTeX with G-BriefPackage Inputenc Error: Unicode char u8: not set up for use with LaTeXPackage inputenc Error: Unicode char ́ (U+301)(inputenc) not set up for use with LaTeX. includePackage inputenc Error: Unicode char ̂ (U+302)(inputenc) not set up for use with LaTeX. … $widehatleft (OA,AA' right )$Package inputenc Error: Unicode char â„¡ (U+2121)(inputenc) not set up for use with LaTeX. printbibliography[heading=bibintoc]Package inputenc Error: Unicode char − (U+2212)(inputenc) not set up for use with LaTeXPackage inputenc Error: Unicode character α (U+3B1) not set up for use with LaTeXPackage inputenc Error: Unicode characterError: ! Package inputenc Error: Unicode char ⊘ (U+2298)(inputenc) not set up for use with LaTeX

            Image
            Monthly twice production release for my software project How to make a software documentation "officially" citable? Was a professor correct to chastise me for writing "Prof. X" rather than "Professor X"? How to safely derail a train during transit? How can I filter or show only one of the duplicates in the results when using Find-Item command? What happens if you roll doubles 3 times then land on "Go to jail?" Clustering points and summing up attributes per cluster in QGIS How easy is it to start Magic from scratch? What can we do to stop prior company from asking us questions? Science fiction short story involving a paper written by a schizophrenic Is a stroke of luck acceptable after a series of unfavorable events? I believe this to be a fraud - hired, then asked to cash check and send cash as Bitcoin Why does standard notation not preserve intervals (visually) How to prepend a string to only the lines of text which ar...
            Read more