eigenvalues and eigenvectors of Diagonalisable matrices












0












$begingroup$



Let $D$:= diag($lambda_1, ldots, lambda_n$), i.e., $D$ is a diagonal matrix in $mathbb{C}^{ntimes n}$ with entries $lambda_1, ldots, lambda_n$ ∈ $mathbb{C}$ on its diagonal.



Furter let $Uinmathbb{C}^{ntimes n}$ be an invertible matrix and define $A$ := $U^{−1} $D$ $U$, i.e. A$ is similar to $D$. Find the eigenvalues and eigenvectors of $A$.




As $U$ is invertible and $D$ is diagonal I think that $A$ should also be a diagonal matrix.



And as $A$ is similar to $D$ then I suppose that $A$ has the same eigenvalues as $D$, namely ${lambda n}^{n}$.



Could someone tell me if I am on the right way?






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    I can tell you that you are not in the right way. For instance, it's almost sure that $A$ will not be diagonal.
    $endgroup$
    – José Carlos Santos
    Jan 6 at 10:26










  • $begingroup$
    See en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization When you write $A=U^{-1}DU$, you have a diagonalization of $A$. The eigenvalues and eigenvectors should be quite obvious.
    $endgroup$
    – Jean-Claude Arbaut
    Jan 6 at 10:28


















0












$begingroup$



Let $D$:= diag($lambda_1, ldots, lambda_n$), i.e., $D$ is a diagonal matrix in $mathbb{C}^{ntimes n}$ with entries $lambda_1, ldots, lambda_n$ ∈ $mathbb{C}$ on its diagonal.



Furter let $Uinmathbb{C}^{ntimes n}$ be an invertible matrix and define $A$ := $U^{−1} $D$ $U$, i.e. A$ is similar to $D$. Find the eigenvalues and eigenvectors of $A$.




As $U$ is invertible and $D$ is diagonal I think that $A$ should also be a diagonal matrix.



And as $A$ is similar to $D$ then I suppose that $A$ has the same eigenvalues as $D$, namely ${lambda n}^{n}$.



Could someone tell me if I am on the right way?






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    I can tell you that you are not in the right way. For instance, it's almost sure that $A$ will not be diagonal.
    $endgroup$
    – José Carlos Santos
    Jan 6 at 10:26










  • $begingroup$
    See en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization When you write $A=U^{-1}DU$, you have a diagonalization of $A$. The eigenvalues and eigenvectors should be quite obvious.
    $endgroup$
    – Jean-Claude Arbaut
    Jan 6 at 10:28
















0












0








0





$begingroup$



Let $D$:= diag($lambda_1, ldots, lambda_n$), i.e., $D$ is a diagonal matrix in $mathbb{C}^{ntimes n}$ with entries $lambda_1, ldots, lambda_n$ ∈ $mathbb{C}$ on its diagonal.



Furter let $Uinmathbb{C}^{ntimes n}$ be an invertible matrix and define $A$ := $U^{−1} $D$ $U$, i.e. A$ is similar to $D$. Find the eigenvalues and eigenvectors of $A$.




As $U$ is invertible and $D$ is diagonal I think that $A$ should also be a diagonal matrix.



And as $A$ is similar to $D$ then I suppose that $A$ has the same eigenvalues as $D$, namely ${lambda n}^{n}$.



Could someone tell me if I am on the right way?






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question









$endgroup$





Let $D$:= diag($lambda_1, ldots, lambda_n$), i.e., $D$ is a diagonal matrix in $mathbb{C}^{ntimes n}$ with entries $lambda_1, ldots, lambda_n$ ∈ $mathbb{C}$ on its diagonal.



Furter let $Uinmathbb{C}^{ntimes n}$ be an invertible matrix and define $A$ := $U^{−1} $D$ $U$, i.e. A$ is similar to $D$. Find the eigenvalues and eigenvectors of $A$.




As $U$ is invertible and $D$ is diagonal I think that $A$ should also be a diagonal matrix.



And as $A$ is similar to $D$ then I suppose that $A$ has the same eigenvalues as $D$, namely ${lambda n}^{n}$.



Could someone tell me if I am on the right way?






linear-algebra eigenvalues-eigenvectors




linear-algebra eigenvalues-eigenvectors






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 10:23









KaiKai

686




686








  • 2




    $begingroup$
    I can tell you that you are not in the right way. For instance, it's almost sure that $A$ will not be diagonal.
    $endgroup$
    – José Carlos Santos
    Jan 6 at 10:26










  • $begingroup$
    See en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization When you write $A=U^{-1}DU$, you have a diagonalization of $A$. The eigenvalues and eigenvectors should be quite obvious.
    $endgroup$
    – Jean-Claude Arbaut
    Jan 6 at 10:28
















  • 2




    $begingroup$
    I can tell you that you are not in the right way. For instance, it's almost sure that $A$ will not be diagonal.
    $endgroup$
    – José Carlos Santos
    Jan 6 at 10:26










  • $begingroup$
    See en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization When you write $A=U^{-1}DU$, you have a diagonalization of $A$. The eigenvalues and eigenvectors should be quite obvious.
    $endgroup$
    – Jean-Claude Arbaut
    Jan 6 at 10:28










2




2




$begingroup$
I can tell you that you are not in the right way. For instance, it's almost sure that $A$ will not be diagonal.
$endgroup$
– José Carlos Santos
Jan 6 at 10:26




$begingroup$
I can tell you that you are not in the right way. For instance, it's almost sure that $A$ will not be diagonal.
$endgroup$
– José Carlos Santos
Jan 6 at 10:26












$begingroup$
See en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization When you write $A=U^{-1}DU$, you have a diagonalization of $A$. The eigenvalues and eigenvectors should be quite obvious.
$endgroup$
– Jean-Claude Arbaut
Jan 6 at 10:28






$begingroup$
See en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization When you write $A=U^{-1}DU$, you have a diagonalization of $A$. The eigenvalues and eigenvectors should be quite obvious.
$endgroup$
– Jean-Claude Arbaut
Jan 6 at 10:28












1 Answer
1






active

oldest

votes


















2












$begingroup$

If $u_i$ is an eigenvector of $D$:
$$
Du_i=lambda_iu_i
$$
than, from:
$$
A=U^{-1}DURightarrow AU^{-1}=U^{-1}D$$



we have:
$$
A(U^{-1}u_i)=U^{-1}Du_i=U^{-1}lambda_iu_i=lambda_i(U^{-1}u_i)
$$



that gives the eigenvectors for the same eigenvalues.



Geometrically you can think at $U$ as a change of basis, so $A$ represents the same linear transformation of $D$ in a new basis, so its eigenvectors are the same, but expressed in the new basis.






share|cite|improve this answer









$endgroup$














    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%2f3063689%2feigenvalues-and-eigenvectors-of-diagonalisable-matrices%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    If $u_i$ is an eigenvector of $D$:
    $$
    Du_i=lambda_iu_i
    $$
    than, from:
    $$
    A=U^{-1}DURightarrow AU^{-1}=U^{-1}D$$



    we have:
    $$
    A(U^{-1}u_i)=U^{-1}Du_i=U^{-1}lambda_iu_i=lambda_i(U^{-1}u_i)
    $$



    that gives the eigenvectors for the same eigenvalues.



    Geometrically you can think at $U$ as a change of basis, so $A$ represents the same linear transformation of $D$ in a new basis, so its eigenvectors are the same, but expressed in the new basis.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      If $u_i$ is an eigenvector of $D$:
      $$
      Du_i=lambda_iu_i
      $$
      than, from:
      $$
      A=U^{-1}DURightarrow AU^{-1}=U^{-1}D$$



      we have:
      $$
      A(U^{-1}u_i)=U^{-1}Du_i=U^{-1}lambda_iu_i=lambda_i(U^{-1}u_i)
      $$



      that gives the eigenvectors for the same eigenvalues.



      Geometrically you can think at $U$ as a change of basis, so $A$ represents the same linear transformation of $D$ in a new basis, so its eigenvectors are the same, but expressed in the new basis.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        If $u_i$ is an eigenvector of $D$:
        $$
        Du_i=lambda_iu_i
        $$
        than, from:
        $$
        A=U^{-1}DURightarrow AU^{-1}=U^{-1}D$$



        we have:
        $$
        A(U^{-1}u_i)=U^{-1}Du_i=U^{-1}lambda_iu_i=lambda_i(U^{-1}u_i)
        $$



        that gives the eigenvectors for the same eigenvalues.



        Geometrically you can think at $U$ as a change of basis, so $A$ represents the same linear transformation of $D$ in a new basis, so its eigenvectors are the same, but expressed in the new basis.






        share|cite|improve this answer









        $endgroup$



        If $u_i$ is an eigenvector of $D$:
        $$
        Du_i=lambda_iu_i
        $$
        than, from:
        $$
        A=U^{-1}DURightarrow AU^{-1}=U^{-1}D$$



        we have:
        $$
        A(U^{-1}u_i)=U^{-1}Du_i=U^{-1}lambda_iu_i=lambda_i(U^{-1}u_i)
        $$



        that gives the eigenvectors for the same eigenvalues.



        Geometrically you can think at $U$ as a change of basis, so $A$ represents the same linear transformation of $D$ in a new basis, so its eigenvectors are the same, but expressed in the new basis.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 10:33









        Emilio NovatiEmilio Novati

        52.2k43574




        52.2k43574






























            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%2f3063689%2feigenvalues-and-eigenvectors-of-diagonalisable-matrices%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

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Image
            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
            Read more

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
            Read more