Is every set of independent eigenvectors of an orthogonally diagonalizable matrix orthogonal?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












Let $f : V rightarrow V$ be a linear operator, where $V$ is a finite dimensional inner product space over $mathbbC$ of dimension $n$.
Suppose we are given that $f$ has $n$ orthogonal eigenvectors $u_1ldots,u_n$, corresponding to the eigenvalues $lambda_1,ldots,lambda_n$.
Now suppose $v_1,ldots,v_n$ is another set of linearly independent eigenvectors of $f$ for the same eigenvalues $lambda_1,ldots,lambda_n$ respectively.



Question: Are these "new" eigenvectors also orthogonal?



If yes, please provide a proof. If not, a counterexample.



To avoid the trivial case, assume that at $lambda_1,ldots,lambda_n$ are NOT pairwise distinct.






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question





















  • This reads like a verbatim homework assignment. Did you try anything at all?
    – Umberto P.
    Sep 6 at 11:55














up vote
1
down vote

favorite












Let $f : V rightarrow V$ be a linear operator, where $V$ is a finite dimensional inner product space over $mathbbC$ of dimension $n$.
Suppose we are given that $f$ has $n$ orthogonal eigenvectors $u_1ldots,u_n$, corresponding to the eigenvalues $lambda_1,ldots,lambda_n$.
Now suppose $v_1,ldots,v_n$ is another set of linearly independent eigenvectors of $f$ for the same eigenvalues $lambda_1,ldots,lambda_n$ respectively.



Question: Are these "new" eigenvectors also orthogonal?



If yes, please provide a proof. If not, a counterexample.



To avoid the trivial case, assume that at $lambda_1,ldots,lambda_n$ are NOT pairwise distinct.






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question





















  • This reads like a verbatim homework assignment. Did you try anything at all?
    – Umberto P.
    Sep 6 at 11:55












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $f : V rightarrow V$ be a linear operator, where $V$ is a finite dimensional inner product space over $mathbbC$ of dimension $n$.
Suppose we are given that $f$ has $n$ orthogonal eigenvectors $u_1ldots,u_n$, corresponding to the eigenvalues $lambda_1,ldots,lambda_n$.
Now suppose $v_1,ldots,v_n$ is another set of linearly independent eigenvectors of $f$ for the same eigenvalues $lambda_1,ldots,lambda_n$ respectively.



Question: Are these "new" eigenvectors also orthogonal?



If yes, please provide a proof. If not, a counterexample.



To avoid the trivial case, assume that at $lambda_1,ldots,lambda_n$ are NOT pairwise distinct.






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question













Let $f : V rightarrow V$ be a linear operator, where $V$ is a finite dimensional inner product space over $mathbbC$ of dimension $n$.
Suppose we are given that $f$ has $n$ orthogonal eigenvectors $u_1ldots,u_n$, corresponding to the eigenvalues $lambda_1,ldots,lambda_n$.
Now suppose $v_1,ldots,v_n$ is another set of linearly independent eigenvectors of $f$ for the same eigenvalues $lambda_1,ldots,lambda_n$ respectively.



Question: Are these "new" eigenvectors also orthogonal?



If yes, please provide a proof. If not, a counterexample.



To avoid the trivial case, assume that at $lambda_1,ldots,lambda_n$ are NOT pairwise distinct.






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 Sep 6 at 11:48









EpsilonDelta

1107




1107











  • This reads like a verbatim homework assignment. Did you try anything at all?
    – Umberto P.
    Sep 6 at 11:55
















  • This reads like a verbatim homework assignment. Did you try anything at all?
    – Umberto P.
    Sep 6 at 11:55















This reads like a verbatim homework assignment. Did you try anything at all?
– Umberto P.
Sep 6 at 11:55




This reads like a verbatim homework assignment. Did you try anything at all?
– Umberto P.
Sep 6 at 11:55










1 Answer
1






active

oldest

votes

















up vote
3
down vote



accepted










Consider the $2 times 2$ identity matrix, and the standard basis. Then look at any other basis.






share|cite|improve this answer




















    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',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    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%2f2907382%2fis-every-set-of-independent-eigenvectors-of-an-orthogonally-diagonalizable-matri%23new-answer', 'question_page');

    );

    Post as a guest

















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    Consider the $2 times 2$ identity matrix, and the standard basis. Then look at any other basis.






    share|cite|improve this answer
























      up vote
      3
      down vote



      accepted










      Consider the $2 times 2$ identity matrix, and the standard basis. Then look at any other basis.






      share|cite|improve this answer






















        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        Consider the $2 times 2$ identity matrix, and the standard basis. Then look at any other basis.






        share|cite|improve this answer












        Consider the $2 times 2$ identity matrix, and the standard basis. Then look at any other basis.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 6 at 11:51









        John Hughes

        60k23987




        60k23987



























             

            draft saved


            draft discarded















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2907382%2fis-every-set-of-independent-eigenvectors-of-an-orthogonally-diagonalizable-matri%23new-answer', 'question_page');

            );

            Post as a guest
































































            -

            這個網誌中的熱門文章

            How to combine Bézier curves to a surface?

            圖片
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
            閱讀完整內容

            Mutual Information Always Non-negative

            圖片
            Clash Royale CLAN TAG #URR8PPP up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 '12 at 15:38 Cameron Buie 83.6k 7 71 154 asked Jun 17 '12 at 15:30 Omri 359 3 13 3 Convexity of the function $tmapsto tlog t$. – Did Jun 17 '12 at 15:33 In addition, the convexity properties require the coefficients in the linear combination sum 1. Then, as p(x,y) is a probability distribution, it fullfits such condition. – Francisco Javier Delgado Ceped Aug 10 at 18:44 add a comment  |  up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 ...
            閱讀完整內容

            Why am i infinitely getting the same tweet with the Twitter Search API?

            圖片
            up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia...
            閱讀完整內容