If $tilde{lambda}$ is an eigenvalue of $A$ of multiplicity $n$, then $A$ is a scalar matrix, where $A$ is $n...












1












$begingroup$



If $tilde{lambda}$ is an eigenvalue of $A$ of multiplicity $n$, then $A$ is a scalar matrix, where $A$ is $n times n$ real symmetric matrix.




So I know that $A = begin{bmatrix}a&b\b&cend{bmatrix} $
and that $$det(A-lambda I_2) = (a-λ)(c-λ) - b^2$$
However I don't really know how to proceed from there.






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    1. Dont assume A is 2$times$2. Also you know that det(A-$lambda$I) = $(x-lambda)^n$.
    $endgroup$
    – Joel Pereira
    Dec 2 '18 at 1:49


















1












$begingroup$



If $tilde{lambda}$ is an eigenvalue of $A$ of multiplicity $n$, then $A$ is a scalar matrix, where $A$ is $n times n$ real symmetric matrix.




So I know that $A = begin{bmatrix}a&b\b&cend{bmatrix} $
and that $$det(A-lambda I_2) = (a-λ)(c-λ) - b^2$$
However I don't really know how to proceed from there.






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    1. Dont assume A is 2$times$2. Also you know that det(A-$lambda$I) = $(x-lambda)^n$.
    $endgroup$
    – Joel Pereira
    Dec 2 '18 at 1:49
















1












1








1





$begingroup$



If $tilde{lambda}$ is an eigenvalue of $A$ of multiplicity $n$, then $A$ is a scalar matrix, where $A$ is $n times n$ real symmetric matrix.




So I know that $A = begin{bmatrix}a&b\b&cend{bmatrix} $
and that $$det(A-lambda I_2) = (a-λ)(c-λ) - b^2$$
However I don't really know how to proceed from there.






linear-algebra







share|cite|improve this question











$endgroup$





If $tilde{lambda}$ is an eigenvalue of $A$ of multiplicity $n$, then $A$ is a scalar matrix, where $A$ is $n times n$ real symmetric matrix.




So I know that $A = begin{bmatrix}a&b\b&cend{bmatrix} $
and that $$det(A-lambda I_2) = (a-λ)(c-λ) - b^2$$
However I don't really know how to proceed from there.






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 5:00









Brahadeesh

6,19742361




6,19742361










asked Dec 2 '18 at 1:43









Zhe TianZhe Tian

132




132












  • $begingroup$
    1. Dont assume A is 2$times$2. Also you know that det(A-$lambda$I) = $(x-lambda)^n$.
    $endgroup$
    – Joel Pereira
    Dec 2 '18 at 1:49




















  • $begingroup$
    1. Dont assume A is 2$times$2. Also you know that det(A-$lambda$I) = $(x-lambda)^n$.
    $endgroup$
    – Joel Pereira
    Dec 2 '18 at 1:49


















$begingroup$
1. Dont assume A is 2$times$2. Also you know that det(A-$lambda$I) = $(x-lambda)^n$.
$endgroup$
– Joel Pereira
Dec 2 '18 at 1:49






$begingroup$
1. Dont assume A is 2$times$2. Also you know that det(A-$lambda$I) = $(x-lambda)^n$.
$endgroup$
– Joel Pereira
Dec 2 '18 at 1:49












2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint:
If $A$ has an eigenvalue of multiplicity $n$, it is the only eigenvalue of $A$, so. Furthermore real symmetric matrices are diagonalisable, so its diagonal form is $lambda I$. Can you deduce that $A$ is $lambda I$?.



More details:
Since is real symmetric, there exists an orthogonal matrix $P$ and a diagonal matrix $D$ such that
$$A=PDP^{-1}=P(lambda I)P^{-1}=lambda(PIP^{-1})=lambda I.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sorry I'm still confused on what you mean?
    $endgroup$
    – Zhe Tian
    Dec 2 '18 at 2:14










  • $begingroup$
    @ZheTian: I've added details. Is that clearer?
    $endgroup$
    – Bernard
    Dec 2 '18 at 12:20



















0












$begingroup$

Your definition of scalar matrix is a diagonal matrix which every element in the diagonal are the same, right?



I assumed that the question should be "... then $A$ is a matrix similar to a scalar matrix". But you also can abuse notation and say that, in fact, $A$ "is'' a scalar matrix by the fact that similiar matrices can represent the same linear transformation...



But, if $A$ is a $ntimes n$ symmetric matrix, then $A$ is diagonalizable and there is a invertible $P$ such that $A=P DP^{-1}$, where $D=diag(lambda,cdots,lambda)$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Hint:
    If $A$ has an eigenvalue of multiplicity $n$, it is the only eigenvalue of $A$, so. Furthermore real symmetric matrices are diagonalisable, so its diagonal form is $lambda I$. Can you deduce that $A$ is $lambda I$?.



    More details:
    Since is real symmetric, there exists an orthogonal matrix $P$ and a diagonal matrix $D$ such that
    $$A=PDP^{-1}=P(lambda I)P^{-1}=lambda(PIP^{-1})=lambda I.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Sorry I'm still confused on what you mean?
      $endgroup$
      – Zhe Tian
      Dec 2 '18 at 2:14










    • $begingroup$
      @ZheTian: I've added details. Is that clearer?
      $endgroup$
      – Bernard
      Dec 2 '18 at 12:20
















    1












    $begingroup$

    Hint:
    If $A$ has an eigenvalue of multiplicity $n$, it is the only eigenvalue of $A$, so. Furthermore real symmetric matrices are diagonalisable, so its diagonal form is $lambda I$. Can you deduce that $A$ is $lambda I$?.



    More details:
    Since is real symmetric, there exists an orthogonal matrix $P$ and a diagonal matrix $D$ such that
    $$A=PDP^{-1}=P(lambda I)P^{-1}=lambda(PIP^{-1})=lambda I.$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Sorry I'm still confused on what you mean?
      $endgroup$
      – Zhe Tian
      Dec 2 '18 at 2:14










    • $begingroup$
      @ZheTian: I've added details. Is that clearer?
      $endgroup$
      – Bernard
      Dec 2 '18 at 12:20














    1












    1








    1





    $begingroup$

    Hint:
    If $A$ has an eigenvalue of multiplicity $n$, it is the only eigenvalue of $A$, so. Furthermore real symmetric matrices are diagonalisable, so its diagonal form is $lambda I$. Can you deduce that $A$ is $lambda I$?.



    More details:
    Since is real symmetric, there exists an orthogonal matrix $P$ and a diagonal matrix $D$ such that
    $$A=PDP^{-1}=P(lambda I)P^{-1}=lambda(PIP^{-1})=lambda I.$$






    share|cite|improve this answer











    $endgroup$



    Hint:
    If $A$ has an eigenvalue of multiplicity $n$, it is the only eigenvalue of $A$, so. Furthermore real symmetric matrices are diagonalisable, so its diagonal form is $lambda I$. Can you deduce that $A$ is $lambda I$?.



    More details:
    Since is real symmetric, there exists an orthogonal matrix $P$ and a diagonal matrix $D$ such that
    $$A=PDP^{-1}=P(lambda I)P^{-1}=lambda(PIP^{-1})=lambda I.$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 2 '18 at 12:20

























    answered Dec 2 '18 at 2:00









    BernardBernard

    119k639112




    119k639112












    • $begingroup$
      Sorry I'm still confused on what you mean?
      $endgroup$
      – Zhe Tian
      Dec 2 '18 at 2:14










    • $begingroup$
      @ZheTian: I've added details. Is that clearer?
      $endgroup$
      – Bernard
      Dec 2 '18 at 12:20


















    • $begingroup$
      Sorry I'm still confused on what you mean?
      $endgroup$
      – Zhe Tian
      Dec 2 '18 at 2:14










    • $begingroup$
      @ZheTian: I've added details. Is that clearer?
      $endgroup$
      – Bernard
      Dec 2 '18 at 12:20
















    $begingroup$
    Sorry I'm still confused on what you mean?
    $endgroup$
    – Zhe Tian
    Dec 2 '18 at 2:14




    $begingroup$
    Sorry I'm still confused on what you mean?
    $endgroup$
    – Zhe Tian
    Dec 2 '18 at 2:14












    $begingroup$
    @ZheTian: I've added details. Is that clearer?
    $endgroup$
    – Bernard
    Dec 2 '18 at 12:20




    $begingroup$
    @ZheTian: I've added details. Is that clearer?
    $endgroup$
    – Bernard
    Dec 2 '18 at 12:20











    0












    $begingroup$

    Your definition of scalar matrix is a diagonal matrix which every element in the diagonal are the same, right?



    I assumed that the question should be "... then $A$ is a matrix similar to a scalar matrix". But you also can abuse notation and say that, in fact, $A$ "is'' a scalar matrix by the fact that similiar matrices can represent the same linear transformation...



    But, if $A$ is a $ntimes n$ symmetric matrix, then $A$ is diagonalizable and there is a invertible $P$ such that $A=P DP^{-1}$, where $D=diag(lambda,cdots,lambda)$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Your definition of scalar matrix is a diagonal matrix which every element in the diagonal are the same, right?



      I assumed that the question should be "... then $A$ is a matrix similar to a scalar matrix". But you also can abuse notation and say that, in fact, $A$ "is'' a scalar matrix by the fact that similiar matrices can represent the same linear transformation...



      But, if $A$ is a $ntimes n$ symmetric matrix, then $A$ is diagonalizable and there is a invertible $P$ such that $A=P DP^{-1}$, where $D=diag(lambda,cdots,lambda)$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your definition of scalar matrix is a diagonal matrix which every element in the diagonal are the same, right?



        I assumed that the question should be "... then $A$ is a matrix similar to a scalar matrix". But you also can abuse notation and say that, in fact, $A$ "is'' a scalar matrix by the fact that similiar matrices can represent the same linear transformation...



        But, if $A$ is a $ntimes n$ symmetric matrix, then $A$ is diagonalizable and there is a invertible $P$ such that $A=P DP^{-1}$, where $D=diag(lambda,cdots,lambda)$.






        share|cite|improve this answer









        $endgroup$



        Your definition of scalar matrix is a diagonal matrix which every element in the diagonal are the same, right?



        I assumed that the question should be "... then $A$ is a matrix similar to a scalar matrix". But you also can abuse notation and say that, in fact, $A$ "is'' a scalar matrix by the fact that similiar matrices can represent the same linear transformation...



        But, if $A$ is a $ntimes n$ symmetric matrix, then $A$ is diagonalizable and there is a invertible $P$ such that $A=P DP^{-1}$, where $D=diag(lambda,cdots,lambda)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 4:43









        RobsonRobson

        769221




        769221






























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