Show y*x=0 by eigenvectors [closed]












0












$begingroup$


Let $A$ be a square matrix, $lambda$ and $mu$ be two distinct eigenvalues of $A$.



If $x$ is an eigenvector corresponding to $lambda$, $y^∗$ is a nonzero row vector such that $y^∗ (A − mu I) = 0$, show that $y^∗ x = 0$.



can someone give me some direction to solve this problem?






linear-algebra







share|cite|improve this question











$endgroup$



closed as off-topic by Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos Nov 29 '18 at 12:03


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    "Can I simply prove it by letting a square matrix?" ... letting a square matrix WHAT?
    $endgroup$
    – David G. Stork
    Nov 29 '18 at 1:53
















0












$begingroup$


Let $A$ be a square matrix, $lambda$ and $mu$ be two distinct eigenvalues of $A$.



If $x$ is an eigenvector corresponding to $lambda$, $y^∗$ is a nonzero row vector such that $y^∗ (A − mu I) = 0$, show that $y^∗ x = 0$.



can someone give me some direction to solve this problem?






linear-algebra







share|cite|improve this question











$endgroup$



closed as off-topic by Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos Nov 29 '18 at 12:03


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    "Can I simply prove it by letting a square matrix?" ... letting a square matrix WHAT?
    $endgroup$
    – David G. Stork
    Nov 29 '18 at 1:53














0












0








0





$begingroup$


Let $A$ be a square matrix, $lambda$ and $mu$ be two distinct eigenvalues of $A$.



If $x$ is an eigenvector corresponding to $lambda$, $y^∗$ is a nonzero row vector such that $y^∗ (A − mu I) = 0$, show that $y^∗ x = 0$.



can someone give me some direction to solve this problem?






linear-algebra







share|cite|improve this question











$endgroup$




Let $A$ be a square matrix, $lambda$ and $mu$ be two distinct eigenvalues of $A$.



If $x$ is an eigenvector corresponding to $lambda$, $y^∗$ is a nonzero row vector such that $y^∗ (A − mu I) = 0$, show that $y^∗ x = 0$.



can someone give me some direction to solve this problem?






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 Nov 29 '18 at 1:59









Siong Thye Goh

101k1466117




101k1466117










asked Nov 29 '18 at 1:52









CColaCCola

346




346




closed as off-topic by Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos Nov 29 '18 at 12:03


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos Nov 29 '18 at 12:03


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, José Carlos Santos, Ali Caglayan, amWhy, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    "Can I simply prove it by letting a square matrix?" ... letting a square matrix WHAT?
    $endgroup$
    – David G. Stork
    Nov 29 '18 at 1:53


















  • $begingroup$
    "Can I simply prove it by letting a square matrix?" ... letting a square matrix WHAT?
    $endgroup$
    – David G. Stork
    Nov 29 '18 at 1:53
















$begingroup$
"Can I simply prove it by letting a square matrix?" ... letting a square matrix WHAT?
$endgroup$
– David G. Stork
Nov 29 '18 at 1:53




$begingroup$
"Can I simply prove it by letting a square matrix?" ... letting a square matrix WHAT?
$endgroup$
– David G. Stork
Nov 29 '18 at 1:53










3 Answers
3






active

oldest

votes


















1












$begingroup$

I will assume I is the identity matrix. Since y*(A−μI)=0 then distributing yields y$^*A$=y$^*μI$ , since x is a eigen vector we have y$^*$Ax-y$^*$μIx=0$
so you need to know the fact that if you have a matrix A and a vector v such that Av=0 then A=0 or v=0.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok thanks! i will try to approach in this way
    $endgroup$
    – CCola
    Nov 29 '18 at 4:11










  • $begingroup$
    No problem, if my answer was helpful I would appreciate it if you click on the check mark
    $endgroup$
    – user612135
    Nov 29 '18 at 4:15



















1












$begingroup$

Hint:



Use the property that
$$y^*(Ax)=(y^*A)x.$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    We may as well start with $y^*(A-mu I) = 0$ and see what happens:
    $$
    begin{align*}
    0 &= y^*(A - mu I)\
    &= y^*(A - mu I)x\
    &= y^*(Ax - mu x)\
    &= ldots
    end{align*}
    $$
    ... can you take it from here?




    spoiler alert




    $$0 = y^*(Ax - mu x) = y^*(lambda x - mu x) = y^*x(lambda - mu)$$ and as $lambda neq mu$ we must have $y^*x = 0$.







    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      so we get y*=0, x=0 or (A-μ)=0?
      $endgroup$
      – CCola
      Nov 29 '18 at 5:37


















    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    I will assume I is the identity matrix. Since y*(A−μI)=0 then distributing yields y$^*A$=y$^*μI$ , since x is a eigen vector we have y$^*$Ax-y$^*$μIx=0$
    so you need to know the fact that if you have a matrix A and a vector v such that Av=0 then A=0 or v=0.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      ok thanks! i will try to approach in this way
      $endgroup$
      – CCola
      Nov 29 '18 at 4:11










    • $begingroup$
      No problem, if my answer was helpful I would appreciate it if you click on the check mark
      $endgroup$
      – user612135
      Nov 29 '18 at 4:15
















    1












    $begingroup$

    I will assume I is the identity matrix. Since y*(A−μI)=0 then distributing yields y$^*A$=y$^*μI$ , since x is a eigen vector we have y$^*$Ax-y$^*$μIx=0$
    so you need to know the fact that if you have a matrix A and a vector v such that Av=0 then A=0 or v=0.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      ok thanks! i will try to approach in this way
      $endgroup$
      – CCola
      Nov 29 '18 at 4:11










    • $begingroup$
      No problem, if my answer was helpful I would appreciate it if you click on the check mark
      $endgroup$
      – user612135
      Nov 29 '18 at 4:15














    1












    1








    1





    $begingroup$

    I will assume I is the identity matrix. Since y*(A−μI)=0 then distributing yields y$^*A$=y$^*μI$ , since x is a eigen vector we have y$^*$Ax-y$^*$μIx=0$
    so you need to know the fact that if you have a matrix A and a vector v such that Av=0 then A=0 or v=0.






    share|cite|improve this answer









    $endgroup$



    I will assume I is the identity matrix. Since y*(A−μI)=0 then distributing yields y$^*A$=y$^*μI$ , since x is a eigen vector we have y$^*$Ax-y$^*$μIx=0$
    so you need to know the fact that if you have a matrix A and a vector v such that Av=0 then A=0 or v=0.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 29 '18 at 2:10







    user612135



















    • $begingroup$
      ok thanks! i will try to approach in this way
      $endgroup$
      – CCola
      Nov 29 '18 at 4:11










    • $begingroup$
      No problem, if my answer was helpful I would appreciate it if you click on the check mark
      $endgroup$
      – user612135
      Nov 29 '18 at 4:15


















    • $begingroup$
      ok thanks! i will try to approach in this way
      $endgroup$
      – CCola
      Nov 29 '18 at 4:11










    • $begingroup$
      No problem, if my answer was helpful I would appreciate it if you click on the check mark
      $endgroup$
      – user612135
      Nov 29 '18 at 4:15
















    $begingroup$
    ok thanks! i will try to approach in this way
    $endgroup$
    – CCola
    Nov 29 '18 at 4:11




    $begingroup$
    ok thanks! i will try to approach in this way
    $endgroup$
    – CCola
    Nov 29 '18 at 4:11












    $begingroup$
    No problem, if my answer was helpful I would appreciate it if you click on the check mark
    $endgroup$
    – user612135
    Nov 29 '18 at 4:15




    $begingroup$
    No problem, if my answer was helpful I would appreciate it if you click on the check mark
    $endgroup$
    – user612135
    Nov 29 '18 at 4:15











    1












    $begingroup$

    Hint:



    Use the property that
    $$y^*(Ax)=(y^*A)x.$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hint:



      Use the property that
      $$y^*(Ax)=(y^*A)x.$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint:



        Use the property that
        $$y^*(Ax)=(y^*A)x.$$






        share|cite|improve this answer









        $endgroup$



        Hint:



        Use the property that
        $$y^*(Ax)=(y^*A)x.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 1:57









        Siong Thye GohSiong Thye Goh

        101k1466117




        101k1466117























            0












            $begingroup$

            We may as well start with $y^*(A-mu I) = 0$ and see what happens:
            $$
            begin{align*}
            0 &= y^*(A - mu I)\
            &= y^*(A - mu I)x\
            &= y^*(Ax - mu x)\
            &= ldots
            end{align*}
            $$
            ... can you take it from here?




            spoiler alert




            $$0 = y^*(Ax - mu x) = y^*(lambda x - mu x) = y^*x(lambda - mu)$$ and as $lambda neq mu$ we must have $y^*x = 0$.







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              so we get y*=0, x=0 or (A-μ)=0?
              $endgroup$
              – CCola
              Nov 29 '18 at 5:37
















            0












            $begingroup$

            We may as well start with $y^*(A-mu I) = 0$ and see what happens:
            $$
            begin{align*}
            0 &= y^*(A - mu I)\
            &= y^*(A - mu I)x\
            &= y^*(Ax - mu x)\
            &= ldots
            end{align*}
            $$
            ... can you take it from here?




            spoiler alert




            $$0 = y^*(Ax - mu x) = y^*(lambda x - mu x) = y^*x(lambda - mu)$$ and as $lambda neq mu$ we must have $y^*x = 0$.







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              so we get y*=0, x=0 or (A-μ)=0?
              $endgroup$
              – CCola
              Nov 29 '18 at 5:37














            0












            0








            0





            $begingroup$

            We may as well start with $y^*(A-mu I) = 0$ and see what happens:
            $$
            begin{align*}
            0 &= y^*(A - mu I)\
            &= y^*(A - mu I)x\
            &= y^*(Ax - mu x)\
            &= ldots
            end{align*}
            $$
            ... can you take it from here?




            spoiler alert




            $$0 = y^*(Ax - mu x) = y^*(lambda x - mu x) = y^*x(lambda - mu)$$ and as $lambda neq mu$ we must have $y^*x = 0$.







            share|cite|improve this answer











            $endgroup$



            We may as well start with $y^*(A-mu I) = 0$ and see what happens:
            $$
            begin{align*}
            0 &= y^*(A - mu I)\
            &= y^*(A - mu I)x\
            &= y^*(Ax - mu x)\
            &= ldots
            end{align*}
            $$
            ... can you take it from here?




            spoiler alert




            $$0 = y^*(Ax - mu x) = y^*(lambda x - mu x) = y^*x(lambda - mu)$$ and as $lambda neq mu$ we must have $y^*x = 0$.








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 29 '18 at 13:16

























            answered Nov 29 '18 at 2:00









            NealNeal

            23.6k23886




            23.6k23886












            • $begingroup$
              so we get y*=0, x=0 or (A-μ)=0?
              $endgroup$
              – CCola
              Nov 29 '18 at 5:37


















            • $begingroup$
              so we get y*=0, x=0 or (A-μ)=0?
              $endgroup$
              – CCola
              Nov 29 '18 at 5:37
















            $begingroup$
            so we get y*=0, x=0 or (A-μ)=0?
            $endgroup$
            – CCola
            Nov 29 '18 at 5:37




            $begingroup$
            so we get y*=0, x=0 or (A-μ)=0?
            $endgroup$
            – CCola
            Nov 29 '18 at 5:37



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