Finding eigenvector only knowing others eigenvectors.












3












$begingroup$


The matrix $A in M_3(mathbb{R})$ satisfy $A^t=A$ and $(1,2,1), (-1,1,0)$ are eigenvectors of $A$. Which vector is also an eigenvector of $A$?
Alternatives: $(0,0,1)$; $(1,1,-3)$; $(1,1,3)$; There is no other eigenvector.



The problem with this exercise is that I don't know the matrix $A$, and I don't have any eigenvalue to start with. I can get a matrix with less variables using $A = A^t$, but there's still 6 variables. Any tips or guidance is appreciated.






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    The matrix $A in M_3(mathbb{R})$ satisfy $A^t=A$ and $(1,2,1), (-1,1,0)$ are eigenvectors of $A$. Which vector is also an eigenvector of $A$?
    Alternatives: $(0,0,1)$; $(1,1,-3)$; $(1,1,3)$; There is no other eigenvector.



    The problem with this exercise is that I don't know the matrix $A$, and I don't have any eigenvalue to start with. I can get a matrix with less variables using $A = A^t$, but there's still 6 variables. Any tips or guidance is appreciated.






    linear-algebra eigenvalues-eigenvectors







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      The matrix $A in M_3(mathbb{R})$ satisfy $A^t=A$ and $(1,2,1), (-1,1,0)$ are eigenvectors of $A$. Which vector is also an eigenvector of $A$?
      Alternatives: $(0,0,1)$; $(1,1,-3)$; $(1,1,3)$; There is no other eigenvector.



      The problem with this exercise is that I don't know the matrix $A$, and I don't have any eigenvalue to start with. I can get a matrix with less variables using $A = A^t$, but there's still 6 variables. Any tips or guidance is appreciated.






      linear-algebra eigenvalues-eigenvectors







      share|cite|improve this question









      $endgroup$




      The matrix $A in M_3(mathbb{R})$ satisfy $A^t=A$ and $(1,2,1), (-1,1,0)$ are eigenvectors of $A$. Which vector is also an eigenvector of $A$?
      Alternatives: $(0,0,1)$; $(1,1,-3)$; $(1,1,3)$; There is no other eigenvector.



      The problem with this exercise is that I don't know the matrix $A$, and I don't have any eigenvalue to start with. I can get a matrix with less variables using $A = A^t$, but there's still 6 variables. Any tips or guidance is appreciated.






      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 1 hour ago









      rodorgasrodorgas

      1915




      1915






















          2 Answers
          2






          active

          oldest

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          2












          $begingroup$

          Since $A$ is symmetric, the eigenvectors (for distinct eigenvalues) are orthogonal.



          So, find which of the vectors is orthogonal to the first two.




          (1,1,-3) is.







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit.
            $endgroup$
            – Chris Custer
            58 mins ago






          • 1




            $begingroup$
            The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this.
            $endgroup$
            – amd
            51 mins ago










          • $begingroup$
            @amd true. Good catch.
            $endgroup$
            – Chris Custer
            45 mins ago










          • $begingroup$
            @amd the orthogonal one must be an eigenvector though, if I'm not mistaken.
            $endgroup$
            – Chris Custer
            35 mins ago












          • $begingroup$
            @ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector"
            $endgroup$
            – angryavian
            14 mins ago



















          4












          $begingroup$

          Hint: the condition $A^t = A$ allows you to use the spectral theorem.




          Hint: Specifically, the spectral theorem implies there is an orthonormal basis of eigenvectors of $A$.







          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
            2






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Since $A$ is symmetric, the eigenvectors (for distinct eigenvalues) are orthogonal.



            So, find which of the vectors is orthogonal to the first two.




            (1,1,-3) is.







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit.
              $endgroup$
              – Chris Custer
              58 mins ago






            • 1




              $begingroup$
              The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this.
              $endgroup$
              – amd
              51 mins ago










            • $begingroup$
              @amd true. Good catch.
              $endgroup$
              – Chris Custer
              45 mins ago










            • $begingroup$
              @amd the orthogonal one must be an eigenvector though, if I'm not mistaken.
              $endgroup$
              – Chris Custer
              35 mins ago












            • $begingroup$
              @ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector"
              $endgroup$
              – angryavian
              14 mins ago
















            2












            $begingroup$

            Since $A$ is symmetric, the eigenvectors (for distinct eigenvalues) are orthogonal.



            So, find which of the vectors is orthogonal to the first two.




            (1,1,-3) is.







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit.
              $endgroup$
              – Chris Custer
              58 mins ago






            • 1




              $begingroup$
              The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this.
              $endgroup$
              – amd
              51 mins ago










            • $begingroup$
              @amd true. Good catch.
              $endgroup$
              – Chris Custer
              45 mins ago










            • $begingroup$
              @amd the orthogonal one must be an eigenvector though, if I'm not mistaken.
              $endgroup$
              – Chris Custer
              35 mins ago












            • $begingroup$
              @ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector"
              $endgroup$
              – angryavian
              14 mins ago














            2












            2








            2





            $begingroup$

            Since $A$ is symmetric, the eigenvectors (for distinct eigenvalues) are orthogonal.



            So, find which of the vectors is orthogonal to the first two.




            (1,1,-3) is.







            share|cite|improve this answer











            $endgroup$



            Since $A$ is symmetric, the eigenvectors (for distinct eigenvalues) are orthogonal.



            So, find which of the vectors is orthogonal to the first two.




            (1,1,-3) is.








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 56 mins ago

























            answered 1 hour ago









            Chris CusterChris Custer

            13.5k3827




            13.5k3827












            • $begingroup$
              I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit.
              $endgroup$
              – Chris Custer
              58 mins ago






            • 1




              $begingroup$
              The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this.
              $endgroup$
              – amd
              51 mins ago










            • $begingroup$
              @amd true. Good catch.
              $endgroup$
              – Chris Custer
              45 mins ago










            • $begingroup$
              @amd the orthogonal one must be an eigenvector though, if I'm not mistaken.
              $endgroup$
              – Chris Custer
              35 mins ago












            • $begingroup$
              @ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector"
              $endgroup$
              – angryavian
              14 mins ago


















            • $begingroup$
              I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit.
              $endgroup$
              – Chris Custer
              58 mins ago






            • 1




              $begingroup$
              The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this.
              $endgroup$
              – amd
              51 mins ago










            • $begingroup$
              @amd true. Good catch.
              $endgroup$
              – Chris Custer
              45 mins ago










            • $begingroup$
              @amd the orthogonal one must be an eigenvector though, if I'm not mistaken.
              $endgroup$
              – Chris Custer
              35 mins ago












            • $begingroup$
              @ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector"
              $endgroup$
              – angryavian
              14 mins ago
















            $begingroup$
            I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit.
            $endgroup$
            – Chris Custer
            58 mins ago




            $begingroup$
            I think I need to rethink it. I'm leaning towards $(1,1,-3)$ because it's orthogonal to both. Will edit.
            $endgroup$
            – Chris Custer
            58 mins ago




            1




            1




            $begingroup$
            The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this.
            $endgroup$
            – amd
            51 mins ago




            $begingroup$
            The other two could be eigenvectors. If so, the matrix would have to be a multiple of the identity, but I don’t see anything in the problem statement that would preclude this.
            $endgroup$
            – amd
            51 mins ago












            $begingroup$
            @amd true. Good catch.
            $endgroup$
            – Chris Custer
            45 mins ago




            $begingroup$
            @amd true. Good catch.
            $endgroup$
            – Chris Custer
            45 mins ago












            $begingroup$
            @amd the orthogonal one must be an eigenvector though, if I'm not mistaken.
            $endgroup$
            – Chris Custer
            35 mins ago






            $begingroup$
            @amd the orthogonal one must be an eigenvector though, if I'm not mistaken.
            $endgroup$
            – Chris Custer
            35 mins ago














            $begingroup$
            @ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector"
            $endgroup$
            – angryavian
            14 mins ago




            $begingroup$
            @ChrisCuster Yeah the problem could probably have been better stated as "which vector is also necessarily an eigenvector"
            $endgroup$
            – angryavian
            14 mins ago











            4












            $begingroup$

            Hint: the condition $A^t = A$ allows you to use the spectral theorem.




            Hint: Specifically, the spectral theorem implies there is an orthonormal basis of eigenvectors of $A$.







            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              Hint: the condition $A^t = A$ allows you to use the spectral theorem.




              Hint: Specifically, the spectral theorem implies there is an orthonormal basis of eigenvectors of $A$.







              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                Hint: the condition $A^t = A$ allows you to use the spectral theorem.




                Hint: Specifically, the spectral theorem implies there is an orthonormal basis of eigenvectors of $A$.







                share|cite|improve this answer









                $endgroup$



                Hint: the condition $A^t = A$ allows you to use the spectral theorem.




                Hint: Specifically, the spectral theorem implies there is an orthonormal basis of eigenvectors of $A$.








                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                angryavianangryavian

                41.2k23380




                41.2k23380






























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