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Using eigenvectors as a basis for space



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Find an orthogonal basis consisting of eigenvectorsLooking for orthogonal basis of eigenvectors using Gram Schmidt processChange of basis help turning a sub-matrix from complex to realConflicting answers for eigenvectorsBasis of the null spaceEigenvectors for prime numbers matricesSymmetric Matrix , Eigenvectors are not orthogonal to the same eigenvalue.diagonalizable matrices using eigenvectorsBasis consisting of eigenvectorsFind basis of fundamental subspaces with given eigenvalues and eigenvectors












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$begingroup$


Consider the following matrix



$A=begin{bmatrix}0&0&0&0\a&0&0&0\0&b&0&0\0&0&c&0end{bmatrix}$



Which $a,b,c$ are real numbers



What conditions are required for $a,b,c$ such that $mathbb{R}^4$ has a basis made of A eigenvectors?










share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Consider the following matrix



    $A=begin{bmatrix}0&0&0&0\a&0&0&0\0&b&0&0\0&0&c&0end{bmatrix}$



    Which $a,b,c$ are real numbers



    What conditions are required for $a,b,c$ such that $mathbb{R}^4$ has a basis made of A eigenvectors?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Consider the following matrix



      $A=begin{bmatrix}0&0&0&0\a&0&0&0\0&b&0&0\0&0&c&0end{bmatrix}$



      Which $a,b,c$ are real numbers



      What conditions are required for $a,b,c$ such that $mathbb{R}^4$ has a basis made of A eigenvectors?










      share|cite|improve this question









      $endgroup$




      Consider the following matrix



      $A=begin{bmatrix}0&0&0&0\a&0&0&0\0&b&0&0\0&0&c&0end{bmatrix}$



      Which $a,b,c$ are real numbers



      What conditions are required for $a,b,c$ such that $mathbb{R}^4$ has a basis made of A eigenvectors?







      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 25 at 9:53









      Alireza HosseiniAlireza Hosseini

      233




      233






















          2 Answers
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          0












          $begingroup$

          The characteristic polynomial of $A$ is $X^4$, so the only eigenvalue is $0$.



          If $A$ is diagonalizable then, it means that $A=0$, so $a=b=c=0$.



          Conversely, if $a=b=c=0$, all vectors are eigenvectors, so of course you can find a basis made of eigenvectors.



          So the necessary and sufficient condition is that $a=b=c=0$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            $lambda = 0$ is the only eigenvalue (it has algebraic multiplicity 4). The only way the system $Av = 0$ has 4 degrees of freedom is if $A=0$, i.e. $a=b=c=0$.






            share|cite|improve this answer









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              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

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              active

              oldest

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              active

              oldest

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              0












              $begingroup$

              The characteristic polynomial of $A$ is $X^4$, so the only eigenvalue is $0$.



              If $A$ is diagonalizable then, it means that $A=0$, so $a=b=c=0$.



              Conversely, if $a=b=c=0$, all vectors are eigenvectors, so of course you can find a basis made of eigenvectors.



              So the necessary and sufficient condition is that $a=b=c=0$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The characteristic polynomial of $A$ is $X^4$, so the only eigenvalue is $0$.



                If $A$ is diagonalizable then, it means that $A=0$, so $a=b=c=0$.



                Conversely, if $a=b=c=0$, all vectors are eigenvectors, so of course you can find a basis made of eigenvectors.



                So the necessary and sufficient condition is that $a=b=c=0$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The characteristic polynomial of $A$ is $X^4$, so the only eigenvalue is $0$.



                  If $A$ is diagonalizable then, it means that $A=0$, so $a=b=c=0$.



                  Conversely, if $a=b=c=0$, all vectors are eigenvectors, so of course you can find a basis made of eigenvectors.



                  So the necessary and sufficient condition is that $a=b=c=0$.






                  share|cite|improve this answer









                  $endgroup$



                  The characteristic polynomial of $A$ is $X^4$, so the only eigenvalue is $0$.



                  If $A$ is diagonalizable then, it means that $A=0$, so $a=b=c=0$.



                  Conversely, if $a=b=c=0$, all vectors are eigenvectors, so of course you can find a basis made of eigenvectors.



                  So the necessary and sufficient condition is that $a=b=c=0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 25 at 9:59









                  TheSilverDoeTheSilverDoe

                  5,608316




                  5,608316























                      0












                      $begingroup$

                      $lambda = 0$ is the only eigenvalue (it has algebraic multiplicity 4). The only way the system $Av = 0$ has 4 degrees of freedom is if $A=0$, i.e. $a=b=c=0$.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        $lambda = 0$ is the only eigenvalue (it has algebraic multiplicity 4). The only way the system $Av = 0$ has 4 degrees of freedom is if $A=0$, i.e. $a=b=c=0$.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          $lambda = 0$ is the only eigenvalue (it has algebraic multiplicity 4). The only way the system $Av = 0$ has 4 degrees of freedom is if $A=0$, i.e. $a=b=c=0$.






                          share|cite|improve this answer









                          $endgroup$



                          $lambda = 0$ is the only eigenvalue (it has algebraic multiplicity 4). The only way the system $Av = 0$ has 4 degrees of freedom is if $A=0$, i.e. $a=b=c=0$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Mar 25 at 10:03









                          PierreCarrePierreCarre

                          2,253215




                          2,253215






























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