sum of two linearly independent eigenvector












0












$begingroup$


'' The sum of two linearly independent eigenvectors is never an eigenvector. ''




  • This is true?

  • if i assume the opposite, then it means the eigenvector can be written as a combination linear of the 2 other eigenvector (coefficient 1 of each), which won't make them linearly independent, which means they wouldn't be eigenvectors in the first place, is this correct?






linear-algebra







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  • $begingroup$
    Rolled back to the original because the edit mangled the question so it no longer made sense.
    $endgroup$
    – Bungo
    Jan 29 at 6:48
















0












$begingroup$


'' The sum of two linearly independent eigenvectors is never an eigenvector. ''




  • This is true?

  • if i assume the opposite, then it means the eigenvector can be written as a combination linear of the 2 other eigenvector (coefficient 1 of each), which won't make them linearly independent, which means they wouldn't be eigenvectors in the first place, is this correct?






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Rolled back to the original because the edit mangled the question so it no longer made sense.
    $endgroup$
    – Bungo
    Jan 29 at 6:48














0












0








0





$begingroup$


'' The sum of two linearly independent eigenvectors is never an eigenvector. ''




  • This is true?

  • if i assume the opposite, then it means the eigenvector can be written as a combination linear of the 2 other eigenvector (coefficient 1 of each), which won't make them linearly independent, which means they wouldn't be eigenvectors in the first place, is this correct?






linear-algebra







share|cite|improve this question











$endgroup$




'' The sum of two linearly independent eigenvectors is never an eigenvector. ''




  • This is true?

  • if i assume the opposite, then it means the eigenvector can be written as a combination linear of the 2 other eigenvector (coefficient 1 of each), which won't make them linearly independent, which means they wouldn't be eigenvectors in the first place, is this correct?






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 Jan 29 at 6:47









Bungo

13.7k22148




13.7k22148










asked Jan 29 at 3:44









ximxim

516




516












  • $begingroup$
    Rolled back to the original because the edit mangled the question so it no longer made sense.
    $endgroup$
    – Bungo
    Jan 29 at 6:48


















  • $begingroup$
    Rolled back to the original because the edit mangled the question so it no longer made sense.
    $endgroup$
    – Bungo
    Jan 29 at 6:48
















$begingroup$
Rolled back to the original because the edit mangled the question so it no longer made sense.
$endgroup$
– Bungo
Jan 29 at 6:48




$begingroup$
Rolled back to the original because the edit mangled the question so it no longer made sense.
$endgroup$
– Bungo
Jan 29 at 6:48










1 Answer
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$begingroup$

Consider the identity matrix, every single non-zero vector of compatible size is an eigenvector.



Hence the statement is false.






share|cite|improve this answer











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    $begingroup$

    Consider the identity matrix, every single non-zero vector of compatible size is an eigenvector.



    Hence the statement is false.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Consider the identity matrix, every single non-zero vector of compatible size is an eigenvector.



      Hence the statement is false.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Consider the identity matrix, every single non-zero vector of compatible size is an eigenvector.



        Hence the statement is false.






        share|cite|improve this answer











        $endgroup$



        Consider the identity matrix, every single non-zero vector of compatible size is an eigenvector.



        Hence the statement is false.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 29 at 6:04

























        answered Jan 29 at 3:48









        Siong Thye GohSiong Thye Goh

        103k1468119




        103k1468119






























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