What is an elegant way to find the third eigenvalue of $A=left[begin{smallmatrix} 51&-12 & -21\ 60...












4












$begingroup$


We had this in an exam today: Given the matrix
$$A=begin{bmatrix}
51&-12 & -21\
60 & -40&28\
57&-68&1
end{bmatrix}$$
Here is the precise question as asked in our exam:




Someone tells you (accurately) that $-48$ and $24$ are eigenvalues of the matrix $A$. Without using a computer, calculator or writing anything down find the third eigenvalue of $A$.




I have used the classical method (through the characteristic polynomial of A) but this leaded to a some tremendous computation and finally got the answer.



But the question intentionally asked to do not make any use of such computation.



How can one elegantly find the third eigenvalue of $A$?










share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    We had this in an exam today: Given the matrix
    $$A=begin{bmatrix}
    51&-12 & -21\
    60 & -40&28\
    57&-68&1
    end{bmatrix}$$
    Here is the precise question as asked in our exam:




    Someone tells you (accurately) that $-48$ and $24$ are eigenvalues of the matrix $A$. Without using a computer, calculator or writing anything down find the third eigenvalue of $A$.




    I have used the classical method (through the characteristic polynomial of A) but this leaded to a some tremendous computation and finally got the answer.



    But the question intentionally asked to do not make any use of such computation.



    How can one elegantly find the third eigenvalue of $A$?










    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      We had this in an exam today: Given the matrix
      $$A=begin{bmatrix}
      51&-12 & -21\
      60 & -40&28\
      57&-68&1
      end{bmatrix}$$
      Here is the precise question as asked in our exam:




      Someone tells you (accurately) that $-48$ and $24$ are eigenvalues of the matrix $A$. Without using a computer, calculator or writing anything down find the third eigenvalue of $A$.




      I have used the classical method (through the characteristic polynomial of A) but this leaded to a some tremendous computation and finally got the answer.



      But the question intentionally asked to do not make any use of such computation.



      How can one elegantly find the third eigenvalue of $A$?










      share|cite|improve this question











      $endgroup$




      We had this in an exam today: Given the matrix
      $$A=begin{bmatrix}
      51&-12 & -21\
      60 & -40&28\
      57&-68&1
      end{bmatrix}$$
      Here is the precise question as asked in our exam:




      Someone tells you (accurately) that $-48$ and $24$ are eigenvalues of the matrix $A$. Without using a computer, calculator or writing anything down find the third eigenvalue of $A$.




      I have used the classical method (through the characteristic polynomial of A) but this leaded to a some tremendous computation and finally got the answer.



      But the question intentionally asked to do not make any use of such computation.



      How can one elegantly find the third eigenvalue of $A$?







      linear-algebra matrices eigenvalues-eigenvectors






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 15 '18 at 12:18









      StubbornAtom

      5,98811238




      5,98811238










      asked Nov 23 '17 at 18:49









      Guy FsoneGuy Fsone

      17.2k42973




      17.2k42973






















          1 Answer
          1






          active

          oldest

          votes


















          14












          $begingroup$

          Use the following property:



          Trace of a matrix is equal to the sum of the eigenvalues.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Excellent!${}{}{}$
            $endgroup$
            – Dave
            Nov 23 '17 at 20:17










          • $begingroup$
            Impressive! Is there a short or intuitive proof for this property?
            $endgroup$
            – Zacky
            Dec 15 '18 at 12:23












          • $begingroup$
            here are some proofs.
            $endgroup$
            – Siong Thye Goh
            Dec 15 '18 at 12:39











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          14












          $begingroup$

          Use the following property:



          Trace of a matrix is equal to the sum of the eigenvalues.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Excellent!${}{}{}$
            $endgroup$
            – Dave
            Nov 23 '17 at 20:17










          • $begingroup$
            Impressive! Is there a short or intuitive proof for this property?
            $endgroup$
            – Zacky
            Dec 15 '18 at 12:23












          • $begingroup$
            here are some proofs.
            $endgroup$
            – Siong Thye Goh
            Dec 15 '18 at 12:39
















          14












          $begingroup$

          Use the following property:



          Trace of a matrix is equal to the sum of the eigenvalues.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Excellent!${}{}{}$
            $endgroup$
            – Dave
            Nov 23 '17 at 20:17










          • $begingroup$
            Impressive! Is there a short or intuitive proof for this property?
            $endgroup$
            – Zacky
            Dec 15 '18 at 12:23












          • $begingroup$
            here are some proofs.
            $endgroup$
            – Siong Thye Goh
            Dec 15 '18 at 12:39














          14












          14








          14





          $begingroup$

          Use the following property:



          Trace of a matrix is equal to the sum of the eigenvalues.






          share|cite|improve this answer









          $endgroup$



          Use the following property:



          Trace of a matrix is equal to the sum of the eigenvalues.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 '17 at 18:50









          Siong Thye GohSiong Thye Goh

          101k1466118




          101k1466118












          • $begingroup$
            Excellent!${}{}{}$
            $endgroup$
            – Dave
            Nov 23 '17 at 20:17










          • $begingroup$
            Impressive! Is there a short or intuitive proof for this property?
            $endgroup$
            – Zacky
            Dec 15 '18 at 12:23












          • $begingroup$
            here are some proofs.
            $endgroup$
            – Siong Thye Goh
            Dec 15 '18 at 12:39


















          • $begingroup$
            Excellent!${}{}{}$
            $endgroup$
            – Dave
            Nov 23 '17 at 20:17










          • $begingroup$
            Impressive! Is there a short or intuitive proof for this property?
            $endgroup$
            – Zacky
            Dec 15 '18 at 12:23












          • $begingroup$
            here are some proofs.
            $endgroup$
            – Siong Thye Goh
            Dec 15 '18 at 12:39
















          $begingroup$
          Excellent!${}{}{}$
          $endgroup$
          – Dave
          Nov 23 '17 at 20:17




          $begingroup$
          Excellent!${}{}{}$
          $endgroup$
          – Dave
          Nov 23 '17 at 20:17












          $begingroup$
          Impressive! Is there a short or intuitive proof for this property?
          $endgroup$
          – Zacky
          Dec 15 '18 at 12:23






          $begingroup$
          Impressive! Is there a short or intuitive proof for this property?
          $endgroup$
          – Zacky
          Dec 15 '18 at 12:23














          $begingroup$
          here are some proofs.
          $endgroup$
          – Siong Thye Goh
          Dec 15 '18 at 12:39




          $begingroup$
          here are some proofs.
          $endgroup$
          – Siong Thye Goh
          Dec 15 '18 at 12:39


















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