Given set M , which of following option is correct?












1












$begingroup$


let $M = { A = begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z} text{and the eigenvalue of A are in }mathbb{Q}}$ .Then



choose the correct option



$1.$ M is empty



$2.$$M = {begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z}}$



$3.$if $A in M$ ,then the eigenvalue of $A$ are in $mathbb{Z}$



$4.$if$ A,B in M$ are such that $AB=I$ then $det A in {+1,-1}$



MY attempt : option $1)$ is obviously false



option $2)$ is True according to the definition of question



option $3)$ is True take $A=I$



option $4)$ is True take $A= B=I$



is my attempt is True



Any hints/solution










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    let $M = { A = begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z} text{and the eigenvalue of A are in }mathbb{Q}}$ .Then



    choose the correct option



    $1.$ M is empty



    $2.$$M = {begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z}}$



    $3.$if $A in M$ ,then the eigenvalue of $A$ are in $mathbb{Z}$



    $4.$if$ A,B in M$ are such that $AB=I$ then $det A in {+1,-1}$



    MY attempt : option $1)$ is obviously false



    option $2)$ is True according to the definition of question



    option $3)$ is True take $A=I$



    option $4)$ is True take $A= B=I$



    is my attempt is True



    Any hints/solution










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      let $M = { A = begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z} text{and the eigenvalue of A are in }mathbb{Q}}$ .Then



      choose the correct option



      $1.$ M is empty



      $2.$$M = {begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z}}$



      $3.$if $A in M$ ,then the eigenvalue of $A$ are in $mathbb{Z}$



      $4.$if$ A,B in M$ are such that $AB=I$ then $det A in {+1,-1}$



      MY attempt : option $1)$ is obviously false



      option $2)$ is True according to the definition of question



      option $3)$ is True take $A=I$



      option $4)$ is True take $A= B=I$



      is my attempt is True



      Any hints/solution










      share|cite|improve this question









      $endgroup$




      let $M = { A = begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z} text{and the eigenvalue of A are in }mathbb{Q}}$ .Then



      choose the correct option



      $1.$ M is empty



      $2.$$M = {begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z}}$



      $3.$if $A in M$ ,then the eigenvalue of $A$ are in $mathbb{Z}$



      $4.$if$ A,B in M$ are such that $AB=I$ then $det A in {+1,-1}$



      MY attempt : option $1)$ is obviously false



      option $2)$ is True according to the definition of question



      option $3)$ is True take $A=I$



      option $4)$ is True take $A= B=I$



      is my attempt is True



      Any hints/solution







      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 15 '18 at 3:55









      jasminejasmine

      1,783417




      1,783417






















          1 Answer
          1






          active

          oldest

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          3












          $begingroup$

          For 1, you should display a matrix that is in $M$. It should be easy to find one.



          For 2, they have removed the requirement that the eigenvalues of $A$ are rational. If you can find a matrix where the eigenvalues are not rational you show that this is not true. Can you find one?



          For 3, finding one matrix is not enough. You are asked to show that if a matrix has integral entries and rational eigenvalues, the eigenvalues are integral. You could think about the rational root theorem and the characteristic polynomial of matrices in $M$.



          4 is the same as 3. In fact it will follow from 3, if 3 is true.






          share|cite|improve this answer











          $endgroup$













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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            For 1, you should display a matrix that is in $M$. It should be easy to find one.



            For 2, they have removed the requirement that the eigenvalues of $A$ are rational. If you can find a matrix where the eigenvalues are not rational you show that this is not true. Can you find one?



            For 3, finding one matrix is not enough. You are asked to show that if a matrix has integral entries and rational eigenvalues, the eigenvalues are integral. You could think about the rational root theorem and the characteristic polynomial of matrices in $M$.



            4 is the same as 3. In fact it will follow from 3, if 3 is true.






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              For 1, you should display a matrix that is in $M$. It should be easy to find one.



              For 2, they have removed the requirement that the eigenvalues of $A$ are rational. If you can find a matrix where the eigenvalues are not rational you show that this is not true. Can you find one?



              For 3, finding one matrix is not enough. You are asked to show that if a matrix has integral entries and rational eigenvalues, the eigenvalues are integral. You could think about the rational root theorem and the characteristic polynomial of matrices in $M$.



              4 is the same as 3. In fact it will follow from 3, if 3 is true.






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                For 1, you should display a matrix that is in $M$. It should be easy to find one.



                For 2, they have removed the requirement that the eigenvalues of $A$ are rational. If you can find a matrix where the eigenvalues are not rational you show that this is not true. Can you find one?



                For 3, finding one matrix is not enough. You are asked to show that if a matrix has integral entries and rational eigenvalues, the eigenvalues are integral. You could think about the rational root theorem and the characteristic polynomial of matrices in $M$.



                4 is the same as 3. In fact it will follow from 3, if 3 is true.






                share|cite|improve this answer











                $endgroup$



                For 1, you should display a matrix that is in $M$. It should be easy to find one.



                For 2, they have removed the requirement that the eigenvalues of $A$ are rational. If you can find a matrix where the eigenvalues are not rational you show that this is not true. Can you find one?



                For 3, finding one matrix is not enough. You are asked to show that if a matrix has integral entries and rational eigenvalues, the eigenvalues are integral. You could think about the rational root theorem and the characteristic polynomial of matrices in $M$.



                4 is the same as 3. In fact it will follow from 3, if 3 is true.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 15 '18 at 4:11

























                answered Dec 15 '18 at 4:03









                Ross MillikanRoss Millikan

                297k23198371




                297k23198371






























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