Do similar matrices have same characteristic equations?












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Since similar matrices have same eigenvalues and characteristic polynomials, then they must have the same characteristic equation, right?






linear-algebra







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    1












    $begingroup$


    Since similar matrices have same eigenvalues and characteristic polynomials, then they must have the same characteristic equation, right?






    linear-algebra







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Since similar matrices have same eigenvalues and characteristic polynomials, then they must have the same characteristic equation, right?






      linear-algebra







      share|cite|improve this question









      $endgroup$




      Since similar matrices have same eigenvalues and characteristic polynomials, then they must have the same characteristic equation, right?






      linear-algebra




      linear-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 hours ago









      Samurai BaleSamurai Bale

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      483






















          2 Answers
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          $begingroup$

          Note that $det (lambda I -A) = det S det (lambda I -A) det S^{-1} = det (lambda I - S A S^{-1})$.






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            3












            $begingroup$

            Suppose $A$ and $B$ are square matrices such that $A = P B P^{-1}$ for some invertible matrix $P$. Then
            begin{align*}text{charpoly}(A,t) & = det(A - tI)\
            & = det(PBP^{-1} - tI)\
            & = det(PBP^{-1}-tPP^{-1})\
            & = det(P(B-tI)P^{-1})\
            & = det(P)det(B - tI) det(P^{-1})\
            & = det(P)det(B - tI) frac{1}{det(P)}\
            & = det(B-tI)\
            & = text{charpoly}(B,t).
            end{align*}



            This shows that similar matrices have the same characteristic polynomial. Note that this proof relies on several facts. In particular, the determinant is multiplicative.






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              2 Answers
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              3












              $begingroup$

              Note that $det (lambda I -A) = det S det (lambda I -A) det S^{-1} = det (lambda I - S A S^{-1})$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Note that $det (lambda I -A) = det S det (lambda I -A) det S^{-1} = det (lambda I - S A S^{-1})$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Note that $det (lambda I -A) = det S det (lambda I -A) det S^{-1} = det (lambda I - S A S^{-1})$.






                  share|cite|improve this answer









                  $endgroup$



                  Note that $det (lambda I -A) = det S det (lambda I -A) det S^{-1} = det (lambda I - S A S^{-1})$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  copper.hatcopper.hat

                  127k559160




                  127k559160























                      3












                      $begingroup$

                      Suppose $A$ and $B$ are square matrices such that $A = P B P^{-1}$ for some invertible matrix $P$. Then
                      begin{align*}text{charpoly}(A,t) & = det(A - tI)\
                      & = det(PBP^{-1} - tI)\
                      & = det(PBP^{-1}-tPP^{-1})\
                      & = det(P(B-tI)P^{-1})\
                      & = det(P)det(B - tI) det(P^{-1})\
                      & = det(P)det(B - tI) frac{1}{det(P)}\
                      & = det(B-tI)\
                      & = text{charpoly}(B,t).
                      end{align*}



                      This shows that similar matrices have the same characteristic polynomial. Note that this proof relies on several facts. In particular, the determinant is multiplicative.






                      share|cite|improve this answer











                      $endgroup$


















                        3












                        $begingroup$

                        Suppose $A$ and $B$ are square matrices such that $A = P B P^{-1}$ for some invertible matrix $P$. Then
                        begin{align*}text{charpoly}(A,t) & = det(A - tI)\
                        & = det(PBP^{-1} - tI)\
                        & = det(PBP^{-1}-tPP^{-1})\
                        & = det(P(B-tI)P^{-1})\
                        & = det(P)det(B - tI) det(P^{-1})\
                        & = det(P)det(B - tI) frac{1}{det(P)}\
                        & = det(B-tI)\
                        & = text{charpoly}(B,t).
                        end{align*}



                        This shows that similar matrices have the same characteristic polynomial. Note that this proof relies on several facts. In particular, the determinant is multiplicative.






                        share|cite|improve this answer











                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          Suppose $A$ and $B$ are square matrices such that $A = P B P^{-1}$ for some invertible matrix $P$. Then
                          begin{align*}text{charpoly}(A,t) & = det(A - tI)\
                          & = det(PBP^{-1} - tI)\
                          & = det(PBP^{-1}-tPP^{-1})\
                          & = det(P(B-tI)P^{-1})\
                          & = det(P)det(B - tI) det(P^{-1})\
                          & = det(P)det(B - tI) frac{1}{det(P)}\
                          & = det(B-tI)\
                          & = text{charpoly}(B,t).
                          end{align*}



                          This shows that similar matrices have the same characteristic polynomial. Note that this proof relies on several facts. In particular, the determinant is multiplicative.






                          share|cite|improve this answer











                          $endgroup$



                          Suppose $A$ and $B$ are square matrices such that $A = P B P^{-1}$ for some invertible matrix $P$. Then
                          begin{align*}text{charpoly}(A,t) & = det(A - tI)\
                          & = det(PBP^{-1} - tI)\
                          & = det(PBP^{-1}-tPP^{-1})\
                          & = det(P(B-tI)P^{-1})\
                          & = det(P)det(B - tI) det(P^{-1})\
                          & = det(P)det(B - tI) frac{1}{det(P)}\
                          & = det(B-tI)\
                          & = text{charpoly}(B,t).
                          end{align*}



                          This shows that similar matrices have the same characteristic polynomial. Note that this proof relies on several facts. In particular, the determinant is multiplicative.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 1 hour ago









                          J. W. Tanner

                          2,9541318




                          2,9541318










                          answered 2 hours ago









                          johnny133253johnny133253

                          384110




                          384110






























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