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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    2












    $begingroup$


    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









    $endgroup$















      2












      2








      2





      $begingroup$


      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 1 hour ago









      Samurai BaleSamurai Bale

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          $begingroup$

          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











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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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              $begingroup$

              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 1 hour ago









                  johnny133253johnny133253

                  384110




                  384110























                      2












                      $begingroup$

                      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









                      $endgroup$


















                        2












                        $begingroup$

                        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









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          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 1 hour ago









                          copper.hatcopper.hat

                          127k559160




                          127k559160






























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