Do similar matrices have same characteristic equations?Jordan Canonical Form - Similar matrices and same...

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Do similar matrices have same characteristic equations?


Jordan Canonical Form - Similar matrices and same minimal polynomialsSimilar Matrices and Characteristic PolynomialsSame characteristic polynomial $iff$ same eigenvalues?If two matrices have the same characteristic polynomials, determinant and trace, are they similar?$3 times 3$ matrices are similar if and only if they have the same characteristic and minimal polynomialHow to prove two diagonalizable matrices are similar iff they have same eigenvalue with same multiplicity.Can two matrices with the same characteristic polynomial have different eigenvalues?Are these $4$ by $4$ matrices similar?are all these matrices similar?Matrices with given characteristic polynomial are similar iff the characteristic polynomial have no repeated roots













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










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    $begingroup$


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










    share|cite|improve this question









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      2












      2








      2





      $begingroup$


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










      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






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      asked 2 hours 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











          $endgroup$





















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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.






              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























                      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 2 hours ago









                          copper.hatcopper.hat

                          127k559160




                          127k559160






























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