Do similar matrices have same characteristic equations? [on hold]












-1












$begingroup$


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






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$



put on hold as off-topic by user21820, mrtaurho, Song, Peter Foreman, Lee David Chung Lin 6 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, mrtaurho, Song, Peter Foreman, Lee David Chung Lin

If this question can be reworded to fit the rules in the help center, please edit the question.





















    -1












    $begingroup$


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






    linear-algebra matrices eigenvalues-eigenvectors







    share|cite|improve this question











    $endgroup$



    put on hold as off-topic by user21820, mrtaurho, Song, Peter Foreman, Lee David Chung Lin 6 hours ago


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, mrtaurho, Song, Peter Foreman, Lee David Chung Lin

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      -1












      -1








      -1





      $begingroup$


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






      linear-algebra matrices eigenvalues-eigenvectors







      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 matrices eigenvalues-eigenvectors




      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 11 hours ago









      Rodrigo de Azevedo

      13k41960




      13k41960










      asked 16 hours ago









      Samurai BaleSamurai Bale

      443




      443




      put on hold as off-topic by user21820, mrtaurho, Song, Peter Foreman, Lee David Chung Lin 6 hours ago


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, mrtaurho, Song, Peter Foreman, Lee David Chung Lin

      If this question can be reworded to fit the rules in the help center, please edit the question.







      put on hold as off-topic by user21820, mrtaurho, Song, Peter Foreman, Lee David Chung Lin 6 hours ago


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, mrtaurho, Song, Peter Foreman, Lee David Chung Lin

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          2 Answers
          2






          active

          oldest

          votes


















          10












          $begingroup$

          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











          $endgroup$





















            8












            $begingroup$

            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 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              10












              $begingroup$

              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











              $endgroup$


















                10












                $begingroup$

                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











                $endgroup$
















                  10












                  10








                  10





                  $begingroup$

                  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 16 hours ago









                  J. W. Tanner

                  3,0121320




                  3,0121320










                  answered 16 hours ago









                  johnny133253johnny133253

                  459111




                  459111























                      8












                      $begingroup$

                      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









                      $endgroup$


















                        8












                        $begingroup$

                        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









                        $endgroup$
















                          8












                          8








                          8





                          $begingroup$

                          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 16 hours ago









                          copper.hatcopper.hat

                          127k559160




                          127k559160















                              -

                              Popular posts from this blog

                              Callistus I

                              Image
                              Callistus I Res apud Vicidata repertae: Nativitas : unknown value ; Roma Obitus : 223 ; Roma Pontificatus 217–222 Praecessor : Zephyrinus Successor : Urbanus I Sanctus Callistus qui ieiunium instituit; e libro manu scripto Francogallico saeculo XIV Callistus (natus die ignoto; obiit anno 222), fuit episcopus Romae et Papa Ecclesiae Catholicae Romanae ab anno 217. Vita | Secundum adversarium suum, presbyterem, fuit Hippolytus Callistus, servus argentarii Christiani. Papae Zephyrino, Callistus Romae arcessivit. Callistus fuit primus diaconus ac dux catacumbae apud Viam Appiam, quae hodie Sancti Callisti appellatur. Pontificatus | Nexus externi | Vicimedia Communia plura habent quae ad Callistus I spectant. Fontes Latinae de papis usque ad annum 530 (Papa Felix IV) Liber pontificalis Opera Omnia .mw-parser-output .stipula{padding:3px;background:#F7F8FF;border:1px solid grey;margin:auto}.mw-parser-output .stipula td.cell1{bac...
                              Read more

                              Tabula Rosettana

                              Image
                              Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
                              Read more

                              How to label and detect the document text images

                              Image
                              1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C...
                              Read more