Do similar matrices have same characteristic equations?












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












    $begingroup$


    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

      503




      503






















          2 Answers
          2






          active

          oldest

          votes


















          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











          $endgroup$





















            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









            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3142004%2fdo-similar-matrices-have-same-characteristic-equations%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              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 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






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3142004%2fdo-similar-matrices-have-same-characteristic-equations%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              SQL Server 17 - Attemping to backup to remote NAS but Access is denied

                              Image
                              .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 0 I was attemping to use Ola Hallengrens scripts to backup to a remote NAS, but it was failing and not giving any direct errors, so I tried doing it manually using Server Management Studio and got the following error: TITLE: Microsoft SQL Server Management Studio Failed to retrieve data for this request. (Microsoft.SqlServer.Management.Sdk.Sfc) For help, click: http://go.microsoft.com/fwlink?ProdName=Microsoft%20SQL%20Server&LinkId=20476 ADDITIONAL INFORMATION: An exception occurred while executing a Transact-SQL statement or batch. (Microsoft.SqlServer.ConnectionInfo) The operating system returned the error '0x80070005(Access is denied.)&#
                              Read more

                              Always On Availability groups resolving state after failover - Remote harden of transaction...

                              Image
                              0 My explanation will be opinion based as it is based on my findings. If more information is needed, I am always happy to add it. The setup We have 4 SQL Server 2016 Instances, and all 4 underlying servers are in a WSFC (In my eyes a strange setup since Always On AG's are 2 by 2. But that is a different question.) 4 nodes + Fileshare witness Version: 13.0.5026.0 (all nodes) The instances that are having the failover issues are instances 1 & 2. They have 3 AG's between them. AG1 is primary on instance 1 and is having the issue. AG2 and AG3 are primary on instance 2. Instances 3 and 4 have their own separate AG's (2). AG1's settings The week before Last week, after checking for long running queries (none) and the synchronization state (green accross the board) we initiated a
                              Read more

                              Restoring from pg_dump with foreign key constraints

                              Image
                              5 2 In restoring a database from a pg_dump , a number of errors are being generated and the whole table is subsequently being ignored. An example: ERROR: insert or update on table "channelproducts" violates foreign key constraint "fk_rails_dfaae373a5" DETAIL: Key (channel_id)=(1) is not present in table "channels". Interestingly enough, I've noted that all these instances are popping up because of the sequence of loading. channels is after channelproducts both alphabetically and in the file and thus I can understand why postgres complains about having to create a child without a parent. caveat: the foreign key is being generated automatically by rails 4.2: I could remove the problem at the source but that still does not really solve the problem... version: PostgreSQL
                              Read more