Showing that is a normal operator












4












$begingroup$


Let $H$ is a Hilbert space



$I$ is unit operator, $T in B(H)$ and $lambda in mathbb C$



$T$ is normal operator $Rightarrow$ $T-lambda I$ is a normal operator too.



I could only write :



I must show that $(T-lambda I)(T-lambda I)^{ast}=(T-lambda I)^{ast}(T-lambda I)$



$TT^{ast}=T^{ast}T$



$I^{ast}=I$



$(T-lambda I)^{ast}=T^{ast}- bar{lambda}I$



(where $ast$ means adjoint and $bar{lambda}I$ means complex conjugate.



I cannot continue. I really stuck



Thanks for any help






functional-analysis operator-theory hilbert-spaces adjoint-operators







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    Let $H$ is a Hilbert space



    $I$ is unit operator, $T in B(H)$ and $lambda in mathbb C$



    $T$ is normal operator $Rightarrow$ $T-lambda I$ is a normal operator too.



    I could only write :



    I must show that $(T-lambda I)(T-lambda I)^{ast}=(T-lambda I)^{ast}(T-lambda I)$



    $TT^{ast}=T^{ast}T$



    $I^{ast}=I$



    $(T-lambda I)^{ast}=T^{ast}- bar{lambda}I$



    (where $ast$ means adjoint and $bar{lambda}I$ means complex conjugate.



    I cannot continue. I really stuck



    Thanks for any help






    functional-analysis operator-theory hilbert-spaces adjoint-operators







    share|cite|improve this question









    $endgroup$















      4












      4








      4


      2



      $begingroup$


      Let $H$ is a Hilbert space



      $I$ is unit operator, $T in B(H)$ and $lambda in mathbb C$



      $T$ is normal operator $Rightarrow$ $T-lambda I$ is a normal operator too.



      I could only write :



      I must show that $(T-lambda I)(T-lambda I)^{ast}=(T-lambda I)^{ast}(T-lambda I)$



      $TT^{ast}=T^{ast}T$



      $I^{ast}=I$



      $(T-lambda I)^{ast}=T^{ast}- bar{lambda}I$



      (where $ast$ means adjoint and $bar{lambda}I$ means complex conjugate.



      I cannot continue. I really stuck



      Thanks for any help






      functional-analysis operator-theory hilbert-spaces adjoint-operators







      share|cite|improve this question









      $endgroup$




      Let $H$ is a Hilbert space



      $I$ is unit operator, $T in B(H)$ and $lambda in mathbb C$



      $T$ is normal operator $Rightarrow$ $T-lambda I$ is a normal operator too.



      I could only write :



      I must show that $(T-lambda I)(T-lambda I)^{ast}=(T-lambda I)^{ast}(T-lambda I)$



      $TT^{ast}=T^{ast}T$



      $I^{ast}=I$



      $(T-lambda I)^{ast}=T^{ast}- bar{lambda}I$



      (where $ast$ means adjoint and $bar{lambda}I$ means complex conjugate.



      I cannot continue. I really stuck



      Thanks for any help






      functional-analysis operator-theory hilbert-spaces adjoint-operators




      functional-analysis operator-theory hilbert-spaces adjoint-operators






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 19 '18 at 19:07









      esrabasaresrabasar

      420110




      420110






















          4 Answers
          4






          active

          oldest

          votes


















          0












          $begingroup$

          Then
          $$(T-lambda I)(T-lambda I)^*=(T-lambda I)(T^*-overlinelambda I)
          =TT^*-lambda T^*-overlinelambda T+|lambda|^2I.$$
          What about $(T-lambda I)^*(T-lambda I)$?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why can we use distribution? It is a composite operator. Could you please explain if it is possible?
            $endgroup$
            – user519955
            Dec 19 '18 at 19:26






          • 1




            $begingroup$
            @user519955 These are all linear operators.
            $endgroup$
            – Lord Shark the Unknown
            Dec 20 '18 at 2:57



















          0












          $begingroup$

          Note that



          $(T - lambda I)^ast = T^ast - bar lambda I; tag 1$



          then, using $TT^ast = T^ast T$ (i.e., the normality of $T$), one has



          $(T - lambda I)(T - lambda I)^ast = (T - lambda I)(T^ast - bar lambda I) = TT^ast - bar lambda T - lambda T^ast + lambda bar lambda I$
          $= T^ast T - bar lambda T - lambda T^ast + lambda bar lambda I = T^ast (T - lambda I) - bar lambda (T - lambda I)$
          $= (T^ast - bar lambda I)(T - lambda I) = (T - lambda I)^ast (T - lambda I), tag 2$



          which asserts that $T - lambda$ is normal.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            You can also use the characterization that $A in B(H)$ is normal if and only if $|Ax| = |A^*x|, forall x in H$.



            For $x in H$ we have



            begin{align}
            |(T - lambda I)x|^2 &= langle Tx - lambda x, Tx - lambda xrangle \
            &= langle Tx,Txrangle - 2operatorname{Re} langle Tx, lambda xrangle + langle lambda x, lambda xrangle\
            &= |Tx|^2 - 2operatorname{Re} overline{lambda}langle Tx, xrangle + |lambda|^2|x|^2\
            &= |T^*x|^2 - 2operatorname{Re} overline{lambda}langle x, T^*xrangle + left|overline{lambda}right|^2|x|^2\
            &= langle overline{lambda} x,overline{lambda}xrangle - 2operatorname{Re} langle overline{lambda}x, T^*xrangle+ langle T^*x,T^*xrangle\
            &= leftlangle overline{lambda}x - T^*x,overline{lambda}x - T^*xrightrangle\
            &= left|left(overline{lambda} I - T^*right)xright|^2\
            &= left|(T-lambda I)^*xright|^2
            end{align}



            so $T - lambda I$ is normal.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              $T$ is normal iff $T$ commutes with its adjoint $T^*$. If $T$ is normal, then $T^*$ commutes with $(T-lambda I)$ and $overline{lambda}I$ commutes with $(T-lambda I)$; hence $T^*-overline{lambda}I=(T-lambda I)^*$ commutes with $T-lambda I$, which makes $T-lambda I$ normal.






              share|cite|improve this answer









              $endgroup$













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                4 Answers
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                4 Answers
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                active

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                0












                $begingroup$

                Then
                $$(T-lambda I)(T-lambda I)^*=(T-lambda I)(T^*-overlinelambda I)
                =TT^*-lambda T^*-overlinelambda T+|lambda|^2I.$$
                What about $(T-lambda I)^*(T-lambda I)$?






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Why can we use distribution? It is a composite operator. Could you please explain if it is possible?
                  $endgroup$
                  – user519955
                  Dec 19 '18 at 19:26






                • 1




                  $begingroup$
                  @user519955 These are all linear operators.
                  $endgroup$
                  – Lord Shark the Unknown
                  Dec 20 '18 at 2:57
















                0












                $begingroup$

                Then
                $$(T-lambda I)(T-lambda I)^*=(T-lambda I)(T^*-overlinelambda I)
                =TT^*-lambda T^*-overlinelambda T+|lambda|^2I.$$
                What about $(T-lambda I)^*(T-lambda I)$?






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Why can we use distribution? It is a composite operator. Could you please explain if it is possible?
                  $endgroup$
                  – user519955
                  Dec 19 '18 at 19:26






                • 1




                  $begingroup$
                  @user519955 These are all linear operators.
                  $endgroup$
                  – Lord Shark the Unknown
                  Dec 20 '18 at 2:57














                0












                0








                0





                $begingroup$

                Then
                $$(T-lambda I)(T-lambda I)^*=(T-lambda I)(T^*-overlinelambda I)
                =TT^*-lambda T^*-overlinelambda T+|lambda|^2I.$$
                What about $(T-lambda I)^*(T-lambda I)$?






                share|cite|improve this answer









                $endgroup$



                Then
                $$(T-lambda I)(T-lambda I)^*=(T-lambda I)(T^*-overlinelambda I)
                =TT^*-lambda T^*-overlinelambda T+|lambda|^2I.$$
                What about $(T-lambda I)^*(T-lambda I)$?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 19 '18 at 19:11









                Lord Shark the UnknownLord Shark the Unknown

                106k1161133




                106k1161133












                • $begingroup$
                  Why can we use distribution? It is a composite operator. Could you please explain if it is possible?
                  $endgroup$
                  – user519955
                  Dec 19 '18 at 19:26






                • 1




                  $begingroup$
                  @user519955 These are all linear operators.
                  $endgroup$
                  – Lord Shark the Unknown
                  Dec 20 '18 at 2:57


















                • $begingroup$
                  Why can we use distribution? It is a composite operator. Could you please explain if it is possible?
                  $endgroup$
                  – user519955
                  Dec 19 '18 at 19:26






                • 1




                  $begingroup$
                  @user519955 These are all linear operators.
                  $endgroup$
                  – Lord Shark the Unknown
                  Dec 20 '18 at 2:57
















                $begingroup$
                Why can we use distribution? It is a composite operator. Could you please explain if it is possible?
                $endgroup$
                – user519955
                Dec 19 '18 at 19:26




                $begingroup$
                Why can we use distribution? It is a composite operator. Could you please explain if it is possible?
                $endgroup$
                – user519955
                Dec 19 '18 at 19:26




                1




                1




                $begingroup$
                @user519955 These are all linear operators.
                $endgroup$
                – Lord Shark the Unknown
                Dec 20 '18 at 2:57




                $begingroup$
                @user519955 These are all linear operators.
                $endgroup$
                – Lord Shark the Unknown
                Dec 20 '18 at 2:57











                0












                $begingroup$

                Note that



                $(T - lambda I)^ast = T^ast - bar lambda I; tag 1$



                then, using $TT^ast = T^ast T$ (i.e., the normality of $T$), one has



                $(T - lambda I)(T - lambda I)^ast = (T - lambda I)(T^ast - bar lambda I) = TT^ast - bar lambda T - lambda T^ast + lambda bar lambda I$
                $= T^ast T - bar lambda T - lambda T^ast + lambda bar lambda I = T^ast (T - lambda I) - bar lambda (T - lambda I)$
                $= (T^ast - bar lambda I)(T - lambda I) = (T - lambda I)^ast (T - lambda I), tag 2$



                which asserts that $T - lambda$ is normal.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Note that



                  $(T - lambda I)^ast = T^ast - bar lambda I; tag 1$



                  then, using $TT^ast = T^ast T$ (i.e., the normality of $T$), one has



                  $(T - lambda I)(T - lambda I)^ast = (T - lambda I)(T^ast - bar lambda I) = TT^ast - bar lambda T - lambda T^ast + lambda bar lambda I$
                  $= T^ast T - bar lambda T - lambda T^ast + lambda bar lambda I = T^ast (T - lambda I) - bar lambda (T - lambda I)$
                  $= (T^ast - bar lambda I)(T - lambda I) = (T - lambda I)^ast (T - lambda I), tag 2$



                  which asserts that $T - lambda$ is normal.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Note that



                    $(T - lambda I)^ast = T^ast - bar lambda I; tag 1$



                    then, using $TT^ast = T^ast T$ (i.e., the normality of $T$), one has



                    $(T - lambda I)(T - lambda I)^ast = (T - lambda I)(T^ast - bar lambda I) = TT^ast - bar lambda T - lambda T^ast + lambda bar lambda I$
                    $= T^ast T - bar lambda T - lambda T^ast + lambda bar lambda I = T^ast (T - lambda I) - bar lambda (T - lambda I)$
                    $= (T^ast - bar lambda I)(T - lambda I) = (T - lambda I)^ast (T - lambda I), tag 2$



                    which asserts that $T - lambda$ is normal.






                    share|cite|improve this answer









                    $endgroup$



                    Note that



                    $(T - lambda I)^ast = T^ast - bar lambda I; tag 1$



                    then, using $TT^ast = T^ast T$ (i.e., the normality of $T$), one has



                    $(T - lambda I)(T - lambda I)^ast = (T - lambda I)(T^ast - bar lambda I) = TT^ast - bar lambda T - lambda T^ast + lambda bar lambda I$
                    $= T^ast T - bar lambda T - lambda T^ast + lambda bar lambda I = T^ast (T - lambda I) - bar lambda (T - lambda I)$
                    $= (T^ast - bar lambda I)(T - lambda I) = (T - lambda I)^ast (T - lambda I), tag 2$



                    which asserts that $T - lambda$ is normal.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 19 '18 at 20:16









                    Robert LewisRobert Lewis

                    48.3k23167




                    48.3k23167























                        0












                        $begingroup$

                        You can also use the characterization that $A in B(H)$ is normal if and only if $|Ax| = |A^*x|, forall x in H$.



                        For $x in H$ we have



                        begin{align}
                        |(T - lambda I)x|^2 &= langle Tx - lambda x, Tx - lambda xrangle \
                        &= langle Tx,Txrangle - 2operatorname{Re} langle Tx, lambda xrangle + langle lambda x, lambda xrangle\
                        &= |Tx|^2 - 2operatorname{Re} overline{lambda}langle Tx, xrangle + |lambda|^2|x|^2\
                        &= |T^*x|^2 - 2operatorname{Re} overline{lambda}langle x, T^*xrangle + left|overline{lambda}right|^2|x|^2\
                        &= langle overline{lambda} x,overline{lambda}xrangle - 2operatorname{Re} langle overline{lambda}x, T^*xrangle+ langle T^*x,T^*xrangle\
                        &= leftlangle overline{lambda}x - T^*x,overline{lambda}x - T^*xrightrangle\
                        &= left|left(overline{lambda} I - T^*right)xright|^2\
                        &= left|(T-lambda I)^*xright|^2
                        end{align}



                        so $T - lambda I$ is normal.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          You can also use the characterization that $A in B(H)$ is normal if and only if $|Ax| = |A^*x|, forall x in H$.



                          For $x in H$ we have



                          begin{align}
                          |(T - lambda I)x|^2 &= langle Tx - lambda x, Tx - lambda xrangle \
                          &= langle Tx,Txrangle - 2operatorname{Re} langle Tx, lambda xrangle + langle lambda x, lambda xrangle\
                          &= |Tx|^2 - 2operatorname{Re} overline{lambda}langle Tx, xrangle + |lambda|^2|x|^2\
                          &= |T^*x|^2 - 2operatorname{Re} overline{lambda}langle x, T^*xrangle + left|overline{lambda}right|^2|x|^2\
                          &= langle overline{lambda} x,overline{lambda}xrangle - 2operatorname{Re} langle overline{lambda}x, T^*xrangle+ langle T^*x,T^*xrangle\
                          &= leftlangle overline{lambda}x - T^*x,overline{lambda}x - T^*xrightrangle\
                          &= left|left(overline{lambda} I - T^*right)xright|^2\
                          &= left|(T-lambda I)^*xright|^2
                          end{align}



                          so $T - lambda I$ is normal.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            You can also use the characterization that $A in B(H)$ is normal if and only if $|Ax| = |A^*x|, forall x in H$.



                            For $x in H$ we have



                            begin{align}
                            |(T - lambda I)x|^2 &= langle Tx - lambda x, Tx - lambda xrangle \
                            &= langle Tx,Txrangle - 2operatorname{Re} langle Tx, lambda xrangle + langle lambda x, lambda xrangle\
                            &= |Tx|^2 - 2operatorname{Re} overline{lambda}langle Tx, xrangle + |lambda|^2|x|^2\
                            &= |T^*x|^2 - 2operatorname{Re} overline{lambda}langle x, T^*xrangle + left|overline{lambda}right|^2|x|^2\
                            &= langle overline{lambda} x,overline{lambda}xrangle - 2operatorname{Re} langle overline{lambda}x, T^*xrangle+ langle T^*x,T^*xrangle\
                            &= leftlangle overline{lambda}x - T^*x,overline{lambda}x - T^*xrightrangle\
                            &= left|left(overline{lambda} I - T^*right)xright|^2\
                            &= left|(T-lambda I)^*xright|^2
                            end{align}



                            so $T - lambda I$ is normal.






                            share|cite|improve this answer









                            $endgroup$



                            You can also use the characterization that $A in B(H)$ is normal if and only if $|Ax| = |A^*x|, forall x in H$.



                            For $x in H$ we have



                            begin{align}
                            |(T - lambda I)x|^2 &= langle Tx - lambda x, Tx - lambda xrangle \
                            &= langle Tx,Txrangle - 2operatorname{Re} langle Tx, lambda xrangle + langle lambda x, lambda xrangle\
                            &= |Tx|^2 - 2operatorname{Re} overline{lambda}langle Tx, xrangle + |lambda|^2|x|^2\
                            &= |T^*x|^2 - 2operatorname{Re} overline{lambda}langle x, T^*xrangle + left|overline{lambda}right|^2|x|^2\
                            &= langle overline{lambda} x,overline{lambda}xrangle - 2operatorname{Re} langle overline{lambda}x, T^*xrangle+ langle T^*x,T^*xrangle\
                            &= leftlangle overline{lambda}x - T^*x,overline{lambda}x - T^*xrightrangle\
                            &= left|left(overline{lambda} I - T^*right)xright|^2\
                            &= left|(T-lambda I)^*xright|^2
                            end{align}



                            so $T - lambda I$ is normal.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 19 '18 at 20:30









                            mechanodroidmechanodroid

                            28.9k62548




                            28.9k62548























                                0












                                $begingroup$

                                $T$ is normal iff $T$ commutes with its adjoint $T^*$. If $T$ is normal, then $T^*$ commutes with $(T-lambda I)$ and $overline{lambda}I$ commutes with $(T-lambda I)$; hence $T^*-overline{lambda}I=(T-lambda I)^*$ commutes with $T-lambda I$, which makes $T-lambda I$ normal.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  $T$ is normal iff $T$ commutes with its adjoint $T^*$. If $T$ is normal, then $T^*$ commutes with $(T-lambda I)$ and $overline{lambda}I$ commutes with $(T-lambda I)$; hence $T^*-overline{lambda}I=(T-lambda I)^*$ commutes with $T-lambda I$, which makes $T-lambda I$ normal.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    $T$ is normal iff $T$ commutes with its adjoint $T^*$. If $T$ is normal, then $T^*$ commutes with $(T-lambda I)$ and $overline{lambda}I$ commutes with $(T-lambda I)$; hence $T^*-overline{lambda}I=(T-lambda I)^*$ commutes with $T-lambda I$, which makes $T-lambda I$ normal.






                                    share|cite|improve this answer









                                    $endgroup$



                                    $T$ is normal iff $T$ commutes with its adjoint $T^*$. If $T$ is normal, then $T^*$ commutes with $(T-lambda I)$ and $overline{lambda}I$ commutes with $(T-lambda I)$; hence $T^*-overline{lambda}I=(T-lambda I)^*$ commutes with $T-lambda I$, which makes $T-lambda I$ normal.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 20 '18 at 2:46









                                    DisintegratingByPartsDisintegratingByParts

                                    59.8k42681




                                    59.8k42681






























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                                        Residenzschloss Arolsen

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                                        Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
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