We have a linear operator T. Show $T^2=Id$ implies $T=T^*$











up vote
1
down vote

favorite












We have a linear operator $T:Vrightarrow V $. V is a finite-dimension inner product space over the field of the complex numbers. Show $T^2=Id$ implies $T=T^*$.



I've tried working with the inner product, trying to get $(Tx,x)=(x,Tx)$ with no luck. Maybe it has something to do with a basis for T (and thus diagonalizability?)






linear-algebra self-adjoint-operators







share|cite|improve this question









New contributor




matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    Suppose $T^2=I$ and $T=T^*$; then $TT^*=I$, so $T$ is unitary and Hermitian. Can you find a nonsymmetric real $2times2$ matrix having trace $0$ and determinant $-1$? Easy one: $begin{bmatrix}0&2\1/2&0end{bmatrix}$; less easy: $begin{bmatrix}2&-3\1&-2end{bmatrix}$.
    – egreg
    5 hours ago















up vote
1
down vote

favorite












We have a linear operator $T:Vrightarrow V $. V is a finite-dimension inner product space over the field of the complex numbers. Show $T^2=Id$ implies $T=T^*$.



I've tried working with the inner product, trying to get $(Tx,x)=(x,Tx)$ with no luck. Maybe it has something to do with a basis for T (and thus diagonalizability?)






linear-algebra self-adjoint-operators







share|cite|improve this question









New contributor




matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    Suppose $T^2=I$ and $T=T^*$; then $TT^*=I$, so $T$ is unitary and Hermitian. Can you find a nonsymmetric real $2times2$ matrix having trace $0$ and determinant $-1$? Easy one: $begin{bmatrix}0&2\1/2&0end{bmatrix}$; less easy: $begin{bmatrix}2&-3\1&-2end{bmatrix}$.
    – egreg
    5 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











We have a linear operator $T:Vrightarrow V $. V is a finite-dimension inner product space over the field of the complex numbers. Show $T^2=Id$ implies $T=T^*$.



I've tried working with the inner product, trying to get $(Tx,x)=(x,Tx)$ with no luck. Maybe it has something to do with a basis for T (and thus diagonalizability?)






linear-algebra self-adjoint-operators







share|cite|improve this question









New contributor




matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











We have a linear operator $T:Vrightarrow V $. V is a finite-dimension inner product space over the field of the complex numbers. Show $T^2=Id$ implies $T=T^*$.



I've tried working with the inner product, trying to get $(Tx,x)=(x,Tx)$ with no luck. Maybe it has something to do with a basis for T (and thus diagonalizability?)






linear-algebra self-adjoint-operators




linear-algebra self-adjoint-operators






share|cite|improve this question









New contributor




matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 15 mins ago









John Doe

9,68811134




9,68811134






New contributor




matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 5 hours ago









matt

62




62




New contributor




matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    Suppose $T^2=I$ and $T=T^*$; then $TT^*=I$, so $T$ is unitary and Hermitian. Can you find a nonsymmetric real $2times2$ matrix having trace $0$ and determinant $-1$? Easy one: $begin{bmatrix}0&2\1/2&0end{bmatrix}$; less easy: $begin{bmatrix}2&-3\1&-2end{bmatrix}$.
    – egreg
    5 hours ago














  • 1




    Suppose $T^2=I$ and $T=T^*$; then $TT^*=I$, so $T$ is unitary and Hermitian. Can you find a nonsymmetric real $2times2$ matrix having trace $0$ and determinant $-1$? Easy one: $begin{bmatrix}0&2\1/2&0end{bmatrix}$; less easy: $begin{bmatrix}2&-3\1&-2end{bmatrix}$.
    – egreg
    5 hours ago








1




1




Suppose $T^2=I$ and $T=T^*$; then $TT^*=I$, so $T$ is unitary and Hermitian. Can you find a nonsymmetric real $2times2$ matrix having trace $0$ and determinant $-1$? Easy one: $begin{bmatrix}0&2\1/2&0end{bmatrix}$; less easy: $begin{bmatrix}2&-3\1&-2end{bmatrix}$.
– egreg
5 hours ago




Suppose $T^2=I$ and $T=T^*$; then $TT^*=I$, so $T$ is unitary and Hermitian. Can you find a nonsymmetric real $2times2$ matrix having trace $0$ and determinant $-1$? Easy one: $begin{bmatrix}0&2\1/2&0end{bmatrix}$; less easy: $begin{bmatrix}2&-3\1&-2end{bmatrix}$.
– egreg
5 hours ago










3 Answers
3






active

oldest

votes

















up vote
4
down vote













You can't prove it, since it is not true. Take$$begin{array}{rccc}Tcolon&mathbb{R}^2&longrightarrow&mathbb{R}\&(x,y)&mapsto&left(2y,frac x2right).end{array}$$Then $T^2=operatorname{Id}$, but $T^*(x,y)=left(frac y2,2xright)$.






share|cite|improve this answer




























    up vote
    1
    down vote













    $(Tx, y)=(Tx, T^2 y)=(T x, T (Ty))=(x,T y)$



    if $T$ is an isometric operator






    share|cite|improve this answer




























      up vote
      0
      down vote













      I will build a basis on which $T$ is diagonal.



      Given a non-zero vector $u$, consider $v=Tu$. We then have $Tu=v$. Let the first two vectors in our basis be $u+v,u-v$. These are eigenvectors of $T$ with eigenvalues $pm1$ (if either of those two vectors is $0$, then don't include that in the basis).



      Repeat the process by picking a $w$ not in the span of $u,v$, look at $Tw$, add $wpm Tw$ to the basis, and so on.



      We get a basis of eigenvectors with real eigenvalues, meaning $T$ is diagonalizable.






      share|cite|improve this answer























        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',
        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
        });


        }
        });






        matt is a new contributor. Be nice, and check out our Code of Conduct.










        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043246%2fwe-have-a-linear-operator-t-show-t2-id-implies-t-t%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        4
        down vote













        You can't prove it, since it is not true. Take$$begin{array}{rccc}Tcolon&mathbb{R}^2&longrightarrow&mathbb{R}\&(x,y)&mapsto&left(2y,frac x2right).end{array}$$Then $T^2=operatorname{Id}$, but $T^*(x,y)=left(frac y2,2xright)$.






        share|cite|improve this answer

























          up vote
          4
          down vote













          You can't prove it, since it is not true. Take$$begin{array}{rccc}Tcolon&mathbb{R}^2&longrightarrow&mathbb{R}\&(x,y)&mapsto&left(2y,frac x2right).end{array}$$Then $T^2=operatorname{Id}$, but $T^*(x,y)=left(frac y2,2xright)$.






          share|cite|improve this answer























            up vote
            4
            down vote










            up vote
            4
            down vote









            You can't prove it, since it is not true. Take$$begin{array}{rccc}Tcolon&mathbb{R}^2&longrightarrow&mathbb{R}\&(x,y)&mapsto&left(2y,frac x2right).end{array}$$Then $T^2=operatorname{Id}$, but $T^*(x,y)=left(frac y2,2xright)$.






            share|cite|improve this answer












            You can't prove it, since it is not true. Take$$begin{array}{rccc}Tcolon&mathbb{R}^2&longrightarrow&mathbb{R}\&(x,y)&mapsto&left(2y,frac x2right).end{array}$$Then $T^2=operatorname{Id}$, but $T^*(x,y)=left(frac y2,2xright)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 5 hours ago









            José Carlos Santos

            147k22117217




            147k22117217






















                up vote
                1
                down vote













                $(Tx, y)=(Tx, T^2 y)=(T x, T (Ty))=(x,T y)$



                if $T$ is an isometric operator






                share|cite|improve this answer

























                  up vote
                  1
                  down vote













                  $(Tx, y)=(Tx, T^2 y)=(T x, T (Ty))=(x,T y)$



                  if $T$ is an isometric operator






                  share|cite|improve this answer























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    $(Tx, y)=(Tx, T^2 y)=(T x, T (Ty))=(x,T y)$



                    if $T$ is an isometric operator






                    share|cite|improve this answer












                    $(Tx, y)=(Tx, T^2 y)=(T x, T (Ty))=(x,T y)$



                    if $T$ is an isometric operator







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 5 hours ago









                    Federico Fallucca

                    1,76318




                    1,76318






















                        up vote
                        0
                        down vote













                        I will build a basis on which $T$ is diagonal.



                        Given a non-zero vector $u$, consider $v=Tu$. We then have $Tu=v$. Let the first two vectors in our basis be $u+v,u-v$. These are eigenvectors of $T$ with eigenvalues $pm1$ (if either of those two vectors is $0$, then don't include that in the basis).



                        Repeat the process by picking a $w$ not in the span of $u,v$, look at $Tw$, add $wpm Tw$ to the basis, and so on.



                        We get a basis of eigenvectors with real eigenvalues, meaning $T$ is diagonalizable.






                        share|cite|improve this answer



























                          up vote
                          0
                          down vote













                          I will build a basis on which $T$ is diagonal.



                          Given a non-zero vector $u$, consider $v=Tu$. We then have $Tu=v$. Let the first two vectors in our basis be $u+v,u-v$. These are eigenvectors of $T$ with eigenvalues $pm1$ (if either of those two vectors is $0$, then don't include that in the basis).



                          Repeat the process by picking a $w$ not in the span of $u,v$, look at $Tw$, add $wpm Tw$ to the basis, and so on.



                          We get a basis of eigenvectors with real eigenvalues, meaning $T$ is diagonalizable.






                          share|cite|improve this answer

























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            I will build a basis on which $T$ is diagonal.



                            Given a non-zero vector $u$, consider $v=Tu$. We then have $Tu=v$. Let the first two vectors in our basis be $u+v,u-v$. These are eigenvectors of $T$ with eigenvalues $pm1$ (if either of those two vectors is $0$, then don't include that in the basis).



                            Repeat the process by picking a $w$ not in the span of $u,v$, look at $Tw$, add $wpm Tw$ to the basis, and so on.



                            We get a basis of eigenvectors with real eigenvalues, meaning $T$ is diagonalizable.






                            share|cite|improve this answer














                            I will build a basis on which $T$ is diagonal.



                            Given a non-zero vector $u$, consider $v=Tu$. We then have $Tu=v$. Let the first two vectors in our basis be $u+v,u-v$. These are eigenvectors of $T$ with eigenvalues $pm1$ (if either of those two vectors is $0$, then don't include that in the basis).



                            Repeat the process by picking a $w$ not in the span of $u,v$, look at $Tw$, add $wpm Tw$ to the basis, and so on.



                            We get a basis of eigenvectors with real eigenvalues, meaning $T$ is diagonalizable.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 5 hours ago

























                            answered 5 hours ago









                            Arthur

                            110k7104186




                            110k7104186






















                                matt is a new contributor. Be nice, and check out our Code of Conduct.










                                draft saved

                                draft discarded


















                                matt is a new contributor. Be nice, and check out our Code of Conduct.













                                matt is a new contributor. Be nice, and check out our Code of Conduct.












                                matt is a new contributor. Be nice, and check out our Code of Conduct.
















                                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.





                                Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                Please pay close attention to the following guidance:


                                • 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.


                                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%2f3043246%2fwe-have-a-linear-operator-t-show-t2-id-implies-t-t%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

                                Ellipse (mathématiques)

                                Image
                                Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                                Read more

                                Quarter-circle Tiles

                                Image
                                up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                                Read more

                                Mont Emei

                                Image
                                Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                                Read more