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












1














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







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    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
    Dec 16 at 22:55
















1














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
    Dec 16 at 22:55














1












1








1







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









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matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Dec 17 at 3:48









John Doe

9,70311134




9,70311134






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asked Dec 16 at 22:15









matt

62




62




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  • 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
    Dec 16 at 22:55














  • 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
    Dec 16 at 22:55








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
Dec 16 at 22:55




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
Dec 16 at 22:55










3 Answers
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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





























    1














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



    if $T$ is an isometric operator






    share|cite|improve this answer





























      0














      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























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        3 Answers
        3






        active

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        3 Answers
        3






        active

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        active

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        active

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        4














        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


























          4














          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
























            4












            4








            4






            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 Dec 16 at 22:29









            José Carlos Santos

            148k22117218




            148k22117218























                1














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



                if $T$ is an isometric operator






                share|cite|improve this answer


























                  1














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



                  if $T$ is an isometric operator






                  share|cite|improve this answer
























                    1












                    1








                    1






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



                    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 Dec 16 at 22:24









                    Federico Fallucca

                    1,79318




                    1,79318























                        0














                        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




























                          0














                          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


























                            0












                            0








                            0






                            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 Dec 16 at 22:37

























                            answered Dec 16 at 22:25









                            Arthur

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                            110k7105186






















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