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











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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
    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









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edited 9 mins ago









John Doe

9,68811134




9,68811134






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asked 5 hours ago









matt

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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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  • 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
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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 $T^*(x,y)=left(frac y2,2xright)$.






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



    if $T$ is an isometric operator






    share|cite|improve this answer




























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      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























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        3 Answers
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        active

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        3 Answers
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        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

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                            110k7104186






















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