We have a linear operator T. Show $T^2=Id$ implies $T=T^*$
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
edited Dec 17 at 3:48
John Doe
9,70311134
asked Dec 16 at 22:15
matt
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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
edited Dec 17 at 3:48
John Doe
9,70311134
asked Dec 16 at 22:15
matt
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.
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
add a comment |
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
edited Dec 17 at 3:48
John Doe
9,70311134
asked Dec 16 at 22:15
matt
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.
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
edited Dec 17 at 3:48
John Doe
9,70311134
asked Dec 16 at 22:15
matt
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.
edited Dec 17 at 3:48
John Doe
9,70311134
asked Dec 16 at 22:15
matt
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.
edited Dec 17 at 3:48
John Doe
9,70311134
edited Dec 17 at 3:48
John Doe
9,70311134
edited Dec 17 at 3:48
John Doe
9,70311134
9,70311134
asked Dec 16 at 22:15
matt
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.
asked Dec 16 at 22:15
matt
62
asked Dec 16 at 22:15
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
Dec 16 at 22:55
add a comment |
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
add a comment |
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 $T^*(x,y)=left(frac y2,2xright)$.
answered Dec 16 at 22:29
José Carlos Santos
148k22117218
add a comment |
$(Tx, y)=(Tx, T^2 y)=(T x, T (Ty))=(x,T y)$
if $T$ is an isometric operator
answered Dec 16 at 22:24
Federico Fallucca
1,79318
add a comment |
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.
answered Dec 16 at 22:25
Arthur
110k7105186
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
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active
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active
oldest
votes
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 $T^*(x,y)=left(frac y2,2xright)$.
answered Dec 16 at 22:29
José Carlos Santos
148k22117218
add a comment |
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 $T^*(x,y)=left(frac y2,2xright)$.
answered Dec 16 at 22:29
José Carlos Santos
148k22117218
add a comment |
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 $T^*(x,y)=left(frac y2,2xright)$.
answered Dec 16 at 22:29
José Carlos Santos
148k22117218
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 $T^*(x,y)=left(frac y2,2xright)$.
answered Dec 16 at 22:29
José Carlos Santos
148k22117218
answered Dec 16 at 22:29
José Carlos Santos
148k22117218
answered Dec 16 at 22:29
José Carlos Santos
148k22117218
answered Dec 16 at 22:29
José Carlos Santos
148k22117218
148k22117218
add a comment |
add a comment |
$(Tx, y)=(Tx, T^2 y)=(T x, T (Ty))=(x,T y)$
if $T$ is an isometric operator
answered Dec 16 at 22:24
Federico Fallucca
1,79318
add a comment |
$(Tx, y)=(Tx, T^2 y)=(T x, T (Ty))=(x,T y)$
if $T$ is an isometric operator
answered Dec 16 at 22:24
Federico Fallucca
1,79318
add a comment |
$(Tx, y)=(Tx, T^2 y)=(T x, T (Ty))=(x,T y)$
if $T$ is an isometric operator
answered Dec 16 at 22:24
Federico Fallucca
1,79318
$(Tx, y)=(Tx, T^2 y)=(T x, T (Ty))=(x,T y)$
if $T$ is an isometric operator
answered Dec 16 at 22:24
Federico Fallucca
1,79318
answered Dec 16 at 22:24
Federico Fallucca
1,79318
answered Dec 16 at 22:24
Federico Fallucca
1,79318
answered Dec 16 at 22:24
Federico Fallucca
1,79318
1,79318
add a comment |
add a comment |
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.
answered Dec 16 at 22:25
Arthur
110k7105186
add a comment |
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.
answered Dec 16 at 22:25
Arthur
110k7105186
add a comment |
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.
answered Dec 16 at 22:25
Arthur
110k7105186
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.
answered Dec 16 at 22:25
Arthur
110k7105186
edited Dec 16 at 22:37
answered Dec 16 at 22:25
Arthur
110k7105186
answered Dec 16 at 22:25
Arthur
110k7105186
answered Dec 16 at 22:25
Arthur
110k7105186
110k7105186
add a comment |
add a comment |
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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