Does squaring all unitary matrices give the identity matrix?
All examples of unitary matrices that I've seen square to give the identity matrix, but I've never seen this explicitly listed as a property of unitary matrices.
Is this true for all unitary matrices?
matrices
asked Oct 12 '15 at 16:43
Alan Wolfe
514424
add a comment |
All examples of unitary matrices that I've seen square to give the identity matrix, but I've never seen this explicitly listed as a property of unitary matrices.
Is this true for all unitary matrices?
matrices
asked Oct 12 '15 at 16:43
Alan Wolfe
514424
add a comment |
All examples of unitary matrices that I've seen square to give the identity matrix, but I've never seen this explicitly listed as a property of unitary matrices.
Is this true for all unitary matrices?
matrices
asked Oct 12 '15 at 16:43
Alan Wolfe
514424
All examples of unitary matrices that I've seen square to give the identity matrix, but I've never seen this explicitly listed as a property of unitary matrices.
Is this true for all unitary matrices?
matrices
matrices
asked Oct 12 '15 at 16:43
Alan Wolfe
514424
asked Oct 12 '15 at 16:43
Alan Wolfe
514424
asked Oct 12 '15 at 16:43
Alan Wolfe
514424
asked Oct 12 '15 at 16:43
Alan Wolfe
514424
asked Oct 12 '15 at 16:43
Alan Wolfe
514424
514424
add a comment |
add a comment |
3 Answers
3
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votes
Certainly not, consider the matrix
$$J := pmatrix{0&-1\1&0}.$$
Since $J^* J = I$, it is unitary, but $J^2 = -I$.
answered Oct 12 '15 at 16:46
Travis
59.3k767145
I think I might be misunderstanding some fundamental thing. What is the difference between $J*J$ and $J^2$?
– Alan Wolfe
Oct 12 '15 at 16:49
Me too. Aren't they the same?
– Ian Miller
Oct 12 '15 at 16:51
$J^2$ is the ordinary matrix product. $J^*$ is the complex conjugate of the transpose of $J$. So $J^*J$ is the ordinary matrix product of $J^*$ times $J$.
– Empy2
Oct 12 '15 at 16:52
Oh i see that the * is elevated, and that your latex has it as ^*, not as *. Thanks!
– Alan Wolfe
Oct 12 '15 at 16:54
Just saw this exchange---thanks for clearing this up, @Michael. (Perhaps a similar issue is the motivation of the downvote?)
– Travis
Oct 12 '15 at 17:01
|
show 2 more comments
The examples you have seen must have been Hermitian; in fact
any two of the properties unitary, involution, Hermitian:
$$
A^*A=I, quad A^2 = I, quad A^* = A
$$
imply the third.
answered Oct 20 '15 at 19:23
Bob Terrell
1,665710
add a comment |
For matrix$A^{-1}=A^T$ and column vectors have unit magnitude, it is unitary matrix. If we could do $A^2=I => A=A^{-1} => A=A^T$ if A is symmetrical and unitary then $A^2 =I$.
answered Nov 29 at 10:32
mahavir
4481320
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Certainly not, consider the matrix
$$J := pmatrix{0&-1\1&0}.$$
Since $J^* J = I$, it is unitary, but $J^2 = -I$.
answered Oct 12 '15 at 16:46
Travis
59.3k767145
I think I might be misunderstanding some fundamental thing. What is the difference between $J*J$ and $J^2$?
– Alan Wolfe
Oct 12 '15 at 16:49
Me too. Aren't they the same?
– Ian Miller
Oct 12 '15 at 16:51
$J^2$ is the ordinary matrix product. $J^*$ is the complex conjugate of the transpose of $J$. So $J^*J$ is the ordinary matrix product of $J^*$ times $J$.
– Empy2
Oct 12 '15 at 16:52
Oh i see that the * is elevated, and that your latex has it as ^*, not as *. Thanks!
– Alan Wolfe
Oct 12 '15 at 16:54
Just saw this exchange---thanks for clearing this up, @Michael. (Perhaps a similar issue is the motivation of the downvote?)
– Travis
Oct 12 '15 at 17:01
|
show 2 more comments
Certainly not, consider the matrix
$$J := pmatrix{0&-1\1&0}.$$
Since $J^* J = I$, it is unitary, but $J^2 = -I$.
answered Oct 12 '15 at 16:46
Travis
59.3k767145
I think I might be misunderstanding some fundamental thing. What is the difference between $J*J$ and $J^2$?
– Alan Wolfe
Oct 12 '15 at 16:49
Me too. Aren't they the same?
– Ian Miller
Oct 12 '15 at 16:51
$J^2$ is the ordinary matrix product. $J^*$ is the complex conjugate of the transpose of $J$. So $J^*J$ is the ordinary matrix product of $J^*$ times $J$.
– Empy2
Oct 12 '15 at 16:52
Oh i see that the * is elevated, and that your latex has it as ^*, not as *. Thanks!
– Alan Wolfe
Oct 12 '15 at 16:54
Just saw this exchange---thanks for clearing this up, @Michael. (Perhaps a similar issue is the motivation of the downvote?)
– Travis
Oct 12 '15 at 17:01
|
show 2 more comments
Certainly not, consider the matrix
$$J := pmatrix{0&-1\1&0}.$$
Since $J^* J = I$, it is unitary, but $J^2 = -I$.
answered Oct 12 '15 at 16:46
Travis
59.3k767145
Certainly not, consider the matrix
$$J := pmatrix{0&-1\1&0}.$$
Since $J^* J = I$, it is unitary, but $J^2 = -I$.
answered Oct 12 '15 at 16:46
Travis
59.3k767145
answered Oct 12 '15 at 16:46
Travis
59.3k767145
answered Oct 12 '15 at 16:46
Travis
59.3k767145
answered Oct 12 '15 at 16:46
Travis
59.3k767145
59.3k767145
I think I might be misunderstanding some fundamental thing. What is the difference between $J*J$ and $J^2$?
– Alan Wolfe
Oct 12 '15 at 16:49
Me too. Aren't they the same?
– Ian Miller
Oct 12 '15 at 16:51
$J^2$ is the ordinary matrix product. $J^*$ is the complex conjugate of the transpose of $J$. So $J^*J$ is the ordinary matrix product of $J^*$ times $J$.
– Empy2
Oct 12 '15 at 16:52
Oh i see that the * is elevated, and that your latex has it as ^*, not as *. Thanks!
– Alan Wolfe
Oct 12 '15 at 16:54
Just saw this exchange---thanks for clearing this up, @Michael. (Perhaps a similar issue is the motivation of the downvote?)
– Travis
Oct 12 '15 at 17:01
|
show 2 more comments
I think I might be misunderstanding some fundamental thing. What is the difference between $J*J$ and $J^2$?
– Alan Wolfe
Oct 12 '15 at 16:49
Me too. Aren't they the same?
– Ian Miller
Oct 12 '15 at 16:51
$J^2$ is the ordinary matrix product. $J^*$ is the complex conjugate of the transpose of $J$. So $J^*J$ is the ordinary matrix product of $J^*$ times $J$.
– Empy2
Oct 12 '15 at 16:52
Oh i see that the * is elevated, and that your latex has it as ^*, not as *. Thanks!
– Alan Wolfe
Oct 12 '15 at 16:54
Just saw this exchange---thanks for clearing this up, @Michael. (Perhaps a similar issue is the motivation of the downvote?)
– Travis
Oct 12 '15 at 17:01
I think I might be misunderstanding some fundamental thing. What is the difference between $J*J$ and $J^2$?
– Alan Wolfe
Oct 12 '15 at 16:49
I think I might be misunderstanding some fundamental thing. What is the difference between $J*J$ and $J^2$?
– Alan Wolfe
Oct 12 '15 at 16:49
Me too. Aren't they the same?
– Ian Miller
Oct 12 '15 at 16:51
Me too. Aren't they the same?
– Ian Miller
Oct 12 '15 at 16:51
$J^2$ is the ordinary matrix product. $J^*$ is the complex conjugate of the transpose of $J$. So $J^*J$ is the ordinary matrix product of $J^*$ times $J$.
– Empy2
Oct 12 '15 at 16:52
$J^2$ is the ordinary matrix product. $J^*$ is the complex conjugate of the transpose of $J$. So $J^*J$ is the ordinary matrix product of $J^*$ times $J$.
– Empy2
Oct 12 '15 at 16:52
Oh i see that the * is elevated, and that your latex has it as ^*, not as *. Thanks!
– Alan Wolfe
Oct 12 '15 at 16:54
Oh i see that the * is elevated, and that your latex has it as ^*, not as *. Thanks!
– Alan Wolfe
Oct 12 '15 at 16:54
Just saw this exchange---thanks for clearing this up, @Michael. (Perhaps a similar issue is the motivation of the downvote?)
– Travis
Oct 12 '15 at 17:01
Just saw this exchange---thanks for clearing this up, @Michael. (Perhaps a similar issue is the motivation of the downvote?)
– Travis
Oct 12 '15 at 17:01
|
show 2 more comments
The examples you have seen must have been Hermitian; in fact
any two of the properties unitary, involution, Hermitian:
$$
A^*A=I, quad A^2 = I, quad A^* = A
$$
imply the third.
answered Oct 20 '15 at 19:23
Bob Terrell
1,665710
add a comment |
The examples you have seen must have been Hermitian; in fact
any two of the properties unitary, involution, Hermitian:
$$
A^*A=I, quad A^2 = I, quad A^* = A
$$
imply the third.
answered Oct 20 '15 at 19:23
Bob Terrell
1,665710
add a comment |
The examples you have seen must have been Hermitian; in fact
any two of the properties unitary, involution, Hermitian:
$$
A^*A=I, quad A^2 = I, quad A^* = A
$$
imply the third.
answered Oct 20 '15 at 19:23
Bob Terrell
1,665710
The examples you have seen must have been Hermitian; in fact
any two of the properties unitary, involution, Hermitian:
$$
A^*A=I, quad A^2 = I, quad A^* = A
$$
imply the third.
answered Oct 20 '15 at 19:23
Bob Terrell
1,665710
answered Oct 20 '15 at 19:23
Bob Terrell
1,665710
answered Oct 20 '15 at 19:23
Bob Terrell
1,665710
answered Oct 20 '15 at 19:23
Bob Terrell
1,665710
1,665710
add a comment |
add a comment |
For matrix$A^{-1}=A^T$ and column vectors have unit magnitude, it is unitary matrix. If we could do $A^2=I => A=A^{-1} => A=A^T$ if A is symmetrical and unitary then $A^2 =I$.
answered Nov 29 at 10:32
mahavir
4481320
add a comment |
For matrix$A^{-1}=A^T$ and column vectors have unit magnitude, it is unitary matrix. If we could do $A^2=I => A=A^{-1} => A=A^T$ if A is symmetrical and unitary then $A^2 =I$.
answered Nov 29 at 10:32
mahavir
4481320
add a comment |
For matrix$A^{-1}=A^T$ and column vectors have unit magnitude, it is unitary matrix. If we could do $A^2=I => A=A^{-1} => A=A^T$ if A is symmetrical and unitary then $A^2 =I$.
answered Nov 29 at 10:32
mahavir
4481320
For matrix$A^{-1}=A^T$ and column vectors have unit magnitude, it is unitary matrix. If we could do $A^2=I => A=A^{-1} => A=A^T$ if A is symmetrical and unitary then $A^2 =I$.
answered Nov 29 at 10:32
mahavir
4481320
answered Nov 29 at 10:32
mahavir
4481320
answered Nov 29 at 10:32
mahavir
4481320
answered Nov 29 at 10:32
mahavir
4481320
4481320
add a comment |
add a comment |
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