Is there an orthogonal matrix that is not unitary?
up vote
3
down vote
favorite
I could find a example of a unitary matrix such that is not orthogonal, thats simple in $mathbb{C}$, but for this exercise of a orthogonal that is not unitary i realize that is possible just on $mathbb{C}$ because all orthogonal matrix on $mathbb{R}$ is unitary, so anyone have a exemple of this case?
linear-algebra matrices
edited Feb 16 '16 at 20:16
John B
12.2k51740
asked Feb 16 '16 at 20:11
Eduardo Silva
68239
add a comment |
up vote
3
down vote
favorite
I could find a example of a unitary matrix such that is not orthogonal, thats simple in $mathbb{C}$, but for this exercise of a orthogonal that is not unitary i realize that is possible just on $mathbb{C}$ because all orthogonal matrix on $mathbb{R}$ is unitary, so anyone have a exemple of this case?
linear-algebra matrices
edited Feb 16 '16 at 20:16
John B
12.2k51740
asked Feb 16 '16 at 20:11
Eduardo Silva
68239
2
Just to be clear: you're asking for a matrix $A$ with complex entries for which $AA^T = I$, but $AA^* neq I$. Is that correct?
– Omnomnomnom
Feb 16 '16 at 20:19
yes, exactly, can u help?
– Eduardo Silva
Feb 16 '16 at 20:23
3
Note that Omnomnomnom's clarification is very important: if a matrix $A$ with real entries satisfies $A A^T=I$ then certainly $A A^*=I$ (since $A^*=A^T$).
– Ian
Feb 16 '16 at 20:38
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I could find a example of a unitary matrix such that is not orthogonal, thats simple in $mathbb{C}$, but for this exercise of a orthogonal that is not unitary i realize that is possible just on $mathbb{C}$ because all orthogonal matrix on $mathbb{R}$ is unitary, so anyone have a exemple of this case?
linear-algebra matrices
edited Feb 16 '16 at 20:16
John B
12.2k51740
asked Feb 16 '16 at 20:11
Eduardo Silva
68239
I could find a example of a unitary matrix such that is not orthogonal, thats simple in $mathbb{C}$, but for this exercise of a orthogonal that is not unitary i realize that is possible just on $mathbb{C}$ because all orthogonal matrix on $mathbb{R}$ is unitary, so anyone have a exemple of this case?
linear-algebra matrices
linear-algebra matrices
edited Feb 16 '16 at 20:16
John B
12.2k51740
asked Feb 16 '16 at 20:11
Eduardo Silva
68239
edited Feb 16 '16 at 20:16
John B
12.2k51740
asked Feb 16 '16 at 20:11
Eduardo Silva
68239
edited Feb 16 '16 at 20:16
John B
12.2k51740
edited Feb 16 '16 at 20:16
John B
12.2k51740
edited Feb 16 '16 at 20:16
John B
12.2k51740
12.2k51740
asked Feb 16 '16 at 20:11
Eduardo Silva
68239
asked Feb 16 '16 at 20:11
Eduardo Silva
68239
asked Feb 16 '16 at 20:11
Eduardo Silva
68239
68239
2
Just to be clear: you're asking for a matrix $A$ with complex entries for which $AA^T = I$, but $AA^* neq I$. Is that correct?
– Omnomnomnom
Feb 16 '16 at 20:19
yes, exactly, can u help?
– Eduardo Silva
Feb 16 '16 at 20:23
3
Note that Omnomnomnom's clarification is very important: if a matrix $A$ with real entries satisfies $A A^T=I$ then certainly $A A^*=I$ (since $A^*=A^T$).
– Ian
Feb 16 '16 at 20:38
add a comment |
2
Just to be clear: you're asking for a matrix $A$ with complex entries for which $AA^T = I$, but $AA^* neq I$. Is that correct?
– Omnomnomnom
Feb 16 '16 at 20:19
yes, exactly, can u help?
– Eduardo Silva
Feb 16 '16 at 20:23
3
Note that Omnomnomnom's clarification is very important: if a matrix $A$ with real entries satisfies $A A^T=I$ then certainly $A A^*=I$ (since $A^*=A^T$).
– Ian
Feb 16 '16 at 20:38
2
2
Just to be clear: you're asking for a matrix $A$ with complex entries for which $AA^T = I$, but $AA^* neq I$. Is that correct?
– Omnomnomnom
Feb 16 '16 at 20:19
Just to be clear: you're asking for a matrix $A$ with complex entries for which $AA^T = I$, but $AA^* neq I$. Is that correct?
– Omnomnomnom
Feb 16 '16 at 20:19
yes, exactly, can u help?
– Eduardo Silva
Feb 16 '16 at 20:23
yes, exactly, can u help?
– Eduardo Silva
Feb 16 '16 at 20:23
3
3
Note that Omnomnomnom's clarification is very important: if a matrix $A$ with real entries satisfies $A A^T=I$ then certainly $A A^*=I$ (since $A^*=A^T$).
– Ian
Feb 16 '16 at 20:38
Note that Omnomnomnom's clarification is very important: if a matrix $A$ with real entries satisfies $A A^T=I$ then certainly $A A^*=I$ (since $A^*=A^T$).
– Ian
Feb 16 '16 at 20:38
add a comment |
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
The matrix
$$
A = pmatrix{
sqrt{2}&i\
i&-sqrt{2}
}
$$
satisfies $AA^T = I$ but
$$
AA^* = pmatrix{5&-2isqrt{2}\2isqrt{2}&5}
$$
answered Feb 16 '16 at 20:36
Omnomnomnom
125k788176
Almost irrelevant, but you missed a factor of $2$ in the terms outside the main diagonal, didn't you? My calculations give $-2isqrt{2}$ and $2isqrt{2}$
– Danilo Gregorin
Nov 20 at 18:46
@DaniloGregorin well spotted
– Omnomnomnom
Nov 20 at 19:51
add a comment |
up vote
0
down vote
Late remark. In general, if $K$ is complex skew-symmetric, then $Q=e^{zK}$ is complex orthogonal for every $zinmathbb C$. When $Kne0$, since $e^{zK}$ is holomorphic and its power series expansion has some nonzero high-order terms, $e^{zK}$ is not a constant function. Therefore it cannot be real all the time (otherwise it will fail to satisfy Cauchy-Riemann equations). Pick a $z$ such that $Q=e^{zK}$ is not real. Then $Q$ is not unitary, because all unitary orthogonal matrices are real.
answered Nov 21 at 5:01
user1551
70.9k566125
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
The matrix
$$
A = pmatrix{
sqrt{2}&i\
i&-sqrt{2}
}
$$
satisfies $AA^T = I$ but
$$
AA^* = pmatrix{5&-2isqrt{2}\2isqrt{2}&5}
$$
answered Feb 16 '16 at 20:36
Omnomnomnom
125k788176
Almost irrelevant, but you missed a factor of $2$ in the terms outside the main diagonal, didn't you? My calculations give $-2isqrt{2}$ and $2isqrt{2}$
– Danilo Gregorin
Nov 20 at 18:46
@DaniloGregorin well spotted
– Omnomnomnom
Nov 20 at 19:51
add a comment |
up vote
7
down vote
accepted
The matrix
$$
A = pmatrix{
sqrt{2}&i\
i&-sqrt{2}
}
$$
satisfies $AA^T = I$ but
$$
AA^* = pmatrix{5&-2isqrt{2}\2isqrt{2}&5}
$$
answered Feb 16 '16 at 20:36
Omnomnomnom
125k788176
Almost irrelevant, but you missed a factor of $2$ in the terms outside the main diagonal, didn't you? My calculations give $-2isqrt{2}$ and $2isqrt{2}$
– Danilo Gregorin
Nov 20 at 18:46
@DaniloGregorin well spotted
– Omnomnomnom
Nov 20 at 19:51
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
The matrix
$$
A = pmatrix{
sqrt{2}&i\
i&-sqrt{2}
}
$$
satisfies $AA^T = I$ but
$$
AA^* = pmatrix{5&-2isqrt{2}\2isqrt{2}&5}
$$
answered Feb 16 '16 at 20:36
Omnomnomnom
125k788176
The matrix
$$
A = pmatrix{
sqrt{2}&i\
i&-sqrt{2}
}
$$
satisfies $AA^T = I$ but
$$
AA^* = pmatrix{5&-2isqrt{2}\2isqrt{2}&5}
$$
answered Feb 16 '16 at 20:36
Omnomnomnom
125k788176
edited Nov 20 at 19:51
answered Feb 16 '16 at 20:36
Omnomnomnom
125k788176
answered Feb 16 '16 at 20:36
Omnomnomnom
125k788176
answered Feb 16 '16 at 20:36
Omnomnomnom
125k788176
125k788176
Almost irrelevant, but you missed a factor of $2$ in the terms outside the main diagonal, didn't you? My calculations give $-2isqrt{2}$ and $2isqrt{2}$
– Danilo Gregorin
Nov 20 at 18:46
@DaniloGregorin well spotted
– Omnomnomnom
Nov 20 at 19:51
add a comment |
Almost irrelevant, but you missed a factor of $2$ in the terms outside the main diagonal, didn't you? My calculations give $-2isqrt{2}$ and $2isqrt{2}$
– Danilo Gregorin
Nov 20 at 18:46
@DaniloGregorin well spotted
– Omnomnomnom
Nov 20 at 19:51
Almost irrelevant, but you missed a factor of $2$ in the terms outside the main diagonal, didn't you? My calculations give $-2isqrt{2}$ and $2isqrt{2}$
– Danilo Gregorin
Nov 20 at 18:46
Almost irrelevant, but you missed a factor of $2$ in the terms outside the main diagonal, didn't you? My calculations give $-2isqrt{2}$ and $2isqrt{2}$
– Danilo Gregorin
Nov 20 at 18:46
@DaniloGregorin well spotted
– Omnomnomnom
Nov 20 at 19:51
@DaniloGregorin well spotted
– Omnomnomnom
Nov 20 at 19:51
add a comment |
up vote
0
down vote
Late remark. In general, if $K$ is complex skew-symmetric, then $Q=e^{zK}$ is complex orthogonal for every $zinmathbb C$. When $Kne0$, since $e^{zK}$ is holomorphic and its power series expansion has some nonzero high-order terms, $e^{zK}$ is not a constant function. Therefore it cannot be real all the time (otherwise it will fail to satisfy Cauchy-Riemann equations). Pick a $z$ such that $Q=e^{zK}$ is not real. Then $Q$ is not unitary, because all unitary orthogonal matrices are real.
answered Nov 21 at 5:01
user1551
70.9k566125
add a comment |
up vote
0
down vote
Late remark. In general, if $K$ is complex skew-symmetric, then $Q=e^{zK}$ is complex orthogonal for every $zinmathbb C$. When $Kne0$, since $e^{zK}$ is holomorphic and its power series expansion has some nonzero high-order terms, $e^{zK}$ is not a constant function. Therefore it cannot be real all the time (otherwise it will fail to satisfy Cauchy-Riemann equations). Pick a $z$ such that $Q=e^{zK}$ is not real. Then $Q$ is not unitary, because all unitary orthogonal matrices are real.
answered Nov 21 at 5:01
user1551
70.9k566125
add a comment |
up vote
0
down vote
up vote
0
down vote
Late remark. In general, if $K$ is complex skew-symmetric, then $Q=e^{zK}$ is complex orthogonal for every $zinmathbb C$. When $Kne0$, since $e^{zK}$ is holomorphic and its power series expansion has some nonzero high-order terms, $e^{zK}$ is not a constant function. Therefore it cannot be real all the time (otherwise it will fail to satisfy Cauchy-Riemann equations). Pick a $z$ such that $Q=e^{zK}$ is not real. Then $Q$ is not unitary, because all unitary orthogonal matrices are real.
answered Nov 21 at 5:01
user1551
70.9k566125
Late remark. In general, if $K$ is complex skew-symmetric, then $Q=e^{zK}$ is complex orthogonal for every $zinmathbb C$. When $Kne0$, since $e^{zK}$ is holomorphic and its power series expansion has some nonzero high-order terms, $e^{zK}$ is not a constant function. Therefore it cannot be real all the time (otherwise it will fail to satisfy Cauchy-Riemann equations). Pick a $z$ such that $Q=e^{zK}$ is not real. Then $Q$ is not unitary, because all unitary orthogonal matrices are real.
answered Nov 21 at 5:01
user1551
70.9k566125
answered Nov 21 at 5:01
user1551
70.9k566125
answered Nov 21 at 5:01
user1551
70.9k566125
answered Nov 21 at 5:01
user1551
70.9k566125
70.9k566125
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1658707%2fis-there-an-orthogonal-matrix-that-is-not-unitary%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
2
Just to be clear: you're asking for a matrix $A$ with complex entries for which $AA^T = I$, but $AA^* neq I$. Is that correct?
– Omnomnomnom
Feb 16 '16 at 20:19
yes, exactly, can u help?
– Eduardo Silva
Feb 16 '16 at 20:23
3
Note that Omnomnomnom's clarification is very important: if a matrix $A$ with real entries satisfies $A A^T=I$ then certainly $A A^*=I$ (since $A^*=A^T$).
– Ian
Feb 16 '16 at 20:38