Is there an orthogonal matrix that is not unitary?
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
12.2k51840
asked Feb 16 '16 at 20:11
Eduardo Silva
68239
add a comment |
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
12.2k51840
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 |
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
12.2k51840
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
12.2k51840
asked Feb 16 '16 at 20:11
Eduardo Silva
68239
edited Feb 16 '16 at 20:16
John B
12.2k51840
asked Feb 16 '16 at 20:11
Eduardo Silva
68239
edited Feb 16 '16 at 20:16
John B
12.2k51840
edited Feb 16 '16 at 20:16
John B
12.2k51840
edited Feb 16 '16 at 20:16
John B
12.2k51840
12.2k51840
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
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
126k788176
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 '18 at 18:46
@DaniloGregorin well spotted
– Omnomnomnom
Nov 20 '18 at 19:51
add a comment |
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 '18 at 5:01
user1551
71.3k566125
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
126k788176
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 '18 at 18:46
@DaniloGregorin well spotted
– Omnomnomnom
Nov 20 '18 at 19:51
add a comment |
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
126k788176
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 '18 at 18:46
@DaniloGregorin well spotted
– Omnomnomnom
Nov 20 '18 at 19:51
add a comment |
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
126k788176
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
126k788176
edited Nov 20 '18 at 19:51
answered Feb 16 '16 at 20:36
Omnomnomnom
126k788176
answered Feb 16 '16 at 20:36
Omnomnomnom
126k788176
answered Feb 16 '16 at 20:36
Omnomnomnom
126k788176
126k788176
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 '18 at 18:46
@DaniloGregorin well spotted
– Omnomnomnom
Nov 20 '18 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 '18 at 18:46
@DaniloGregorin well spotted
– Omnomnomnom
Nov 20 '18 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 '18 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 '18 at 18:46
@DaniloGregorin well spotted
– Omnomnomnom
Nov 20 '18 at 19:51
@DaniloGregorin well spotted
– Omnomnomnom
Nov 20 '18 at 19:51
add a comment |
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 '18 at 5:01
user1551
71.3k566125
add a comment |
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 '18 at 5:01
user1551
71.3k566125
add a comment |
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 '18 at 5:01
user1551
71.3k566125
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 '18 at 5:01
user1551
71.3k566125
answered Nov 21 '18 at 5:01
user1551
71.3k566125
answered Nov 21 '18 at 5:01
user1551
71.3k566125
answered Nov 21 '18 at 5:01
user1551
71.3k566125
71.3k566125
add a comment |
add a comment |
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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