Is $X$ always diagonal matrix when $e^{iX} = 1$?
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Suppose $e^{iX} = 1$. Then is $X$ always a diagonal matrix? What happens if $X$ is constrained to be a hermitian matrix?
linear-algebra
asked Dec 4 '18 at 7:22
Mark VanderbiltMark Vanderbilt
284
$endgroup$
add a comment |
$begingroup$
Suppose $e^{iX} = 1$. Then is $X$ always a diagonal matrix? What happens if $X$ is constrained to be a hermitian matrix?
linear-algebra
asked Dec 4 '18 at 7:22
Mark VanderbiltMark Vanderbilt
284
$endgroup$
$begingroup$
For $e^{iX}=1$, $X$ has to be hermitian. In general if $A^2=1$, then $X=2 pi A$ will work. $A$ doesn't have to be diagonal, see the answers.
$endgroup$
– lisyarus
Dec 4 '18 at 7:36
add a comment |
$begingroup$
Suppose $e^{iX} = 1$. Then is $X$ always a diagonal matrix? What happens if $X$ is constrained to be a hermitian matrix?
linear-algebra
asked Dec 4 '18 at 7:22
Mark VanderbiltMark Vanderbilt
284
$endgroup$
Suppose $e^{iX} = 1$. Then is $X$ always a diagonal matrix? What happens if $X$ is constrained to be a hermitian matrix?
linear-algebra
linear-algebra
asked Dec 4 '18 at 7:22
Mark VanderbiltMark Vanderbilt
284
asked Dec 4 '18 at 7:22
Mark VanderbiltMark Vanderbilt
284
asked Dec 4 '18 at 7:22
Mark VanderbiltMark Vanderbilt
284
asked Dec 4 '18 at 7:22
Mark VanderbiltMark Vanderbilt
284
asked Dec 4 '18 at 7:22
Mark VanderbiltMark Vanderbilt
284
284
$begingroup$
For $e^{iX}=1$, $X$ has to be hermitian. In general if $A^2=1$, then $X=2 pi A$ will work. $A$ doesn't have to be diagonal, see the answers.
$endgroup$
– lisyarus
Dec 4 '18 at 7:36
add a comment |
$begingroup$
For $e^{iX}=1$, $X$ has to be hermitian. In general if $A^2=1$, then $X=2 pi A$ will work. $A$ doesn't have to be diagonal, see the answers.
$endgroup$
– lisyarus
Dec 4 '18 at 7:36
$begingroup$
For $e^{iX}=1$, $X$ has to be hermitian. In general if $A^2=1$, then $X=2 pi A$ will work. $A$ doesn't have to be diagonal, see the answers.
$endgroup$
– lisyarus
Dec 4 '18 at 7:36
$begingroup$
For $e^{iX}=1$, $X$ has to be hermitian. In general if $A^2=1$, then $X=2 pi A$ will work. $A$ doesn't have to be diagonal, see the answers.
$endgroup$
– lisyarus
Dec 4 '18 at 7:36
add a comment |
1 Answer
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$begingroup$
We could have
$$X=2pipmatrix{1&0\0&-1}.$$
That is diagonal! But if we take a conjugate of $X$ that will still work, say
$$X=2pipmatrix{0&1\1&0}$$
instead.
answered Dec 4 '18 at 7:28
Lord Shark the UnknownLord Shark the Unknown
102k1059132
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We could have
$$X=2pipmatrix{1&0\0&-1}.$$
That is diagonal! But if we take a conjugate of $X$ that will still work, say
$$X=2pipmatrix{0&1\1&0}$$
instead.
answered Dec 4 '18 at 7:28
Lord Shark the UnknownLord Shark the Unknown
102k1059132
$endgroup$
add a comment |
$begingroup$
We could have
$$X=2pipmatrix{1&0\0&-1}.$$
That is diagonal! But if we take a conjugate of $X$ that will still work, say
$$X=2pipmatrix{0&1\1&0}$$
instead.
answered Dec 4 '18 at 7:28
Lord Shark the UnknownLord Shark the Unknown
102k1059132
$endgroup$
add a comment |
$begingroup$
We could have
$$X=2pipmatrix{1&0\0&-1}.$$
That is diagonal! But if we take a conjugate of $X$ that will still work, say
$$X=2pipmatrix{0&1\1&0}$$
instead.
answered Dec 4 '18 at 7:28
Lord Shark the UnknownLord Shark the Unknown
102k1059132
$endgroup$
We could have
$$X=2pipmatrix{1&0\0&-1}.$$
That is diagonal! But if we take a conjugate of $X$ that will still work, say
$$X=2pipmatrix{0&1\1&0}$$
instead.
answered Dec 4 '18 at 7:28
Lord Shark the UnknownLord Shark the Unknown
102k1059132
answered Dec 4 '18 at 7:28
Lord Shark the UnknownLord Shark the Unknown
102k1059132
answered Dec 4 '18 at 7:28
Lord Shark the UnknownLord Shark the Unknown
102k1059132
answered Dec 4 '18 at 7:28
Lord Shark the UnknownLord Shark the Unknown
102k1059132
102k1059132
add a comment |
add a comment |
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$begingroup$
For $e^{iX}=1$, $X$ has to be hermitian. In general if $A^2=1$, then $X=2 pi A$ will work. $A$ doesn't have to be diagonal, see the answers.
$endgroup$
– lisyarus
Dec 4 '18 at 7:36