Mixing
Showing that $−i(U − E)(U + E)^-1$ is a Hermitian matrix.
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Given $U in mathbbC^n times n$ a unitary Matrix that doesn't have $-1$ as an Eigenvalue.
I am trying show that $U + I$ is invertiable and that $−i(U − E)(U + E)^-1$ is a Hermitian matrix.
Let $lambda$ be an Eigenvalue of $U$
It follows that the Eigenvalues of $U + I$ have the form $lambda+1$. Since $lambda neq -1$ then $U + I$ doesn't have zero as an Eigenvalue and therefore invertiable. (Is this correct?)
Let $B = −i(U − I)(U + I)^-1 = (-iU + iI)(U + I)^-1$
We can show that $B^H = B$ or $B^T = overlineB$
$$overlineB = overline(-iU + iI)(U + I)^-1 = (-i overlineU - i I)(overlineU + I)^-1$$
$$B^T = (U^T + I)^-1(-i U^T + i I)$$
There is where I get stuck. Not sure how to relate $U^T$ and $overlineU$
linear-algebra matrices
asked Jul 16 at 23:24
yarafoudah
846
add a comment |Â
up vote
1
down vote
favorite
Given $U in mathbbC^n times n$ a unitary Matrix that doesn't have $-1$ as an Eigenvalue.
I am trying show that $U + I$ is invertiable and that $−i(U − E)(U + E)^-1$ is a Hermitian matrix.
Let $lambda$ be an Eigenvalue of $U$
It follows that the Eigenvalues of $U + I$ have the form $lambda+1$. Since $lambda neq -1$ then $U + I$ doesn't have zero as an Eigenvalue and therefore invertiable. (Is this correct?)
Let $B = −i(U − I)(U + I)^-1 = (-iU + iI)(U + I)^-1$
We can show that $B^H = B$ or $B^T = overlineB$
$$overlineB = overline(-iU + iI)(U + I)^-1 = (-i overlineU - i I)(overlineU + I)^-1$$
$$B^T = (U^T + I)^-1(-i U^T + i I)$$
There is where I get stuck. Not sure how to relate $U^T$ and $overlineU$
linear-algebra matrices
asked Jul 16 at 23:24
yarafoudah
846
$U$ is unitary. What does that tell you about $U^T$ and $overlineU$?
– Kolja
Jul 16 at 23:38
I am not really sure. Is $U^T = U$ ?
– yarafoudah
Jul 16 at 23:47
It's $U^H=U^-1$.
– Kolja
Jul 17 at 10:12
What are the constraints on $E$ as a matrix?
– Xoque55
Jul 17 at 15:03
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given $U in mathbbC^n times n$ a unitary Matrix that doesn't have $-1$ as an Eigenvalue.
I am trying show that $U + I$ is invertiable and that $−i(U − E)(U + E)^-1$ is a Hermitian matrix.
Let $lambda$ be an Eigenvalue of $U$
It follows that the Eigenvalues of $U + I$ have the form $lambda+1$. Since $lambda neq -1$ then $U + I$ doesn't have zero as an Eigenvalue and therefore invertiable. (Is this correct?)
Let $B = −i(U − I)(U + I)^-1 = (-iU + iI)(U + I)^-1$
We can show that $B^H = B$ or $B^T = overlineB$
$$overlineB = overline(-iU + iI)(U + I)^-1 = (-i overlineU - i I)(overlineU + I)^-1$$
$$B^T = (U^T + I)^-1(-i U^T + i I)$$
There is where I get stuck. Not sure how to relate $U^T$ and $overlineU$
linear-algebra matrices
asked Jul 16 at 23:24
yarafoudah
846
Given $U in mathbbC^n times n$ a unitary Matrix that doesn't have $-1$ as an Eigenvalue.
I am trying show that $U + I$ is invertiable and that $−i(U − E)(U + E)^-1$ is a Hermitian matrix.
Let $lambda$ be an Eigenvalue of $U$
It follows that the Eigenvalues of $U + I$ have the form $lambda+1$. Since $lambda neq -1$ then $U + I$ doesn't have zero as an Eigenvalue and therefore invertiable. (Is this correct?)
Let $B = −i(U − I)(U + I)^-1 = (-iU + iI)(U + I)^-1$
We can show that $B^H = B$ or $B^T = overlineB$
$$overlineB = overline(-iU + iI)(U + I)^-1 = (-i overlineU - i I)(overlineU + I)^-1$$
$$B^T = (U^T + I)^-1(-i U^T + i I)$$
There is where I get stuck. Not sure how to relate $U^T$ and $overlineU$
linear-algebra matrices
asked Jul 16 at 23:24
yarafoudah
846
asked Jul 16 at 23:24
yarafoudah
846
asked Jul 16 at 23:24
yarafoudah
846
asked Jul 16 at 23:24
yarafoudah
846
846
$U$ is unitary. What does that tell you about $U^T$ and $overlineU$?
– Kolja
Jul 16 at 23:38
I am not really sure. Is $U^T = U$ ?
– yarafoudah
Jul 16 at 23:47
It's $U^H=U^-1$.
– Kolja
Jul 17 at 10:12
What are the constraints on $E$ as a matrix?
– Xoque55
Jul 17 at 15:03
add a comment |Â
$U$ is unitary. What does that tell you about $U^T$ and $overlineU$?
– Kolja
Jul 16 at 23:38
I am not really sure. Is $U^T = U$ ?
– yarafoudah
Jul 16 at 23:47
It's $U^H=U^-1$.
– Kolja
Jul 17 at 10:12
What are the constraints on $E$ as a matrix?
– Xoque55
Jul 17 at 15:03
$U$ is unitary. What does that tell you about $U^T$ and $overlineU$?
– Kolja
Jul 16 at 23:38
$U$ is unitary. What does that tell you about $U^T$ and $overlineU$?
– Kolja
Jul 16 at 23:38
I am not really sure. Is $U^T = U$ ?
– yarafoudah
Jul 16 at 23:47
I am not really sure. Is $U^T = U$ ?
– yarafoudah
Jul 16 at 23:47
It's $U^H=U^-1$.
– Kolja
Jul 17 at 10:12
It's $U^H=U^-1$.
– Kolja
Jul 17 at 10:12
What are the constraints on $E$ as a matrix?
– Xoque55
Jul 17 at 15:03
What are the constraints on $E$ as a matrix?
– Xoque55
Jul 17 at 15:03
add a comment |Â
1 Answer
1
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oldest
votes
up vote
0
down vote
Since a unitary matrix is normal, it is unitarily diagonalizable. So all that needs to be shown is that if $|lambda|=1$, $lambdane-1$, then
$$
-i,fraclambda-1lambda+1in mathbb R.
$$
For this,
$$
-i,fraclambda-1lambda+1
=-i,frac(lambda-1)(barlambda+1)lambda+1,
$$
so we only care about the numerator now. And
$$
-i(lambda-1)(barlambda+1)=-i(|lambda|^2-1+lambda-barlambda)=-i(lambda-barlambda)=2operatornameImlambdainmathbb R.
$$
answered Jul 17 at 15:00
Martin Argerami
116k1071164
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Since a unitary matrix is normal, it is unitarily diagonalizable. So all that needs to be shown is that if $|lambda|=1$, $lambdane-1$, then
$$
-i,fraclambda-1lambda+1in mathbb R.
$$
For this,
$$
-i,fraclambda-1lambda+1
=-i,frac(lambda-1)(barlambda+1)lambda+1,
$$
so we only care about the numerator now. And
$$
-i(lambda-1)(barlambda+1)=-i(|lambda|^2-1+lambda-barlambda)=-i(lambda-barlambda)=2operatornameImlambdainmathbb R.
$$
answered Jul 17 at 15:00
Martin Argerami
116k1071164
add a comment |Â
up vote
0
down vote
Since a unitary matrix is normal, it is unitarily diagonalizable. So all that needs to be shown is that if $|lambda|=1$, $lambdane-1$, then
$$
-i,fraclambda-1lambda+1in mathbb R.
$$
For this,
$$
-i,fraclambda-1lambda+1
=-i,frac(lambda-1)(barlambda+1)lambda+1,
$$
so we only care about the numerator now. And
$$
-i(lambda-1)(barlambda+1)=-i(|lambda|^2-1+lambda-barlambda)=-i(lambda-barlambda)=2operatornameImlambdainmathbb R.
$$
answered Jul 17 at 15:00
Martin Argerami
116k1071164
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since a unitary matrix is normal, it is unitarily diagonalizable. So all that needs to be shown is that if $|lambda|=1$, $lambdane-1$, then
$$
-i,fraclambda-1lambda+1in mathbb R.
$$
For this,
$$
-i,fraclambda-1lambda+1
=-i,frac(lambda-1)(barlambda+1)lambda+1,
$$
so we only care about the numerator now. And
$$
-i(lambda-1)(barlambda+1)=-i(|lambda|^2-1+lambda-barlambda)=-i(lambda-barlambda)=2operatornameImlambdainmathbb R.
$$
answered Jul 17 at 15:00
Martin Argerami
116k1071164
Since a unitary matrix is normal, it is unitarily diagonalizable. So all that needs to be shown is that if $|lambda|=1$, $lambdane-1$, then
$$
-i,fraclambda-1lambda+1in mathbb R.
$$
For this,
$$
-i,fraclambda-1lambda+1
=-i,frac(lambda-1)(barlambda+1)lambda+1,
$$
so we only care about the numerator now. And
$$
-i(lambda-1)(barlambda+1)=-i(|lambda|^2-1+lambda-barlambda)=-i(lambda-barlambda)=2operatornameImlambdainmathbb R.
$$
answered Jul 17 at 15:00
Martin Argerami
116k1071164
answered Jul 17 at 15:00
Martin Argerami
116k1071164
answered Jul 17 at 15:00
Martin Argerami
116k1071164
answered Jul 17 at 15:00
Martin Argerami
116k1071164
116k1071164
add a comment |Â
add a comment |Â
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$U$ is unitary. What does that tell you about $U^T$ and $overlineU$?
– Kolja
Jul 16 at 23:38
I am not really sure. Is $U^T = U$ ?
– yarafoudah
Jul 16 at 23:47
It's $U^H=U^-1$.
– Kolja
Jul 17 at 10:12
What are the constraints on $E$ as a matrix?
– Xoque55
Jul 17 at 15:03