Hermitian Property of a Householder Transform on a Complex Field
up vote
2
down vote
favorite
Let $mathbf{H}$ be a Householder (i.e. elementary reflector), such that $mathbf{Hx} = mathbf{e}_1$, for an $mathbf{x} in Bbb{C}^n$, having $|mathbf{x}|_2 = 1$.
For this I have defined $mathbf{H} = e^{itheta}left(mathbf{I} - 2frac{mathbf{u}mathbf{u}^*}{mathbf{u}^*mathbf{u}}right)$, where $e^{itheta} = frac{overline{x_1}}{|x_1|}$ and $mathbf{u} = mathbf{x} - e^{itheta}|mathbf{x}|_2mathbf{e}_1$. I am indeed getting the expected result of $mathbf{Hx} = mathbf{e}_1$, but my trouble is figuring out how to enforce the property that $mathbf{H} = mathbf{H}^*$. Can anyone please point me in the right direction?
I have also checked that this $mathbf{H}$ is unitary, since $mathbf{H}^* = mathbf{H^{-1}} = e^{-itheta}left(mathbf{I} - 2frac{mathbf{u}mathbf{u}^*}{mathbf{u}^*mathbf{u}}right)$ and $mathbf{H}^*mathbf{H} = mathbf{I}$, but again, I don't see how $mathbf{H} = mathbf{H^{-1}}$. I even get that $mathbf{H}^{-1}mathbf{e}_1 = mathbf{x}$, which should be expected.
Any help is really well appreciated.
linear-algebra transformation reflection
add a comment |
up vote
2
down vote
favorite
Let $mathbf{H}$ be a Householder (i.e. elementary reflector), such that $mathbf{Hx} = mathbf{e}_1$, for an $mathbf{x} in Bbb{C}^n$, having $|mathbf{x}|_2 = 1$.
For this I have defined $mathbf{H} = e^{itheta}left(mathbf{I} - 2frac{mathbf{u}mathbf{u}^*}{mathbf{u}^*mathbf{u}}right)$, where $e^{itheta} = frac{overline{x_1}}{|x_1|}$ and $mathbf{u} = mathbf{x} - e^{itheta}|mathbf{x}|_2mathbf{e}_1$. I am indeed getting the expected result of $mathbf{Hx} = mathbf{e}_1$, but my trouble is figuring out how to enforce the property that $mathbf{H} = mathbf{H}^*$. Can anyone please point me in the right direction?
I have also checked that this $mathbf{H}$ is unitary, since $mathbf{H}^* = mathbf{H^{-1}} = e^{-itheta}left(mathbf{I} - 2frac{mathbf{u}mathbf{u}^*}{mathbf{u}^*mathbf{u}}right)$ and $mathbf{H}^*mathbf{H} = mathbf{I}$, but again, I don't see how $mathbf{H} = mathbf{H^{-1}}$. I even get that $mathbf{H}^{-1}mathbf{e}_1 = mathbf{x}$, which should be expected.
Any help is really well appreciated.
linear-algebra transformation reflection
A Householder reflection $H$ by definition is a linear operator that is a reflection ($xmapsto-x$) on some one-dimensional subspace and an identity map on the orthogonal hyperplane. So, by definition, it must be Hermitian. It follows that if $Hx=y$, then $x^ast y (=x^ast Hx)$ must be real. In your case, if $x_1$ is not real, $H$ simply does not exist. If $x_1$ is real instead, then $e^{itheta}=1$ and your problem is automatically resolved.
– user1551
Nov 15 at 12:08
@user1551 To corroborate what you have kindly provided, then, I can't build an $mathbf{H}$ with the Hermitian property if the working field is complex, right?
– Im YoungMin
Nov 15 at 18:47
Sometimes you can and sometimes you cannot. That depends on whether $x_1$ (the first element of $x$ is real). You can always construct $H$ in the way outlined in your question, but unless $e^{itheta}$ is real, the resulting $H$ is not Hermitian (and strictly speaking, it is not a reflection either). Rather it is a complex scalar multiple of a Hermitian unitary matrix.
– user1551
Nov 16 at 1:52
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $mathbf{H}$ be a Householder (i.e. elementary reflector), such that $mathbf{Hx} = mathbf{e}_1$, for an $mathbf{x} in Bbb{C}^n$, having $|mathbf{x}|_2 = 1$.
For this I have defined $mathbf{H} = e^{itheta}left(mathbf{I} - 2frac{mathbf{u}mathbf{u}^*}{mathbf{u}^*mathbf{u}}right)$, where $e^{itheta} = frac{overline{x_1}}{|x_1|}$ and $mathbf{u} = mathbf{x} - e^{itheta}|mathbf{x}|_2mathbf{e}_1$. I am indeed getting the expected result of $mathbf{Hx} = mathbf{e}_1$, but my trouble is figuring out how to enforce the property that $mathbf{H} = mathbf{H}^*$. Can anyone please point me in the right direction?
I have also checked that this $mathbf{H}$ is unitary, since $mathbf{H}^* = mathbf{H^{-1}} = e^{-itheta}left(mathbf{I} - 2frac{mathbf{u}mathbf{u}^*}{mathbf{u}^*mathbf{u}}right)$ and $mathbf{H}^*mathbf{H} = mathbf{I}$, but again, I don't see how $mathbf{H} = mathbf{H^{-1}}$. I even get that $mathbf{H}^{-1}mathbf{e}_1 = mathbf{x}$, which should be expected.
Any help is really well appreciated.
linear-algebra transformation reflection
Let $mathbf{H}$ be a Householder (i.e. elementary reflector), such that $mathbf{Hx} = mathbf{e}_1$, for an $mathbf{x} in Bbb{C}^n$, having $|mathbf{x}|_2 = 1$.
For this I have defined $mathbf{H} = e^{itheta}left(mathbf{I} - 2frac{mathbf{u}mathbf{u}^*}{mathbf{u}^*mathbf{u}}right)$, where $e^{itheta} = frac{overline{x_1}}{|x_1|}$ and $mathbf{u} = mathbf{x} - e^{itheta}|mathbf{x}|_2mathbf{e}_1$. I am indeed getting the expected result of $mathbf{Hx} = mathbf{e}_1$, but my trouble is figuring out how to enforce the property that $mathbf{H} = mathbf{H}^*$. Can anyone please point me in the right direction?
I have also checked that this $mathbf{H}$ is unitary, since $mathbf{H}^* = mathbf{H^{-1}} = e^{-itheta}left(mathbf{I} - 2frac{mathbf{u}mathbf{u}^*}{mathbf{u}^*mathbf{u}}right)$ and $mathbf{H}^*mathbf{H} = mathbf{I}$, but again, I don't see how $mathbf{H} = mathbf{H^{-1}}$. I even get that $mathbf{H}^{-1}mathbf{e}_1 = mathbf{x}$, which should be expected.
Any help is really well appreciated.
linear-algebra transformation reflection
linear-algebra transformation reflection
edited Nov 15 at 18:39
asked Nov 15 at 8:53
Im YoungMin
134
134
A Householder reflection $H$ by definition is a linear operator that is a reflection ($xmapsto-x$) on some one-dimensional subspace and an identity map on the orthogonal hyperplane. So, by definition, it must be Hermitian. It follows that if $Hx=y$, then $x^ast y (=x^ast Hx)$ must be real. In your case, if $x_1$ is not real, $H$ simply does not exist. If $x_1$ is real instead, then $e^{itheta}=1$ and your problem is automatically resolved.
– user1551
Nov 15 at 12:08
@user1551 To corroborate what you have kindly provided, then, I can't build an $mathbf{H}$ with the Hermitian property if the working field is complex, right?
– Im YoungMin
Nov 15 at 18:47
Sometimes you can and sometimes you cannot. That depends on whether $x_1$ (the first element of $x$ is real). You can always construct $H$ in the way outlined in your question, but unless $e^{itheta}$ is real, the resulting $H$ is not Hermitian (and strictly speaking, it is not a reflection either). Rather it is a complex scalar multiple of a Hermitian unitary matrix.
– user1551
Nov 16 at 1:52
add a comment |
A Householder reflection $H$ by definition is a linear operator that is a reflection ($xmapsto-x$) on some one-dimensional subspace and an identity map on the orthogonal hyperplane. So, by definition, it must be Hermitian. It follows that if $Hx=y$, then $x^ast y (=x^ast Hx)$ must be real. In your case, if $x_1$ is not real, $H$ simply does not exist. If $x_1$ is real instead, then $e^{itheta}=1$ and your problem is automatically resolved.
– user1551
Nov 15 at 12:08
@user1551 To corroborate what you have kindly provided, then, I can't build an $mathbf{H}$ with the Hermitian property if the working field is complex, right?
– Im YoungMin
Nov 15 at 18:47
Sometimes you can and sometimes you cannot. That depends on whether $x_1$ (the first element of $x$ is real). You can always construct $H$ in the way outlined in your question, but unless $e^{itheta}$ is real, the resulting $H$ is not Hermitian (and strictly speaking, it is not a reflection either). Rather it is a complex scalar multiple of a Hermitian unitary matrix.
– user1551
Nov 16 at 1:52
A Householder reflection $H$ by definition is a linear operator that is a reflection ($xmapsto-x$) on some one-dimensional subspace and an identity map on the orthogonal hyperplane. So, by definition, it must be Hermitian. It follows that if $Hx=y$, then $x^ast y (=x^ast Hx)$ must be real. In your case, if $x_1$ is not real, $H$ simply does not exist. If $x_1$ is real instead, then $e^{itheta}=1$ and your problem is automatically resolved.
– user1551
Nov 15 at 12:08
A Householder reflection $H$ by definition is a linear operator that is a reflection ($xmapsto-x$) on some one-dimensional subspace and an identity map on the orthogonal hyperplane. So, by definition, it must be Hermitian. It follows that if $Hx=y$, then $x^ast y (=x^ast Hx)$ must be real. In your case, if $x_1$ is not real, $H$ simply does not exist. If $x_1$ is real instead, then $e^{itheta}=1$ and your problem is automatically resolved.
– user1551
Nov 15 at 12:08
@user1551 To corroborate what you have kindly provided, then, I can't build an $mathbf{H}$ with the Hermitian property if the working field is complex, right?
– Im YoungMin
Nov 15 at 18:47
@user1551 To corroborate what you have kindly provided, then, I can't build an $mathbf{H}$ with the Hermitian property if the working field is complex, right?
– Im YoungMin
Nov 15 at 18:47
Sometimes you can and sometimes you cannot. That depends on whether $x_1$ (the first element of $x$ is real). You can always construct $H$ in the way outlined in your question, but unless $e^{itheta}$ is real, the resulting $H$ is not Hermitian (and strictly speaking, it is not a reflection either). Rather it is a complex scalar multiple of a Hermitian unitary matrix.
– user1551
Nov 16 at 1:52
Sometimes you can and sometimes you cannot. That depends on whether $x_1$ (the first element of $x$ is real). You can always construct $H$ in the way outlined in your question, but unless $e^{itheta}$ is real, the resulting $H$ is not Hermitian (and strictly speaking, it is not a reflection either). Rather it is a complex scalar multiple of a Hermitian unitary matrix.
– user1551
Nov 16 at 1:52
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
It is not always possible to construct a Householder reflection that maps a prespecified unit vector in $mathbb C^n$ to another.
A Householder reflection $H$, by definition, is a linear operator whose restrictions are $-operatorname{Id}$ on $V$ and $operatorname{Id}$ on $V^perp$ for some one-dimensional subspace $Vsubseteqmathbb C^n$. For any $x,yinmathbb C^n$, if we write $x=u_x+v_x$ and $y=u_y+v_y$ where $u_x,u_yin V^perp$ and $v_x,v_yin V$, then
$$
langle Hx,yrangle
=langle u_x-v_x,u_y+v_yrangle
=langle u_x,u_yrangle - langle v_x,v_yrangle
=langle u_x+v_x,u_y-v_yrangle
=langle x,Hyrangle.
$$
It follows that $H$ is necessarily Hermitian. Being Hermitian is not an optional property, but a must.
Consequently, if $y=Hx$ for some unit vectors $x$ and $y$, then $langle x,yrangle=langle x,Hxrangle$ must be real. In other words, you can construct a Householder reflection that maps $x$ to $y$ only when $langle x,yrangle$ is real.
This is an often neglected point. People usually only deal with Householder reflection over the reals. Since $langle x,yrangle$ is real in this case, they don't realise that the story is a bit different over $mathbb C$.
In your case, if you want $Hx=e_1$, we need $langle x,e_1rangle=x_1$ to be real. If $x_1$ is not real, you cannot construct $H$. If $x_1$ is real instead, then $e^{itheta}=1$ and all apparent difficulties in your question vanish. The construction is then analogous to that over $mathbb R$. Just replace any matrix/vector transpose by conjugate transpose.
Some people do try to generalise Householder reflection in the real case to the complex case in a similar manner to your question. Now the construction is always possible. However, the resulting $H$ in general is neither Hermitian nor a reflection, but a complex scalar multiple of a Hermitian Householder reflection.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
It is not always possible to construct a Householder reflection that maps a prespecified unit vector in $mathbb C^n$ to another.
A Householder reflection $H$, by definition, is a linear operator whose restrictions are $-operatorname{Id}$ on $V$ and $operatorname{Id}$ on $V^perp$ for some one-dimensional subspace $Vsubseteqmathbb C^n$. For any $x,yinmathbb C^n$, if we write $x=u_x+v_x$ and $y=u_y+v_y$ where $u_x,u_yin V^perp$ and $v_x,v_yin V$, then
$$
langle Hx,yrangle
=langle u_x-v_x,u_y+v_yrangle
=langle u_x,u_yrangle - langle v_x,v_yrangle
=langle u_x+v_x,u_y-v_yrangle
=langle x,Hyrangle.
$$
It follows that $H$ is necessarily Hermitian. Being Hermitian is not an optional property, but a must.
Consequently, if $y=Hx$ for some unit vectors $x$ and $y$, then $langle x,yrangle=langle x,Hxrangle$ must be real. In other words, you can construct a Householder reflection that maps $x$ to $y$ only when $langle x,yrangle$ is real.
This is an often neglected point. People usually only deal with Householder reflection over the reals. Since $langle x,yrangle$ is real in this case, they don't realise that the story is a bit different over $mathbb C$.
In your case, if you want $Hx=e_1$, we need $langle x,e_1rangle=x_1$ to be real. If $x_1$ is not real, you cannot construct $H$. If $x_1$ is real instead, then $e^{itheta}=1$ and all apparent difficulties in your question vanish. The construction is then analogous to that over $mathbb R$. Just replace any matrix/vector transpose by conjugate transpose.
Some people do try to generalise Householder reflection in the real case to the complex case in a similar manner to your question. Now the construction is always possible. However, the resulting $H$ in general is neither Hermitian nor a reflection, but a complex scalar multiple of a Hermitian Householder reflection.
add a comment |
up vote
0
down vote
accepted
It is not always possible to construct a Householder reflection that maps a prespecified unit vector in $mathbb C^n$ to another.
A Householder reflection $H$, by definition, is a linear operator whose restrictions are $-operatorname{Id}$ on $V$ and $operatorname{Id}$ on $V^perp$ for some one-dimensional subspace $Vsubseteqmathbb C^n$. For any $x,yinmathbb C^n$, if we write $x=u_x+v_x$ and $y=u_y+v_y$ where $u_x,u_yin V^perp$ and $v_x,v_yin V$, then
$$
langle Hx,yrangle
=langle u_x-v_x,u_y+v_yrangle
=langle u_x,u_yrangle - langle v_x,v_yrangle
=langle u_x+v_x,u_y-v_yrangle
=langle x,Hyrangle.
$$
It follows that $H$ is necessarily Hermitian. Being Hermitian is not an optional property, but a must.
Consequently, if $y=Hx$ for some unit vectors $x$ and $y$, then $langle x,yrangle=langle x,Hxrangle$ must be real. In other words, you can construct a Householder reflection that maps $x$ to $y$ only when $langle x,yrangle$ is real.
This is an often neglected point. People usually only deal with Householder reflection over the reals. Since $langle x,yrangle$ is real in this case, they don't realise that the story is a bit different over $mathbb C$.
In your case, if you want $Hx=e_1$, we need $langle x,e_1rangle=x_1$ to be real. If $x_1$ is not real, you cannot construct $H$. If $x_1$ is real instead, then $e^{itheta}=1$ and all apparent difficulties in your question vanish. The construction is then analogous to that over $mathbb R$. Just replace any matrix/vector transpose by conjugate transpose.
Some people do try to generalise Householder reflection in the real case to the complex case in a similar manner to your question. Now the construction is always possible. However, the resulting $H$ in general is neither Hermitian nor a reflection, but a complex scalar multiple of a Hermitian Householder reflection.
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
It is not always possible to construct a Householder reflection that maps a prespecified unit vector in $mathbb C^n$ to another.
A Householder reflection $H$, by definition, is a linear operator whose restrictions are $-operatorname{Id}$ on $V$ and $operatorname{Id}$ on $V^perp$ for some one-dimensional subspace $Vsubseteqmathbb C^n$. For any $x,yinmathbb C^n$, if we write $x=u_x+v_x$ and $y=u_y+v_y$ where $u_x,u_yin V^perp$ and $v_x,v_yin V$, then
$$
langle Hx,yrangle
=langle u_x-v_x,u_y+v_yrangle
=langle u_x,u_yrangle - langle v_x,v_yrangle
=langle u_x+v_x,u_y-v_yrangle
=langle x,Hyrangle.
$$
It follows that $H$ is necessarily Hermitian. Being Hermitian is not an optional property, but a must.
Consequently, if $y=Hx$ for some unit vectors $x$ and $y$, then $langle x,yrangle=langle x,Hxrangle$ must be real. In other words, you can construct a Householder reflection that maps $x$ to $y$ only when $langle x,yrangle$ is real.
This is an often neglected point. People usually only deal with Householder reflection over the reals. Since $langle x,yrangle$ is real in this case, they don't realise that the story is a bit different over $mathbb C$.
In your case, if you want $Hx=e_1$, we need $langle x,e_1rangle=x_1$ to be real. If $x_1$ is not real, you cannot construct $H$. If $x_1$ is real instead, then $e^{itheta}=1$ and all apparent difficulties in your question vanish. The construction is then analogous to that over $mathbb R$. Just replace any matrix/vector transpose by conjugate transpose.
Some people do try to generalise Householder reflection in the real case to the complex case in a similar manner to your question. Now the construction is always possible. However, the resulting $H$ in general is neither Hermitian nor a reflection, but a complex scalar multiple of a Hermitian Householder reflection.
It is not always possible to construct a Householder reflection that maps a prespecified unit vector in $mathbb C^n$ to another.
A Householder reflection $H$, by definition, is a linear operator whose restrictions are $-operatorname{Id}$ on $V$ and $operatorname{Id}$ on $V^perp$ for some one-dimensional subspace $Vsubseteqmathbb C^n$. For any $x,yinmathbb C^n$, if we write $x=u_x+v_x$ and $y=u_y+v_y$ where $u_x,u_yin V^perp$ and $v_x,v_yin V$, then
$$
langle Hx,yrangle
=langle u_x-v_x,u_y+v_yrangle
=langle u_x,u_yrangle - langle v_x,v_yrangle
=langle u_x+v_x,u_y-v_yrangle
=langle x,Hyrangle.
$$
It follows that $H$ is necessarily Hermitian. Being Hermitian is not an optional property, but a must.
Consequently, if $y=Hx$ for some unit vectors $x$ and $y$, then $langle x,yrangle=langle x,Hxrangle$ must be real. In other words, you can construct a Householder reflection that maps $x$ to $y$ only when $langle x,yrangle$ is real.
This is an often neglected point. People usually only deal with Householder reflection over the reals. Since $langle x,yrangle$ is real in this case, they don't realise that the story is a bit different over $mathbb C$.
In your case, if you want $Hx=e_1$, we need $langle x,e_1rangle=x_1$ to be real. If $x_1$ is not real, you cannot construct $H$. If $x_1$ is real instead, then $e^{itheta}=1$ and all apparent difficulties in your question vanish. The construction is then analogous to that over $mathbb R$. Just replace any matrix/vector transpose by conjugate transpose.
Some people do try to generalise Householder reflection in the real case to the complex case in a similar manner to your question. Now the construction is always possible. However, the resulting $H$ in general is neither Hermitian nor a reflection, but a complex scalar multiple of a Hermitian Householder reflection.
edited Nov 16 at 1:41
answered Nov 16 at 1:28
user1551
70.3k566125
70.3k566125
add a comment |
add a comment |
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%2f2999435%2fhermitian-property-of-a-householder-transform-on-a-complex-field%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
A Householder reflection $H$ by definition is a linear operator that is a reflection ($xmapsto-x$) on some one-dimensional subspace and an identity map on the orthogonal hyperplane. So, by definition, it must be Hermitian. It follows that if $Hx=y$, then $x^ast y (=x^ast Hx)$ must be real. In your case, if $x_1$ is not real, $H$ simply does not exist. If $x_1$ is real instead, then $e^{itheta}=1$ and your problem is automatically resolved.
– user1551
Nov 15 at 12:08
@user1551 To corroborate what you have kindly provided, then, I can't build an $mathbf{H}$ with the Hermitian property if the working field is complex, right?
– Im YoungMin
Nov 15 at 18:47
Sometimes you can and sometimes you cannot. That depends on whether $x_1$ (the first element of $x$ is real). You can always construct $H$ in the way outlined in your question, but unless $e^{itheta}$ is real, the resulting $H$ is not Hermitian (and strictly speaking, it is not a reflection either). Rather it is a complex scalar multiple of a Hermitian unitary matrix.
– user1551
Nov 16 at 1:52