Difference between an isometric operator and a unitary operator on a Hilbert space The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)For complex matrices, if $langle Ax,xrangle=langle Bx,xrangle$ for all $x$, then $langle Ax,yrangle=langle Bx,yrangle$ for all $x$ and $y$?How can we prove that Unitary Transformation is isometric?A Finite Dimensional non-Unitary Isometry?Examples of non-unitary isometries on finite dimensional Hilbert spaces?Hilbert vs Inner Product SpaceFor a Hilbert space $mathcalH$, is every bounded linear operator on $mathcalH$ a linear combination of unitary operators?A operator is unitary if and only if it is a surjective isometryExistence of Unitary MapPolar decomposition of Bounded Normal Operator on Hilbert SpaceProving an isometric dilation of a non unitary operator on Hilbert space implies infinite dimensional space involving matricesunitary operator between two Hilbert subspacesDon't unitary operators have the same domain and codomain?What is the difference between a *surjective* unitary operator and a *bijective* unitary transformationsExamples of non-unitary isometries on finite dimensional Hilbert spaces?
How to handle characters who are more educated than the author?
Simulating Exploding Dice
should truth entail possible truth
Could an empire control the whole planet with today's comunication methods?
How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time
Do working physicists consider Newtonian mechanics to be "falsified"?
How to support a colleague who finds meetings extremely tiring?
What's the point in a preamp?
Didn't get enough time to take a Coding Test - what to do now?
US Healthcare consultation for visitors
Is an up-to-date browser secure on an out-of-date OS?
Why did Peik Lin say, "I'm not an animal"?
Can I visit the Trinity College (Cambridge) library and see some of their rare books
Do I have Disadvantage attacking with an off-hand weapon?
Is it ok to offer lower paid work as a trial period before negotiating for a full-time job?
Loose spokes after only a few rides
How to type a long/em dash `—`
Mortgage adviser recommends a longer term than necessary combined with overpayments
Keeping a retro style to sci-fi spaceships?
How did passengers keep warm on sail ships?
Presidential Pardon
Does Parliament hold absolute power in the UK?
Why can't wing-mounted spoilers be used to steepen approaches?
Single author papers against my advisor's will?
Difference between an isometric operator and a unitary operator on a Hilbert space
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)For complex matrices, if $langle Ax,xrangle=langle Bx,xrangle$ for all $x$, then $langle Ax,yrangle=langle Bx,yrangle$ for all $x$ and $y$?How can we prove that Unitary Transformation is isometric?A Finite Dimensional non-Unitary Isometry?Examples of non-unitary isometries on finite dimensional Hilbert spaces?Hilbert vs Inner Product SpaceFor a Hilbert space $mathcalH$, is every bounded linear operator on $mathcalH$ a linear combination of unitary operators?A operator is unitary if and only if it is a surjective isometryExistence of Unitary MapPolar decomposition of Bounded Normal Operator on Hilbert SpaceProving an isometric dilation of a non unitary operator on Hilbert space implies infinite dimensional space involving matricesunitary operator between two Hilbert subspacesDon't unitary operators have the same domain and codomain?What is the difference between a *surjective* unitary operator and a *bijective* unitary transformationsExamples of non-unitary isometries on finite dimensional Hilbert spaces?
$begingroup$
It seems that both isometric and unitary operators on a Hilbert space have the following property:
$U^*U = I$ ($U$ is an operator and $I$ is an identity operator, $^*$ is a binary operation.)
What is the difference between isometry and unitary? Which one is more general,
or are they the same? Are they isomorphic?
metric-spaces operator-theory hilbert-spaces inner-product-space
asked Aug 16 '14 at 14:39
sleeve chensleeve chen
3,18142256
$endgroup$
add a comment |
$begingroup$
It seems that both isometric and unitary operators on a Hilbert space have the following property:
$U^*U = I$ ($U$ is an operator and $I$ is an identity operator, $^*$ is a binary operation.)
What is the difference between isometry and unitary? Which one is more general,
or are they the same? Are they isomorphic?
metric-spaces operator-theory hilbert-spaces inner-product-space
asked Aug 16 '14 at 14:39
sleeve chensleeve chen
3,18142256
$endgroup$
$begingroup$
Unitary is complex isometry.
$endgroup$
– Shuchang
Aug 16 '14 at 14:43
4
$begingroup$
I think unitary operator guarantees invertibility, i.e. U* = inv(U); however, isometric operator not.
$endgroup$
– sleeve chen
Aug 16 '14 at 17:26
$begingroup$
@sleeve: That's right. You succinctly answered your own question a few minutes before I posted.
$endgroup$
– Jonas Meyer
Aug 16 '14 at 17:35
add a comment |
$begingroup$
It seems that both isometric and unitary operators on a Hilbert space have the following property:
$U^*U = I$ ($U$ is an operator and $I$ is an identity operator, $^*$ is a binary operation.)
What is the difference between isometry and unitary? Which one is more general,
or are they the same? Are they isomorphic?
metric-spaces operator-theory hilbert-spaces inner-product-space
asked Aug 16 '14 at 14:39
sleeve chensleeve chen
3,18142256
$endgroup$
It seems that both isometric and unitary operators on a Hilbert space have the following property:
$U^*U = I$ ($U$ is an operator and $I$ is an identity operator, $^*$ is a binary operation.)
What is the difference between isometry and unitary? Which one is more general,
or are they the same? Are they isomorphic?
metric-spaces operator-theory hilbert-spaces inner-product-space
metric-spaces operator-theory hilbert-spaces inner-product-space
asked Aug 16 '14 at 14:39
sleeve chensleeve chen
3,18142256
asked Aug 16 '14 at 14:39
sleeve chensleeve chen
3,18142256
edited Mar 31 at 6:59
sleeve chen
asked Aug 16 '14 at 14:39
sleeve chensleeve chen
3,18142256
asked Aug 16 '14 at 14:39
sleeve chensleeve chen
3,18142256
asked Aug 16 '14 at 14:39
sleeve chensleeve chen
3,18142256
3,18142256
$begingroup$
Unitary is complex isometry.
$endgroup$
– Shuchang
Aug 16 '14 at 14:43
4
$begingroup$
I think unitary operator guarantees invertibility, i.e. U* = inv(U); however, isometric operator not.
$endgroup$
– sleeve chen
Aug 16 '14 at 17:26
$begingroup$
@sleeve: That's right. You succinctly answered your own question a few minutes before I posted.
$endgroup$
– Jonas Meyer
Aug 16 '14 at 17:35
add a comment |
$begingroup$
Unitary is complex isometry.
$endgroup$
– Shuchang
Aug 16 '14 at 14:43
4
$begingroup$
I think unitary operator guarantees invertibility, i.e. U* = inv(U); however, isometric operator not.
$endgroup$
– sleeve chen
Aug 16 '14 at 17:26
$begingroup$
@sleeve: That's right. You succinctly answered your own question a few minutes before I posted.
$endgroup$
– Jonas Meyer
Aug 16 '14 at 17:35
$begingroup$
Unitary is complex isometry.
$endgroup$
– Shuchang
Aug 16 '14 at 14:43
$begingroup$
Unitary is complex isometry.
$endgroup$
– Shuchang
Aug 16 '14 at 14:43
4
4
$begingroup$
I think unitary operator guarantees invertibility, i.e. U* = inv(U); however, isometric operator not.
$endgroup$
– sleeve chen
Aug 16 '14 at 17:26
$begingroup$
I think unitary operator guarantees invertibility, i.e. U* = inv(U); however, isometric operator not.
$endgroup$
– sleeve chen
Aug 16 '14 at 17:26
$begingroup$
@sleeve: That's right. You succinctly answered your own question a few minutes before I posted.
$endgroup$
– Jonas Meyer
Aug 16 '14 at 17:35
$begingroup$
@sleeve: That's right. You succinctly answered your own question a few minutes before I posted.
$endgroup$
– Jonas Meyer
Aug 16 '14 at 17:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
An isometric operator on a (complex) Hilbert space is a linear operator that preserves distances. That is, $T$ is an isometry if (by definition) $|Tx-Ty|=|x-y|$ for all $x$ and $y$ in the space. By linearity, this is equivalent to $|Tx|=|x|$ for all $x$. Because of the definition of the norm in terms of the inner product and the definition of adjoint operators, this is equivalent to $langle T^*Tx,xrangle=langle x,xrangle$ for all $x$. This implies that $T^*T=I$. Conversely, if $T^*T=I$, you can show that $T$ is an isometry (this direction is easier).
A unitary operator $U$ does indeed satisfy $U^*U=I$, and therefore in particular is an isometry. However, unitary operators must also be surjective (by definition), and are therefore isometric and invertible. They are the isometric isomorphisms on Hilbert space. One way to characterize them algebraically is to say that $U$ is a unitary if $U^*U=UU^*=I$.
On infinite dimensional Hilbert spaces (unlike in finite dimensional cases), there are always nonunitary isometries. For example, on $ell^2$, the operator sending $(a_0,a_1,a_2,a_3,ldots)$ to $(0,a_0,a_1,a_2,ldots)$ is a nonunitary isometry.
I'm not sure what you mean by "isomorphic". One notion of equivalence of linear transformations is similarity; but a surjective operator is never similar to a nonsurjective operator. A stronger notion is unitary equivalence, i.e., similarity induced by a unitary transformation (since these are the isometric isomorphisms of Hilbert space), which again cannot happen between a nonunitary isometry and a unitary operator (or between any nonunitary operator and a unitary operator).
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 16 '14 at 17:30
Jonas MeyerJonas Meyer
41.1k6149259
$endgroup$
1
$begingroup$
I cited your answer in my question here math.stackexchange.com/q/2091816. I'd appreciate your feedback.
$endgroup$
– Tom Collinge
Jan 10 '17 at 13:39
1
$begingroup$
Interestingly, in the real case the argument in your first paragraph needs to be (slightly) more complicated: going from $⟨O^T O x, x⟩=⟨x, x⟩$ to $O^T O = I$ requires a stronger argument than in the complex case (e.g., using the fact that $O^T O$ is symmetric and therefore diagonalizable, and going into an eigenbasis).
$endgroup$
– tparker
Jul 7 '18 at 6:09
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
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%2f899230%2fdifference-between-an-isometric-operator-and-a-unitary-operator-on-a-hilbert-spa%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An isometric operator on a (complex) Hilbert space is a linear operator that preserves distances. That is, $T$ is an isometry if (by definition) $|Tx-Ty|=|x-y|$ for all $x$ and $y$ in the space. By linearity, this is equivalent to $|Tx|=|x|$ for all $x$. Because of the definition of the norm in terms of the inner product and the definition of adjoint operators, this is equivalent to $langle T^*Tx,xrangle=langle x,xrangle$ for all $x$. This implies that $T^*T=I$. Conversely, if $T^*T=I$, you can show that $T$ is an isometry (this direction is easier).
A unitary operator $U$ does indeed satisfy $U^*U=I$, and therefore in particular is an isometry. However, unitary operators must also be surjective (by definition), and are therefore isometric and invertible. They are the isometric isomorphisms on Hilbert space. One way to characterize them algebraically is to say that $U$ is a unitary if $U^*U=UU^*=I$.
On infinite dimensional Hilbert spaces (unlike in finite dimensional cases), there are always nonunitary isometries. For example, on $ell^2$, the operator sending $(a_0,a_1,a_2,a_3,ldots)$ to $(0,a_0,a_1,a_2,ldots)$ is a nonunitary isometry.
I'm not sure what you mean by "isomorphic". One notion of equivalence of linear transformations is similarity; but a surjective operator is never similar to a nonsurjective operator. A stronger notion is unitary equivalence, i.e., similarity induced by a unitary transformation (since these are the isometric isomorphisms of Hilbert space), which again cannot happen between a nonunitary isometry and a unitary operator (or between any nonunitary operator and a unitary operator).
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 16 '14 at 17:30
Jonas MeyerJonas Meyer
41.1k6149259
$endgroup$
1
$begingroup$
I cited your answer in my question here math.stackexchange.com/q/2091816. I'd appreciate your feedback.
$endgroup$
– Tom Collinge
Jan 10 '17 at 13:39
1
$begingroup$
Interestingly, in the real case the argument in your first paragraph needs to be (slightly) more complicated: going from $⟨O^T O x, x⟩=⟨x, x⟩$ to $O^T O = I$ requires a stronger argument than in the complex case (e.g., using the fact that $O^T O$ is symmetric and therefore diagonalizable, and going into an eigenbasis).
$endgroup$
– tparker
Jul 7 '18 at 6:09
add a comment |
$begingroup$
An isometric operator on a (complex) Hilbert space is a linear operator that preserves distances. That is, $T$ is an isometry if (by definition) $|Tx-Ty|=|x-y|$ for all $x$ and $y$ in the space. By linearity, this is equivalent to $|Tx|=|x|$ for all $x$. Because of the definition of the norm in terms of the inner product and the definition of adjoint operators, this is equivalent to $langle T^*Tx,xrangle=langle x,xrangle$ for all $x$. This implies that $T^*T=I$. Conversely, if $T^*T=I$, you can show that $T$ is an isometry (this direction is easier).
A unitary operator $U$ does indeed satisfy $U^*U=I$, and therefore in particular is an isometry. However, unitary operators must also be surjective (by definition), and are therefore isometric and invertible. They are the isometric isomorphisms on Hilbert space. One way to characterize them algebraically is to say that $U$ is a unitary if $U^*U=UU^*=I$.
On infinite dimensional Hilbert spaces (unlike in finite dimensional cases), there are always nonunitary isometries. For example, on $ell^2$, the operator sending $(a_0,a_1,a_2,a_3,ldots)$ to $(0,a_0,a_1,a_2,ldots)$ is a nonunitary isometry.
I'm not sure what you mean by "isomorphic". One notion of equivalence of linear transformations is similarity; but a surjective operator is never similar to a nonsurjective operator. A stronger notion is unitary equivalence, i.e., similarity induced by a unitary transformation (since these are the isometric isomorphisms of Hilbert space), which again cannot happen between a nonunitary isometry and a unitary operator (or between any nonunitary operator and a unitary operator).
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 16 '14 at 17:30
Jonas MeyerJonas Meyer
41.1k6149259
$endgroup$
1
$begingroup$
I cited your answer in my question here math.stackexchange.com/q/2091816. I'd appreciate your feedback.
$endgroup$
– Tom Collinge
Jan 10 '17 at 13:39
1
$begingroup$
Interestingly, in the real case the argument in your first paragraph needs to be (slightly) more complicated: going from $⟨O^T O x, x⟩=⟨x, x⟩$ to $O^T O = I$ requires a stronger argument than in the complex case (e.g., using the fact that $O^T O$ is symmetric and therefore diagonalizable, and going into an eigenbasis).
$endgroup$
– tparker
Jul 7 '18 at 6:09
add a comment |
$begingroup$
An isometric operator on a (complex) Hilbert space is a linear operator that preserves distances. That is, $T$ is an isometry if (by definition) $|Tx-Ty|=|x-y|$ for all $x$ and $y$ in the space. By linearity, this is equivalent to $|Tx|=|x|$ for all $x$. Because of the definition of the norm in terms of the inner product and the definition of adjoint operators, this is equivalent to $langle T^*Tx,xrangle=langle x,xrangle$ for all $x$. This implies that $T^*T=I$. Conversely, if $T^*T=I$, you can show that $T$ is an isometry (this direction is easier).
A unitary operator $U$ does indeed satisfy $U^*U=I$, and therefore in particular is an isometry. However, unitary operators must also be surjective (by definition), and are therefore isometric and invertible. They are the isometric isomorphisms on Hilbert space. One way to characterize them algebraically is to say that $U$ is a unitary if $U^*U=UU^*=I$.
On infinite dimensional Hilbert spaces (unlike in finite dimensional cases), there are always nonunitary isometries. For example, on $ell^2$, the operator sending $(a_0,a_1,a_2,a_3,ldots)$ to $(0,a_0,a_1,a_2,ldots)$ is a nonunitary isometry.
I'm not sure what you mean by "isomorphic". One notion of equivalence of linear transformations is similarity; but a surjective operator is never similar to a nonsurjective operator. A stronger notion is unitary equivalence, i.e., similarity induced by a unitary transformation (since these are the isometric isomorphisms of Hilbert space), which again cannot happen between a nonunitary isometry and a unitary operator (or between any nonunitary operator and a unitary operator).
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 16 '14 at 17:30
Jonas MeyerJonas Meyer
41.1k6149259
$endgroup$
An isometric operator on a (complex) Hilbert space is a linear operator that preserves distances. That is, $T$ is an isometry if (by definition) $|Tx-Ty|=|x-y|$ for all $x$ and $y$ in the space. By linearity, this is equivalent to $|Tx|=|x|$ for all $x$. Because of the definition of the norm in terms of the inner product and the definition of adjoint operators, this is equivalent to $langle T^*Tx,xrangle=langle x,xrangle$ for all $x$. This implies that $T^*T=I$. Conversely, if $T^*T=I$, you can show that $T$ is an isometry (this direction is easier).
A unitary operator $U$ does indeed satisfy $U^*U=I$, and therefore in particular is an isometry. However, unitary operators must also be surjective (by definition), and are therefore isometric and invertible. They are the isometric isomorphisms on Hilbert space. One way to characterize them algebraically is to say that $U$ is a unitary if $U^*U=UU^*=I$.
On infinite dimensional Hilbert spaces (unlike in finite dimensional cases), there are always nonunitary isometries. For example, on $ell^2$, the operator sending $(a_0,a_1,a_2,a_3,ldots)$ to $(0,a_0,a_1,a_2,ldots)$ is a nonunitary isometry.
I'm not sure what you mean by "isomorphic". One notion of equivalence of linear transformations is similarity; but a surjective operator is never similar to a nonsurjective operator. A stronger notion is unitary equivalence, i.e., similarity induced by a unitary transformation (since these are the isometric isomorphisms of Hilbert space), which again cannot happen between a nonunitary isometry and a unitary operator (or between any nonunitary operator and a unitary operator).
edited Apr 13 '17 at 12:21
Community♦
1
answered Aug 16 '14 at 17:30
Jonas MeyerJonas Meyer
41.1k6149259
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered Aug 16 '14 at 17:30
Jonas MeyerJonas Meyer
41.1k6149259
answered Aug 16 '14 at 17:30
Jonas MeyerJonas Meyer
41.1k6149259
answered Aug 16 '14 at 17:30
Jonas MeyerJonas Meyer
41.1k6149259
41.1k6149259
1
$begingroup$
I cited your answer in my question here math.stackexchange.com/q/2091816. I'd appreciate your feedback.
$endgroup$
– Tom Collinge
Jan 10 '17 at 13:39
1
$begingroup$
Interestingly, in the real case the argument in your first paragraph needs to be (slightly) more complicated: going from $⟨O^T O x, x⟩=⟨x, x⟩$ to $O^T O = I$ requires a stronger argument than in the complex case (e.g., using the fact that $O^T O$ is symmetric and therefore diagonalizable, and going into an eigenbasis).
$endgroup$
– tparker
Jul 7 '18 at 6:09
add a comment |
1
$begingroup$
I cited your answer in my question here math.stackexchange.com/q/2091816. I'd appreciate your feedback.
$endgroup$
– Tom Collinge
Jan 10 '17 at 13:39
1
$begingroup$
Interestingly, in the real case the argument in your first paragraph needs to be (slightly) more complicated: going from $⟨O^T O x, x⟩=⟨x, x⟩$ to $O^T O = I$ requires a stronger argument than in the complex case (e.g., using the fact that $O^T O$ is symmetric and therefore diagonalizable, and going into an eigenbasis).
$endgroup$
– tparker
Jul 7 '18 at 6:09
1
1
$begingroup$
I cited your answer in my question here math.stackexchange.com/q/2091816. I'd appreciate your feedback.
$endgroup$
– Tom Collinge
Jan 10 '17 at 13:39
$begingroup$
I cited your answer in my question here math.stackexchange.com/q/2091816. I'd appreciate your feedback.
$endgroup$
– Tom Collinge
Jan 10 '17 at 13:39
1
1
$begingroup$
Interestingly, in the real case the argument in your first paragraph needs to be (slightly) more complicated: going from $⟨O^T O x, x⟩=⟨x, x⟩$ to $O^T O = I$ requires a stronger argument than in the complex case (e.g., using the fact that $O^T O$ is symmetric and therefore diagonalizable, and going into an eigenbasis).
$endgroup$
– tparker
Jul 7 '18 at 6:09
$begingroup$
Interestingly, in the real case the argument in your first paragraph needs to be (slightly) more complicated: going from $⟨O^T O x, x⟩=⟨x, x⟩$ to $O^T O = I$ requires a stronger argument than in the complex case (e.g., using the fact that $O^T O$ is symmetric and therefore diagonalizable, and going into an eigenbasis).
$endgroup$
– tparker
Jul 7 '18 at 6:09
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.
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%2f899230%2fdifference-between-an-isometric-operator-and-a-unitary-operator-on-a-hilbert-spa%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
$begingroup$
Unitary is complex isometry.
$endgroup$
– Shuchang
Aug 16 '14 at 14:43
4
$begingroup$
I think unitary operator guarantees invertibility, i.e. U* = inv(U); however, isometric operator not.
$endgroup$
– sleeve chen
Aug 16 '14 at 17:26
$begingroup$
@sleeve: That's right. You succinctly answered your own question a few minutes before I posted.
$endgroup$
– Jonas Meyer
Aug 16 '14 at 17:35