Mixing
Skew symmetric subtracted from Identity
0
$begingroup$
How can I prove that for any skew-symmetric matrix $S$ with $S^T = -S$, $I - S$ is non-singular and $(I-S)^{-1}(I+S)$ is an orthogonal Matrix (cayley transform of S).
linear-algebra numerical-linear-algebra
asked Jan 27 at 15:30
Usman AshrafUsman Ashraf
13
$endgroup$
add a comment |
0
$begingroup$
How can I prove that for any skew-symmetric matrix $S$ with $S^T = -S$, $I - S$ is non-singular and $(I-S)^{-1}(I+S)$ is an orthogonal Matrix (cayley transform of S).
linear-algebra numerical-linear-algebra
asked Jan 27 at 15:30
Usman AshrafUsman Ashraf
13
$endgroup$
2
$begingroup$
Are you asking if $(I-S)^{-1}(I-S)=I$ is orthogonal?
$endgroup$
– Song
Jan 27 at 15:43
$begingroup$
Yeah, how can I prove it's orthogonality?
$endgroup$
– Usman Ashraf
Jan 27 at 15:44
$begingroup$
It is identity matrix ...
$endgroup$
– Song
Jan 27 at 15:45
$begingroup$
Thanks. So, just like $A^{-1}A$
$endgroup$
– Usman Ashraf
Jan 27 at 15:47
1
$begingroup$
What can I say about the singularity of $I-S$?
$endgroup$
– Usman Ashraf
Jan 27 at 15:48
add a comment |
0
0
0
$begingroup$
How can I prove that for any skew-symmetric matrix $S$ with $S^T = -S$, $I - S$ is non-singular and $(I-S)^{-1}(I+S)$ is an orthogonal Matrix (cayley transform of S).
linear-algebra numerical-linear-algebra
asked Jan 27 at 15:30
Usman AshrafUsman Ashraf
13
$endgroup$
How can I prove that for any skew-symmetric matrix $S$ with $S^T = -S$, $I - S$ is non-singular and $(I-S)^{-1}(I+S)$ is an orthogonal Matrix (cayley transform of S).
linear-algebra numerical-linear-algebra
linear-algebra numerical-linear-algebra
asked Jan 27 at 15:30
Usman AshrafUsman Ashraf
13
asked Jan 27 at 15:30
Usman AshrafUsman Ashraf
13
edited Jan 29 at 6:35
Usman Ashraf
asked Jan 27 at 15:30
Usman AshrafUsman Ashraf
13
asked Jan 27 at 15:30
Usman AshrafUsman Ashraf
13
asked Jan 27 at 15:30
Usman AshrafUsman Ashraf
13
13
2
$begingroup$
Are you asking if $(I-S)^{-1}(I-S)=I$ is orthogonal?
$endgroup$
– Song
Jan 27 at 15:43
$begingroup$
Yeah, how can I prove it's orthogonality?
$endgroup$
– Usman Ashraf
Jan 27 at 15:44
$begingroup$
It is identity matrix ...
$endgroup$
– Song
Jan 27 at 15:45
$begingroup$
Thanks. So, just like $A^{-1}A$
$endgroup$
– Usman Ashraf
Jan 27 at 15:47
1
$begingroup$
What can I say about the singularity of $I-S$?
$endgroup$
– Usman Ashraf
Jan 27 at 15:48
add a comment |
2
$begingroup$
Are you asking if $(I-S)^{-1}(I-S)=I$ is orthogonal?
$endgroup$
– Song
Jan 27 at 15:43
$begingroup$
Yeah, how can I prove it's orthogonality?
$endgroup$
– Usman Ashraf
Jan 27 at 15:44
$begingroup$
It is identity matrix ...
$endgroup$
– Song
Jan 27 at 15:45
$begingroup$
Thanks. So, just like $A^{-1}A$
$endgroup$
– Usman Ashraf
Jan 27 at 15:47
1
$begingroup$
What can I say about the singularity of $I-S$?
$endgroup$
– Usman Ashraf
Jan 27 at 15:48
2
2
$begingroup$
Are you asking if $(I-S)^{-1}(I-S)=I$ is orthogonal?
$endgroup$
– Song
Jan 27 at 15:43
$begingroup$
Are you asking if $(I-S)^{-1}(I-S)=I$ is orthogonal?
$endgroup$
– Song
Jan 27 at 15:43
$begingroup$
Yeah, how can I prove it's orthogonality?
$endgroup$
– Usman Ashraf
Jan 27 at 15:44
$begingroup$
Yeah, how can I prove it's orthogonality?
$endgroup$
– Usman Ashraf
Jan 27 at 15:44
$begingroup$
It is identity matrix ...
$endgroup$
– Song
Jan 27 at 15:45
$begingroup$
It is identity matrix ...
$endgroup$
– Song
Jan 27 at 15:45
$begingroup$
Thanks. So, just like $A^{-1}A$
$endgroup$
– Usman Ashraf
Jan 27 at 15:47
$begingroup$
Thanks. So, just like $A^{-1}A$
$endgroup$
– Usman Ashraf
Jan 27 at 15:47
1
1
$begingroup$
What can I say about the singularity of $I-S$?
$endgroup$
– Usman Ashraf
Jan 27 at 15:48
$begingroup$
What can I say about the singularity of $I-S$?
$endgroup$
– Usman Ashraf
Jan 27 at 15:48
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
Let $Sx=x $,then $S^TSx=S^Tx=-Sx=-x $. Since $S^TS $ is positive semidefinite, $-1$ is not an eigenvalue of $S^TS $, hence $x=0$.
answered Jan 28 at 4:39
FredFred
48.7k11849
$endgroup$
$begingroup$
Can you explain how you equated $-S^Tx$ to -x?
$endgroup$
– Usman Ashraf
Jan 28 at 4:52
$begingroup$
$Sx=x$, hence $S^Tx=-Sx=-x$.
$endgroup$
– Fred
Jan 28 at 6:13
$begingroup$
Can you also say why can you assume that Sx = x in the first place? Also, how x = 0 should that I-S is non-singular?
$endgroup$
– Usman Ashraf
Jan 29 at 6:25
$begingroup$
$I-S$ is non-singular $ iff ker(I-S)={0}$. Now let $x in ker(I-S)$, then $Sx=x$. Above I have shown that this implies $x=0$. Hence $ ker(I-S)={0}$.
$endgroup$
– Fred
Jan 29 at 6:30
$begingroup$
Alright, that makes sense! Thanks. Can you say, how can I prove orthogonality for (I-S)^-1 * (I+S). I just added this in the question.
$endgroup$
– Usman Ashraf
Jan 29 at 6:37
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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
});
}
});
draft saved
draft discarded
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%2f3089702%2fskew-symmetric-subtracted-from-identity%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
0
$begingroup$
Let $Sx=x $,then $S^TSx=S^Tx=-Sx=-x $. Since $S^TS $ is positive semidefinite, $-1$ is not an eigenvalue of $S^TS $, hence $x=0$.
answered Jan 28 at 4:39
FredFred
48.7k11849
$endgroup$
$begingroup$
Can you explain how you equated $-S^Tx$ to -x?
$endgroup$
– Usman Ashraf
Jan 28 at 4:52
$begingroup$
$Sx=x$, hence $S^Tx=-Sx=-x$.
$endgroup$
– Fred
Jan 28 at 6:13
$begingroup$
Can you also say why can you assume that Sx = x in the first place? Also, how x = 0 should that I-S is non-singular?
$endgroup$
– Usman Ashraf
Jan 29 at 6:25
$begingroup$
$I-S$ is non-singular $ iff ker(I-S)={0}$. Now let $x in ker(I-S)$, then $Sx=x$. Above I have shown that this implies $x=0$. Hence $ ker(I-S)={0}$.
$endgroup$
– Fred
Jan 29 at 6:30
$begingroup$
Alright, that makes sense! Thanks. Can you say, how can I prove orthogonality for (I-S)^-1 * (I+S). I just added this in the question.
$endgroup$
– Usman Ashraf
Jan 29 at 6:37
add a comment |
0
$begingroup$
Let $Sx=x $,then $S^TSx=S^Tx=-Sx=-x $. Since $S^TS $ is positive semidefinite, $-1$ is not an eigenvalue of $S^TS $, hence $x=0$.
answered Jan 28 at 4:39
FredFred
48.7k11849
$endgroup$
$begingroup$
Can you explain how you equated $-S^Tx$ to -x?
$endgroup$
– Usman Ashraf
Jan 28 at 4:52
$begingroup$
$Sx=x$, hence $S^Tx=-Sx=-x$.
$endgroup$
– Fred
Jan 28 at 6:13
$begingroup$
Can you also say why can you assume that Sx = x in the first place? Also, how x = 0 should that I-S is non-singular?
$endgroup$
– Usman Ashraf
Jan 29 at 6:25
$begingroup$
$I-S$ is non-singular $ iff ker(I-S)={0}$. Now let $x in ker(I-S)$, then $Sx=x$. Above I have shown that this implies $x=0$. Hence $ ker(I-S)={0}$.
$endgroup$
– Fred
Jan 29 at 6:30
$begingroup$
Alright, that makes sense! Thanks. Can you say, how can I prove orthogonality for (I-S)^-1 * (I+S). I just added this in the question.
$endgroup$
– Usman Ashraf
Jan 29 at 6:37
add a comment |
0
0
0
$begingroup$
Let $Sx=x $,then $S^TSx=S^Tx=-Sx=-x $. Since $S^TS $ is positive semidefinite, $-1$ is not an eigenvalue of $S^TS $, hence $x=0$.
answered Jan 28 at 4:39
FredFred
48.7k11849
$endgroup$
Let $Sx=x $,then $S^TSx=S^Tx=-Sx=-x $. Since $S^TS $ is positive semidefinite, $-1$ is not an eigenvalue of $S^TS $, hence $x=0$.
answered Jan 28 at 4:39
FredFred
48.7k11849
answered Jan 28 at 4:39
FredFred
48.7k11849
answered Jan 28 at 4:39
FredFred
48.7k11849
answered Jan 28 at 4:39
FredFred
48.7k11849
48.7k11849
$begingroup$
Can you explain how you equated $-S^Tx$ to -x?
$endgroup$
– Usman Ashraf
Jan 28 at 4:52
$begingroup$
$Sx=x$, hence $S^Tx=-Sx=-x$.
$endgroup$
– Fred
Jan 28 at 6:13
$begingroup$
Can you also say why can you assume that Sx = x in the first place? Also, how x = 0 should that I-S is non-singular?
$endgroup$
– Usman Ashraf
Jan 29 at 6:25
$begingroup$
$I-S$ is non-singular $ iff ker(I-S)={0}$. Now let $x in ker(I-S)$, then $Sx=x$. Above I have shown that this implies $x=0$. Hence $ ker(I-S)={0}$.
$endgroup$
– Fred
Jan 29 at 6:30
$begingroup$
Alright, that makes sense! Thanks. Can you say, how can I prove orthogonality for (I-S)^-1 * (I+S). I just added this in the question.
$endgroup$
– Usman Ashraf
Jan 29 at 6:37
add a comment |
$begingroup$
Can you explain how you equated $-S^Tx$ to -x?
$endgroup$
– Usman Ashraf
Jan 28 at 4:52
$begingroup$
$Sx=x$, hence $S^Tx=-Sx=-x$.
$endgroup$
– Fred
Jan 28 at 6:13
$begingroup$
Can you also say why can you assume that Sx = x in the first place? Also, how x = 0 should that I-S is non-singular?
$endgroup$
– Usman Ashraf
Jan 29 at 6:25
$begingroup$
$I-S$ is non-singular $ iff ker(I-S)={0}$. Now let $x in ker(I-S)$, then $Sx=x$. Above I have shown that this implies $x=0$. Hence $ ker(I-S)={0}$.
$endgroup$
– Fred
Jan 29 at 6:30
$begingroup$
Alright, that makes sense! Thanks. Can you say, how can I prove orthogonality for (I-S)^-1 * (I+S). I just added this in the question.
$endgroup$
– Usman Ashraf
Jan 29 at 6:37
$begingroup$
Can you explain how you equated $-S^Tx$ to -x?
$endgroup$
– Usman Ashraf
Jan 28 at 4:52
$begingroup$
Can you explain how you equated $-S^Tx$ to -x?
$endgroup$
– Usman Ashraf
Jan 28 at 4:52
$begingroup$
$Sx=x$, hence $S^Tx=-Sx=-x$.
$endgroup$
– Fred
Jan 28 at 6:13
$begingroup$
$Sx=x$, hence $S^Tx=-Sx=-x$.
$endgroup$
– Fred
Jan 28 at 6:13
$begingroup$
Can you also say why can you assume that Sx = x in the first place? Also, how x = 0 should that I-S is non-singular?
$endgroup$
– Usman Ashraf
Jan 29 at 6:25
$begingroup$
Can you also say why can you assume that Sx = x in the first place? Also, how x = 0 should that I-S is non-singular?
$endgroup$
– Usman Ashraf
Jan 29 at 6:25
$begingroup$
$I-S$ is non-singular $ iff ker(I-S)={0}$. Now let $x in ker(I-S)$, then $Sx=x$. Above I have shown that this implies $x=0$. Hence $ ker(I-S)={0}$.
$endgroup$
– Fred
Jan 29 at 6:30
$begingroup$
$I-S$ is non-singular $ iff ker(I-S)={0}$. Now let $x in ker(I-S)$, then $Sx=x$. Above I have shown that this implies $x=0$. Hence $ ker(I-S)={0}$.
$endgroup$
– Fred
Jan 29 at 6:30
$begingroup$
Alright, that makes sense! Thanks. Can you say, how can I prove orthogonality for (I-S)^-1 * (I+S). I just added this in the question.
$endgroup$
– Usman Ashraf
Jan 29 at 6:37
$begingroup$
Alright, that makes sense! Thanks. Can you say, how can I prove orthogonality for (I-S)^-1 * (I+S). I just added this in the question.
$endgroup$
– Usman Ashraf
Jan 29 at 6:37
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3089702%2fskew-symmetric-subtracted-from-identity%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
2
$begingroup$
Are you asking if $(I-S)^{-1}(I-S)=I$ is orthogonal?
$endgroup$
– Song
Jan 27 at 15:43
$begingroup$
Yeah, how can I prove it's orthogonality?
$endgroup$
– Usman Ashraf
Jan 27 at 15:44
$begingroup$
It is identity matrix ...
$endgroup$
– Song
Jan 27 at 15:45
$begingroup$
Thanks. So, just like $A^{-1}A$
$endgroup$
– Usman Ashraf
Jan 27 at 15:47
1
$begingroup$
What can I say about the singularity of $I-S$?
$endgroup$
– Usman Ashraf
Jan 27 at 15:48