When is the nullspace unique in this case?
up vote
0
down vote
favorite
I am given the following:
$$
I - AA^T
$$
is a projection matrix onto the orthogonal complement of $< A >$.
So the nullspace of $I-AA^T$ is the subspace spanned by the set of vectors $x$ such that $(I-AA^T)x = 0$. Since we know the projection matrix projects x onto the orthogonal complement of $<A>$, then I believe this means the nullspace of $I-AA^T$ is $<A>$?
Based on this information, is it possible to determine when $<A>$ is unique?
linear-algebra matrices projection projection-matrices
asked Nov 21 at 15:56
Iamanon
1157
add a comment |
up vote
0
down vote
favorite
I am given the following:
$$
I - AA^T
$$
is a projection matrix onto the orthogonal complement of $< A >$.
So the nullspace of $I-AA^T$ is the subspace spanned by the set of vectors $x$ such that $(I-AA^T)x = 0$. Since we know the projection matrix projects x onto the orthogonal complement of $<A>$, then I believe this means the nullspace of $I-AA^T$ is $<A>$?
Based on this information, is it possible to determine when $<A>$ is unique?
linear-algebra matrices projection projection-matrices
asked Nov 21 at 15:56
Iamanon
1157
1
Are some conditions missing? Unless all the singular values of $A$ are zeros or ones, $I-AA^T$ is not a projection.
– user1551
Nov 21 at 15:59
1
It is not a projection since $(I-AA^T)^2 neq I-AA^T$.
– LinAlg
Nov 21 at 16:01
Hmmmm. I am initially not given any information about $A$. In a later problem, I see that $A$ has orthonormal columns and is not necessarily square.
– Iamanon
Nov 21 at 16:09
After consulting with a colleague, it seems that we are supposed to assume A has orthonormal columns. So it is a projection matrix based on this condition.
– Iamanon
Nov 21 at 16:17
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am given the following:
$$
I - AA^T
$$
is a projection matrix onto the orthogonal complement of $< A >$.
So the nullspace of $I-AA^T$ is the subspace spanned by the set of vectors $x$ such that $(I-AA^T)x = 0$. Since we know the projection matrix projects x onto the orthogonal complement of $<A>$, then I believe this means the nullspace of $I-AA^T$ is $<A>$?
Based on this information, is it possible to determine when $<A>$ is unique?
linear-algebra matrices projection projection-matrices
asked Nov 21 at 15:56
Iamanon
1157
I am given the following:
$$
I - AA^T
$$
is a projection matrix onto the orthogonal complement of $< A >$.
So the nullspace of $I-AA^T$ is the subspace spanned by the set of vectors $x$ such that $(I-AA^T)x = 0$. Since we know the projection matrix projects x onto the orthogonal complement of $<A>$, then I believe this means the nullspace of $I-AA^T$ is $<A>$?
Based on this information, is it possible to determine when $<A>$ is unique?
linear-algebra matrices projection projection-matrices
linear-algebra matrices projection projection-matrices
asked Nov 21 at 15:56
Iamanon
1157
asked Nov 21 at 15:56
Iamanon
1157
asked Nov 21 at 15:56
Iamanon
1157
asked Nov 21 at 15:56
Iamanon
1157
asked Nov 21 at 15:56
Iamanon
1157
1157
1
Are some conditions missing? Unless all the singular values of $A$ are zeros or ones, $I-AA^T$ is not a projection.
– user1551
Nov 21 at 15:59
1
It is not a projection since $(I-AA^T)^2 neq I-AA^T$.
– LinAlg
Nov 21 at 16:01
Hmmmm. I am initially not given any information about $A$. In a later problem, I see that $A$ has orthonormal columns and is not necessarily square.
– Iamanon
Nov 21 at 16:09
After consulting with a colleague, it seems that we are supposed to assume A has orthonormal columns. So it is a projection matrix based on this condition.
– Iamanon
Nov 21 at 16:17
add a comment |
1
Are some conditions missing? Unless all the singular values of $A$ are zeros or ones, $I-AA^T$ is not a projection.
– user1551
Nov 21 at 15:59
1
It is not a projection since $(I-AA^T)^2 neq I-AA^T$.
– LinAlg
Nov 21 at 16:01
Hmmmm. I am initially not given any information about $A$. In a later problem, I see that $A$ has orthonormal columns and is not necessarily square.
– Iamanon
Nov 21 at 16:09
After consulting with a colleague, it seems that we are supposed to assume A has orthonormal columns. So it is a projection matrix based on this condition.
– Iamanon
Nov 21 at 16:17
1
1
Are some conditions missing? Unless all the singular values of $A$ are zeros or ones, $I-AA^T$ is not a projection.
– user1551
Nov 21 at 15:59
Are some conditions missing? Unless all the singular values of $A$ are zeros or ones, $I-AA^T$ is not a projection.
– user1551
Nov 21 at 15:59
1
1
It is not a projection since $(I-AA^T)^2 neq I-AA^T$.
– LinAlg
Nov 21 at 16:01
It is not a projection since $(I-AA^T)^2 neq I-AA^T$.
– LinAlg
Nov 21 at 16:01
Hmmmm. I am initially not given any information about $A$. In a later problem, I see that $A$ has orthonormal columns and is not necessarily square.
– Iamanon
Nov 21 at 16:09
Hmmmm. I am initially not given any information about $A$. In a later problem, I see that $A$ has orthonormal columns and is not necessarily square.
– Iamanon
Nov 21 at 16:09
After consulting with a colleague, it seems that we are supposed to assume A has orthonormal columns. So it is a projection matrix based on this condition.
– Iamanon
Nov 21 at 16:17
After consulting with a colleague, it seems that we are supposed to assume A has orthonormal columns. So it is a projection matrix based on this condition.
– Iamanon
Nov 21 at 16:17
add a comment |
active
oldest
votes
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
});
}
});
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%2f3007912%2fwhen-is-the-nullspace-unique-in-this-case%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3007912%2fwhen-is-the-nullspace-unique-in-this-case%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
1
Are some conditions missing? Unless all the singular values of $A$ are zeros or ones, $I-AA^T$ is not a projection.
– user1551
Nov 21 at 15:59
1
It is not a projection since $(I-AA^T)^2 neq I-AA^T$.
– LinAlg
Nov 21 at 16:01
Hmmmm. I am initially not given any information about $A$. In a later problem, I see that $A$ has orthonormal columns and is not necessarily square.
– Iamanon
Nov 21 at 16:09
After consulting with a colleague, it seems that we are supposed to assume A has orthonormal columns. So it is a projection matrix based on this condition.
– Iamanon
Nov 21 at 16:17