Mixing
do eigenvectors correspond to direction of maximum scaling?
$begingroup$
Does the eigenvector correspond to a direction in which maximum scaling occurs by a given transformation matrix (A) acting upon this vector.
I quote from :
https://math.stackexchange.com/q/243553
No other vector when acted by this matrix will get stretched as much
as this eigenvector.
Is the above statement always true?... For example let
$$
A = left( begin{array}{ccc}
0.578385540014544 & 0.703045745965410 \
0.477513363789115 & 0.922698950982510 \
end{array} right)
$$
The largest eigenvalue is 1.35 (approx.)
Now, consider the vector (not eigenvector)
$$
v = left( begin{array}{ccc}
-0.538656963091298 \
-0.842525178326001 \
end{array} right)
$$
magnitude(v) = 1.0
magnitude(A*v) = 1.373
So this vector(v), which is not the eigenvector of A is scaled by a larger amount (x1.373), compared to the eigenvector which is scaled by x 1.35 (approx.)
Is this just an artifact of numerical precision ? I can easily create more examples of random square transformation matrices (A) where the eigenvector does not correspond to the direction of maximum scaling.
eigenvalues-eigenvectors
edited Apr 13 '17 at 12:19
Community♦
1
asked Sep 16 '15 at 6:03
user1837245user1837245
162
$endgroup$
add a comment |
$begingroup$
Does the eigenvector correspond to a direction in which maximum scaling occurs by a given transformation matrix (A) acting upon this vector.
I quote from :
https://math.stackexchange.com/q/243553
No other vector when acted by this matrix will get stretched as much
as this eigenvector.
Is the above statement always true?... For example let
$$
A = left( begin{array}{ccc}
0.578385540014544 & 0.703045745965410 \
0.477513363789115 & 0.922698950982510 \
end{array} right)
$$
The largest eigenvalue is 1.35 (approx.)
Now, consider the vector (not eigenvector)
$$
v = left( begin{array}{ccc}
-0.538656963091298 \
-0.842525178326001 \
end{array} right)
$$
magnitude(v) = 1.0
magnitude(A*v) = 1.373
So this vector(v), which is not the eigenvector of A is scaled by a larger amount (x1.373), compared to the eigenvector which is scaled by x 1.35 (approx.)
Is this just an artifact of numerical precision ? I can easily create more examples of random square transformation matrices (A) where the eigenvector does not correspond to the direction of maximum scaling.
eigenvalues-eigenvectors
edited Apr 13 '17 at 12:19
Community♦
1
asked Sep 16 '15 at 6:03
user1837245user1837245
162
$endgroup$
1
$begingroup$
The quote is a false statement. It is true for a symmetric matrix that the largest eigenvalue (in absolute value) equals the largest stretch factor. Your example shows that it's not necessarily true in general.
$endgroup$
– Greg Martin
Sep 16 '15 at 6:59
$begingroup$
Thank you for the answer Greg.
$endgroup$
– user1837245
Sep 16 '15 at 19:58
add a comment |
$begingroup$
Does the eigenvector correspond to a direction in which maximum scaling occurs by a given transformation matrix (A) acting upon this vector.
I quote from :
https://math.stackexchange.com/q/243553
No other vector when acted by this matrix will get stretched as much
as this eigenvector.
Is the above statement always true?... For example let
$$
A = left( begin{array}{ccc}
0.578385540014544 & 0.703045745965410 \
0.477513363789115 & 0.922698950982510 \
end{array} right)
$$
The largest eigenvalue is 1.35 (approx.)
Now, consider the vector (not eigenvector)
$$
v = left( begin{array}{ccc}
-0.538656963091298 \
-0.842525178326001 \
end{array} right)
$$
magnitude(v) = 1.0
magnitude(A*v) = 1.373
So this vector(v), which is not the eigenvector of A is scaled by a larger amount (x1.373), compared to the eigenvector which is scaled by x 1.35 (approx.)
Is this just an artifact of numerical precision ? I can easily create more examples of random square transformation matrices (A) where the eigenvector does not correspond to the direction of maximum scaling.
eigenvalues-eigenvectors
edited Apr 13 '17 at 12:19
Community♦
1
asked Sep 16 '15 at 6:03
user1837245user1837245
162
$endgroup$
Does the eigenvector correspond to a direction in which maximum scaling occurs by a given transformation matrix (A) acting upon this vector.
I quote from :
https://math.stackexchange.com/q/243553
No other vector when acted by this matrix will get stretched as much
as this eigenvector.
Is the above statement always true?... For example let
$$
A = left( begin{array}{ccc}
0.578385540014544 & 0.703045745965410 \
0.477513363789115 & 0.922698950982510 \
end{array} right)
$$
The largest eigenvalue is 1.35 (approx.)
Now, consider the vector (not eigenvector)
$$
v = left( begin{array}{ccc}
-0.538656963091298 \
-0.842525178326001 \
end{array} right)
$$
magnitude(v) = 1.0
magnitude(A*v) = 1.373
So this vector(v), which is not the eigenvector of A is scaled by a larger amount (x1.373), compared to the eigenvector which is scaled by x 1.35 (approx.)
Is this just an artifact of numerical precision ? I can easily create more examples of random square transformation matrices (A) where the eigenvector does not correspond to the direction of maximum scaling.
eigenvalues-eigenvectors
eigenvalues-eigenvectors
edited Apr 13 '17 at 12:19
Community♦
1
asked Sep 16 '15 at 6:03
user1837245user1837245
162
edited Apr 13 '17 at 12:19
Community♦
1
asked Sep 16 '15 at 6:03
user1837245user1837245
162
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
asked Sep 16 '15 at 6:03
user1837245user1837245
162
asked Sep 16 '15 at 6:03
user1837245user1837245
162
asked Sep 16 '15 at 6:03
user1837245user1837245
162
162
1
$begingroup$
The quote is a false statement. It is true for a symmetric matrix that the largest eigenvalue (in absolute value) equals the largest stretch factor. Your example shows that it's not necessarily true in general.
$endgroup$
– Greg Martin
Sep 16 '15 at 6:59
$begingroup$
Thank you for the answer Greg.
$endgroup$
– user1837245
Sep 16 '15 at 19:58
add a comment |
1
$begingroup$
The quote is a false statement. It is true for a symmetric matrix that the largest eigenvalue (in absolute value) equals the largest stretch factor. Your example shows that it's not necessarily true in general.
$endgroup$
– Greg Martin
Sep 16 '15 at 6:59
$begingroup$
Thank you for the answer Greg.
$endgroup$
– user1837245
Sep 16 '15 at 19:58
1
1
$begingroup$
The quote is a false statement. It is true for a symmetric matrix that the largest eigenvalue (in absolute value) equals the largest stretch factor. Your example shows that it's not necessarily true in general.
$endgroup$
– Greg Martin
Sep 16 '15 at 6:59
$begingroup$
The quote is a false statement. It is true for a symmetric matrix that the largest eigenvalue (in absolute value) equals the largest stretch factor. Your example shows that it's not necessarily true in general.
$endgroup$
– Greg Martin
Sep 16 '15 at 6:59
$begingroup$
Thank you for the answer Greg.
$endgroup$
– user1837245
Sep 16 '15 at 19:58
$begingroup$
Thank you for the answer Greg.
$endgroup$
– user1837245
Sep 16 '15 at 19:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No - An eigenvector is the "input vector" of a matrix that "outputs" a vector in the same direction. If you really want an intuitive understanding of Eigenvalues and Eigenvectors, follow the link. (I don't have the reputation needed to post images.)
Linear Transformations & Eigenvectors Explained
edited Jan 5 at 14:59
Bhavishya Desai
32
answered Oct 12 '15 at 21:55
user3527854user3527854
111
$endgroup$
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
});
}
});
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%2f1437569%2fdo-eigenvectors-correspond-to-direction-of-maximum-scaling%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
$begingroup$
No - An eigenvector is the "input vector" of a matrix that "outputs" a vector in the same direction. If you really want an intuitive understanding of Eigenvalues and Eigenvectors, follow the link. (I don't have the reputation needed to post images.)
Linear Transformations & Eigenvectors Explained
edited Jan 5 at 14:59
Bhavishya Desai
32
answered Oct 12 '15 at 21:55
user3527854user3527854
111
$endgroup$
add a comment |
$begingroup$
No - An eigenvector is the "input vector" of a matrix that "outputs" a vector in the same direction. If you really want an intuitive understanding of Eigenvalues and Eigenvectors, follow the link. (I don't have the reputation needed to post images.)
Linear Transformations & Eigenvectors Explained
edited Jan 5 at 14:59
Bhavishya Desai
32
answered Oct 12 '15 at 21:55
user3527854user3527854
111
$endgroup$
add a comment |
$begingroup$
No - An eigenvector is the "input vector" of a matrix that "outputs" a vector in the same direction. If you really want an intuitive understanding of Eigenvalues and Eigenvectors, follow the link. (I don't have the reputation needed to post images.)
Linear Transformations & Eigenvectors Explained
edited Jan 5 at 14:59
Bhavishya Desai
32
answered Oct 12 '15 at 21:55
user3527854user3527854
111
$endgroup$
No - An eigenvector is the "input vector" of a matrix that "outputs" a vector in the same direction. If you really want an intuitive understanding of Eigenvalues and Eigenvectors, follow the link. (I don't have the reputation needed to post images.)
Linear Transformations & Eigenvectors Explained
edited Jan 5 at 14:59
Bhavishya Desai
32
answered Oct 12 '15 at 21:55
user3527854user3527854
111
edited Jan 5 at 14:59
Bhavishya Desai
32
edited Jan 5 at 14:59
Bhavishya Desai
32
edited Jan 5 at 14:59
Bhavishya Desai
32
32
answered Oct 12 '15 at 21:55
user3527854user3527854
111
answered Oct 12 '15 at 21:55
user3527854user3527854
111
answered Oct 12 '15 at 21:55
user3527854user3527854
111
111
add a comment |
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%2f1437569%2fdo-eigenvectors-correspond-to-direction-of-maximum-scaling%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
$begingroup$
The quote is a false statement. It is true for a symmetric matrix that the largest eigenvalue (in absolute value) equals the largest stretch factor. Your example shows that it's not necessarily true in general.
$endgroup$
– Greg Martin
Sep 16 '15 at 6:59
$begingroup$
Thank you for the answer Greg.
$endgroup$
– user1837245
Sep 16 '15 at 19:58