Suppose A is a square matrix and v is an eigenvector of A with associated eigenvalue 23. If x is a real...
$begingroup$
Suppose A is a square matrix and v is an eigenvector of A with associated eigenvalue 23. If x is a real number, compute A(xv) in terms of x and v.
A(xv) = (?)
Not sure where to go with this question, is there a rule or condition that I am missing?
So far I thought it would be something using the following;
det(A - xI) = 0 and Ax = vx and solving for A but any simplification using these ideas have been wrong so far. Any suggestions?
linear-algebra matrices eigenvalues-eigenvectors vectors
asked Nov 20 '18 at 0:52
S. SnakeS. Snake
485
$endgroup$
add a comment |
$begingroup$
Suppose A is a square matrix and v is an eigenvector of A with associated eigenvalue 23. If x is a real number, compute A(xv) in terms of x and v.
A(xv) = (?)
Not sure where to go with this question, is there a rule or condition that I am missing?
So far I thought it would be something using the following;
det(A - xI) = 0 and Ax = vx and solving for A but any simplification using these ideas have been wrong so far. Any suggestions?
linear-algebra matrices eigenvalues-eigenvectors vectors
asked Nov 20 '18 at 0:52
S. SnakeS. Snake
485
$endgroup$
2
$begingroup$
It is only a matter of really understanding the basic definitions to conclude that $$A(xv)=xAv=xcdot23 v=23xv$$
$endgroup$
– DonAntonio
Nov 20 '18 at 0:54
$begingroup$
I mistakenly thought x=23. Thank you!
$endgroup$
– S. Snake
Nov 20 '18 at 0:55
add a comment |
$begingroup$
Suppose A is a square matrix and v is an eigenvector of A with associated eigenvalue 23. If x is a real number, compute A(xv) in terms of x and v.
A(xv) = (?)
Not sure where to go with this question, is there a rule or condition that I am missing?
So far I thought it would be something using the following;
det(A - xI) = 0 and Ax = vx and solving for A but any simplification using these ideas have been wrong so far. Any suggestions?
linear-algebra matrices eigenvalues-eigenvectors vectors
asked Nov 20 '18 at 0:52
S. SnakeS. Snake
485
$endgroup$
Suppose A is a square matrix and v is an eigenvector of A with associated eigenvalue 23. If x is a real number, compute A(xv) in terms of x and v.
A(xv) = (?)
Not sure where to go with this question, is there a rule or condition that I am missing?
So far I thought it would be something using the following;
det(A - xI) = 0 and Ax = vx and solving for A but any simplification using these ideas have been wrong so far. Any suggestions?
linear-algebra matrices eigenvalues-eigenvectors vectors
linear-algebra matrices eigenvalues-eigenvectors vectors
asked Nov 20 '18 at 0:52
S. SnakeS. Snake
485
asked Nov 20 '18 at 0:52
S. SnakeS. Snake
485
asked Nov 20 '18 at 0:52
S. SnakeS. Snake
485
asked Nov 20 '18 at 0:52
S. SnakeS. Snake
485
asked Nov 20 '18 at 0:52
S. SnakeS. Snake
485
485
2
$begingroup$
It is only a matter of really understanding the basic definitions to conclude that $$A(xv)=xAv=xcdot23 v=23xv$$
$endgroup$
– DonAntonio
Nov 20 '18 at 0:54
$begingroup$
I mistakenly thought x=23. Thank you!
$endgroup$
– S. Snake
Nov 20 '18 at 0:55
add a comment |
2
$begingroup$
It is only a matter of really understanding the basic definitions to conclude that $$A(xv)=xAv=xcdot23 v=23xv$$
$endgroup$
– DonAntonio
Nov 20 '18 at 0:54
$begingroup$
I mistakenly thought x=23. Thank you!
$endgroup$
– S. Snake
Nov 20 '18 at 0:55
2
2
$begingroup$
It is only a matter of really understanding the basic definitions to conclude that $$A(xv)=xAv=xcdot23 v=23xv$$
$endgroup$
– DonAntonio
Nov 20 '18 at 0:54
$begingroup$
It is only a matter of really understanding the basic definitions to conclude that $$A(xv)=xAv=xcdot23 v=23xv$$
$endgroup$
– DonAntonio
Nov 20 '18 at 0:54
$begingroup$
I mistakenly thought x=23. Thank you!
$endgroup$
– S. Snake
Nov 20 '18 at 0:55
$begingroup$
I mistakenly thought x=23. Thank you!
$endgroup$
– S. Snake
Nov 20 '18 at 0:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $v$ is an eigenvector with eigenvalue $23$, this means $Av=23v$. So for any real number $x$, we have
$$A(xv)=xAv=xcdot23 v=23(xv).$$
$endgroup$
$begingroup$
This answer is essentially just repeating DonAntonio's comment, and exists primarily to remove this question from the Unanswered queue. Please upvote or choose as Best Answer to complete this process.
$endgroup$
– aleph_two
Dec 31 '18 at 4:45
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%2f3005775%2fsuppose-a-is-a-square-matrix-and-v-is-an-eigenvector-of-a-with-associated-eigenv%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
$begingroup$
If $v$ is an eigenvector with eigenvalue $23$, this means $Av=23v$. So for any real number $x$, we have
$$A(xv)=xAv=xcdot23 v=23(xv).$$
$endgroup$
$begingroup$
This answer is essentially just repeating DonAntonio's comment, and exists primarily to remove this question from the Unanswered queue. Please upvote or choose as Best Answer to complete this process.
$endgroup$
– aleph_two
Dec 31 '18 at 4:45
add a comment |
$begingroup$
If $v$ is an eigenvector with eigenvalue $23$, this means $Av=23v$. So for any real number $x$, we have
$$A(xv)=xAv=xcdot23 v=23(xv).$$
$endgroup$
$begingroup$
This answer is essentially just repeating DonAntonio's comment, and exists primarily to remove this question from the Unanswered queue. Please upvote or choose as Best Answer to complete this process.
$endgroup$
– aleph_two
Dec 31 '18 at 4:45
add a comment |
$begingroup$
If $v$ is an eigenvector with eigenvalue $23$, this means $Av=23v$. So for any real number $x$, we have
$$A(xv)=xAv=xcdot23 v=23(xv).$$
$endgroup$
If $v$ is an eigenvector with eigenvalue $23$, this means $Av=23v$. So for any real number $x$, we have
$$A(xv)=xAv=xcdot23 v=23(xv).$$
answered Dec 31 '18 at 4:44
community wiki
aleph_two
$begingroup$
This answer is essentially just repeating DonAntonio's comment, and exists primarily to remove this question from the Unanswered queue. Please upvote or choose as Best Answer to complete this process.
$endgroup$
– aleph_two
Dec 31 '18 at 4:45
add a comment |
$begingroup$
This answer is essentially just repeating DonAntonio's comment, and exists primarily to remove this question from the Unanswered queue. Please upvote or choose as Best Answer to complete this process.
$endgroup$
– aleph_two
Dec 31 '18 at 4:45
$begingroup$
This answer is essentially just repeating DonAntonio's comment, and exists primarily to remove this question from the Unanswered queue. Please upvote or choose as Best Answer to complete this process.
$endgroup$
– aleph_two
Dec 31 '18 at 4:45
$begingroup$
This answer is essentially just repeating DonAntonio's comment, and exists primarily to remove this question from the Unanswered queue. Please upvote or choose as Best Answer to complete this process.
$endgroup$
– aleph_two
Dec 31 '18 at 4:45
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%2f3005775%2fsuppose-a-is-a-square-matrix-and-v-is-an-eigenvector-of-a-with-associated-eigenv%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
$begingroup$
It is only a matter of really understanding the basic definitions to conclude that $$A(xv)=xAv=xcdot23 v=23xv$$
$endgroup$
– DonAntonio
Nov 20 '18 at 0:54
$begingroup$
I mistakenly thought x=23. Thank you!
$endgroup$
– S. Snake
Nov 20 '18 at 0:55