How to deduce third eigenvalue/eigenvector pair of a matrix from first two pairs and determinant
$begingroup$
Suppose you have a 3 x 3 matrix 'A' with unknown values whose determinant is known and non - zero. If you know two of its eigenvalues and their corresponding eigenvectors, how can you find the third eigenvalue and eigenvector?
linear-algebra matrices
asked Dec 11 '18 at 2:25
BuluBestTapuBuluBestTapu
778
$endgroup$
|
show 1 more comment
$begingroup$
Suppose you have a 3 x 3 matrix 'A' with unknown values whose determinant is known and non - zero. If you know two of its eigenvalues and their corresponding eigenvectors, how can you find the third eigenvalue and eigenvector?
linear-algebra matrices
asked Dec 11 '18 at 2:25
BuluBestTapuBuluBestTapu
778
$endgroup$
1
$begingroup$
The product of the three eigenvalues is the determinant.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:25
$begingroup$
I know that, but how do you find its corresponding eigenvector?
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:26
$begingroup$
Same as you find any eigenvector: Null space of $A-lambda I$. There's not a special trick (unless $A$ is symmetric, then you can find the O. Complement of the first two evectors)
$endgroup$
– Morgan Rodgers
Dec 11 '18 at 2:28
$begingroup$
You do not know what the matrix values of 'A' are. You only know its determinant and that it's 3x3. I have updated the question to reflect this.
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:31
$begingroup$
There’s no way to determine the missing eigenspace without more information about $A$.
$endgroup$
– amd
Dec 11 '18 at 2:50
|
show 1 more comment
$begingroup$
Suppose you have a 3 x 3 matrix 'A' with unknown values whose determinant is known and non - zero. If you know two of its eigenvalues and their corresponding eigenvectors, how can you find the third eigenvalue and eigenvector?
linear-algebra matrices
asked Dec 11 '18 at 2:25
BuluBestTapuBuluBestTapu
778
$endgroup$
Suppose you have a 3 x 3 matrix 'A' with unknown values whose determinant is known and non - zero. If you know two of its eigenvalues and their corresponding eigenvectors, how can you find the third eigenvalue and eigenvector?
linear-algebra matrices
linear-algebra matrices
asked Dec 11 '18 at 2:25
BuluBestTapuBuluBestTapu
778
asked Dec 11 '18 at 2:25
BuluBestTapuBuluBestTapu
778
edited Dec 11 '18 at 2:31
BuluBestTapu
asked Dec 11 '18 at 2:25
BuluBestTapuBuluBestTapu
778
asked Dec 11 '18 at 2:25
BuluBestTapuBuluBestTapu
778
asked Dec 11 '18 at 2:25
BuluBestTapuBuluBestTapu
778
778
1
$begingroup$
The product of the three eigenvalues is the determinant.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:25
$begingroup$
I know that, but how do you find its corresponding eigenvector?
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:26
$begingroup$
Same as you find any eigenvector: Null space of $A-lambda I$. There's not a special trick (unless $A$ is symmetric, then you can find the O. Complement of the first two evectors)
$endgroup$
– Morgan Rodgers
Dec 11 '18 at 2:28
$begingroup$
You do not know what the matrix values of 'A' are. You only know its determinant and that it's 3x3. I have updated the question to reflect this.
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:31
$begingroup$
There’s no way to determine the missing eigenspace without more information about $A$.
$endgroup$
– amd
Dec 11 '18 at 2:50
|
show 1 more comment
1
$begingroup$
The product of the three eigenvalues is the determinant.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:25
$begingroup$
I know that, but how do you find its corresponding eigenvector?
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:26
$begingroup$
Same as you find any eigenvector: Null space of $A-lambda I$. There's not a special trick (unless $A$ is symmetric, then you can find the O. Complement of the first two evectors)
$endgroup$
– Morgan Rodgers
Dec 11 '18 at 2:28
$begingroup$
You do not know what the matrix values of 'A' are. You only know its determinant and that it's 3x3. I have updated the question to reflect this.
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:31
$begingroup$
There’s no way to determine the missing eigenspace without more information about $A$.
$endgroup$
– amd
Dec 11 '18 at 2:50
1
1
$begingroup$
The product of the three eigenvalues is the determinant.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:25
$begingroup$
The product of the three eigenvalues is the determinant.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:25
$begingroup$
I know that, but how do you find its corresponding eigenvector?
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:26
$begingroup$
I know that, but how do you find its corresponding eigenvector?
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:26
$begingroup$
Same as you find any eigenvector: Null space of $A-lambda I$. There's not a special trick (unless $A$ is symmetric, then you can find the O. Complement of the first two evectors)
$endgroup$
– Morgan Rodgers
Dec 11 '18 at 2:28
$begingroup$
Same as you find any eigenvector: Null space of $A-lambda I$. There's not a special trick (unless $A$ is symmetric, then you can find the O. Complement of the first two evectors)
$endgroup$
– Morgan Rodgers
Dec 11 '18 at 2:28
$begingroup$
You do not know what the matrix values of 'A' are. You only know its determinant and that it's 3x3. I have updated the question to reflect this.
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:31
$begingroup$
You do not know what the matrix values of 'A' are. You only know its determinant and that it's 3x3. I have updated the question to reflect this.
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:31
$begingroup$
There’s no way to determine the missing eigenspace without more information about $A$.
$endgroup$
– amd
Dec 11 '18 at 2:50
$begingroup$
There’s no way to determine the missing eigenspace without more information about $A$.
$endgroup$
– amd
Dec 11 '18 at 2:50
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Let $A=[a_{i,j}]$. You know $spectrum(A)={lambda,mu,nu}$ and $u,vinmathbb{R}^3$ s.t. $Au=lambda u,Av=mu v$; that is, (since $u,v$ are known up to a factor) $3+2+2$ independent algebraic relations linking the $9$ unknowns $(a_{i,j})$.
So, you miss 2 relations.
EDIT. That follows is false: "you could conclude if you were given -for example- in addition, an eigenvector of $A^T$".
Indeed, the eigenvector of $A^T$ associated to $nu$ is orthogonal to the plane $span(u,v)$.
Finally, I think the right question is: "find the eigenvector of $A^T$ associated with the third eigenvalue".
answered Dec 11 '18 at 10:41
loup blancloup blanc
23.4k21851
$endgroup$
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%2f3034790%2fhow-to-deduce-third-eigenvalue-eigenvector-pair-of-a-matrix-from-first-two-pairs%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
$begingroup$
Let $A=[a_{i,j}]$. You know $spectrum(A)={lambda,mu,nu}$ and $u,vinmathbb{R}^3$ s.t. $Au=lambda u,Av=mu v$; that is, (since $u,v$ are known up to a factor) $3+2+2$ independent algebraic relations linking the $9$ unknowns $(a_{i,j})$.
So, you miss 2 relations.
EDIT. That follows is false: "you could conclude if you were given -for example- in addition, an eigenvector of $A^T$".
Indeed, the eigenvector of $A^T$ associated to $nu$ is orthogonal to the plane $span(u,v)$.
Finally, I think the right question is: "find the eigenvector of $A^T$ associated with the third eigenvalue".
answered Dec 11 '18 at 10:41
loup blancloup blanc
23.4k21851
$endgroup$
add a comment |
$begingroup$
Let $A=[a_{i,j}]$. You know $spectrum(A)={lambda,mu,nu}$ and $u,vinmathbb{R}^3$ s.t. $Au=lambda u,Av=mu v$; that is, (since $u,v$ are known up to a factor) $3+2+2$ independent algebraic relations linking the $9$ unknowns $(a_{i,j})$.
So, you miss 2 relations.
EDIT. That follows is false: "you could conclude if you were given -for example- in addition, an eigenvector of $A^T$".
Indeed, the eigenvector of $A^T$ associated to $nu$ is orthogonal to the plane $span(u,v)$.
Finally, I think the right question is: "find the eigenvector of $A^T$ associated with the third eigenvalue".
answered Dec 11 '18 at 10:41
loup blancloup blanc
23.4k21851
$endgroup$
add a comment |
$begingroup$
Let $A=[a_{i,j}]$. You know $spectrum(A)={lambda,mu,nu}$ and $u,vinmathbb{R}^3$ s.t. $Au=lambda u,Av=mu v$; that is, (since $u,v$ are known up to a factor) $3+2+2$ independent algebraic relations linking the $9$ unknowns $(a_{i,j})$.
So, you miss 2 relations.
EDIT. That follows is false: "you could conclude if you were given -for example- in addition, an eigenvector of $A^T$".
Indeed, the eigenvector of $A^T$ associated to $nu$ is orthogonal to the plane $span(u,v)$.
Finally, I think the right question is: "find the eigenvector of $A^T$ associated with the third eigenvalue".
answered Dec 11 '18 at 10:41
loup blancloup blanc
23.4k21851
$endgroup$
Let $A=[a_{i,j}]$. You know $spectrum(A)={lambda,mu,nu}$ and $u,vinmathbb{R}^3$ s.t. $Au=lambda u,Av=mu v$; that is, (since $u,v$ are known up to a factor) $3+2+2$ independent algebraic relations linking the $9$ unknowns $(a_{i,j})$.
So, you miss 2 relations.
EDIT. That follows is false: "you could conclude if you were given -for example- in addition, an eigenvector of $A^T$".
Indeed, the eigenvector of $A^T$ associated to $nu$ is orthogonal to the plane $span(u,v)$.
Finally, I think the right question is: "find the eigenvector of $A^T$ associated with the third eigenvalue".
answered Dec 11 '18 at 10:41
loup blancloup blanc
23.4k21851
edited Dec 11 '18 at 11:17
answered Dec 11 '18 at 10:41
loup blancloup blanc
23.4k21851
answered Dec 11 '18 at 10:41
loup blancloup blanc
23.4k21851
answered Dec 11 '18 at 10:41
loup blancloup blanc
23.4k21851
23.4k21851
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%2f3034790%2fhow-to-deduce-third-eigenvalue-eigenvector-pair-of-a-matrix-from-first-two-pairs%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 product of the three eigenvalues is the determinant.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:25
$begingroup$
I know that, but how do you find its corresponding eigenvector?
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:26
$begingroup$
Same as you find any eigenvector: Null space of $A-lambda I$. There's not a special trick (unless $A$ is symmetric, then you can find the O. Complement of the first two evectors)
$endgroup$
– Morgan Rodgers
Dec 11 '18 at 2:28
$begingroup$
You do not know what the matrix values of 'A' are. You only know its determinant and that it's 3x3. I have updated the question to reflect this.
$endgroup$
– BuluBestTapu
Dec 11 '18 at 2:31
$begingroup$
There’s no way to determine the missing eigenspace without more information about $A$.
$endgroup$
– amd
Dec 11 '18 at 2:50