Eigenvalue of Matrix: Determinant equals to the product of all its eigenvalue
Question:
The solution of (a) let the $lambda = 0$, l do not understand why.
Isn't that $lambda$ can only have the value which is the same as each eigenvalue of the matrix? why $lambda$ can be substituted with $0$ ?
Does $lambda$ still have the meaning of eigenvalue?
matrices eigenvalues-eigenvectors
asked Oct 28 at 7:38
Vincent Li
62
add a comment |
Question:
The solution of (a) let the $lambda = 0$, l do not understand why.
Isn't that $lambda$ can only have the value which is the same as each eigenvalue of the matrix? why $lambda$ can be substituted with $0$ ?
Does $lambda$ still have the meaning of eigenvalue?
matrices eigenvalues-eigenvectors
asked Oct 28 at 7:38
Vincent Li
62
1
The choice of $lambda$ for the variable of the characteristic polynomial is perhaps confusing here. We have $det (t I - B) = cdots - lambda_1 lambda_2lambda_3$, and at $t = 0$, $cdots = 0$, so evaluating both sides there gives $det (-B) = -lambda_1 lambda_2 lambda_3$.
– Travis
Oct 28 at 7:47
add a comment |
Question:
The solution of (a) let the $lambda = 0$, l do not understand why.
Isn't that $lambda$ can only have the value which is the same as each eigenvalue of the matrix? why $lambda$ can be substituted with $0$ ?
Does $lambda$ still have the meaning of eigenvalue?
matrices eigenvalues-eigenvectors
asked Oct 28 at 7:38
Vincent Li
62
Question:
The solution of (a) let the $lambda = 0$, l do not understand why.
Isn't that $lambda$ can only have the value which is the same as each eigenvalue of the matrix? why $lambda$ can be substituted with $0$ ?
Does $lambda$ still have the meaning of eigenvalue?
matrices eigenvalues-eigenvectors
matrices eigenvalues-eigenvectors
asked Oct 28 at 7:38
Vincent Li
62
asked Oct 28 at 7:38
Vincent Li
62
edited Oct 28 at 7:43
asked Oct 28 at 7:38
Vincent Li
62
asked Oct 28 at 7:38
Vincent Li
62
asked Oct 28 at 7:38
Vincent Li
62
62
1
The choice of $lambda$ for the variable of the characteristic polynomial is perhaps confusing here. We have $det (t I - B) = cdots - lambda_1 lambda_2lambda_3$, and at $t = 0$, $cdots = 0$, so evaluating both sides there gives $det (-B) = -lambda_1 lambda_2 lambda_3$.
– Travis
Oct 28 at 7:47
add a comment |
1
The choice of $lambda$ for the variable of the characteristic polynomial is perhaps confusing here. We have $det (t I - B) = cdots - lambda_1 lambda_2lambda_3$, and at $t = 0$, $cdots = 0$, so evaluating both sides there gives $det (-B) = -lambda_1 lambda_2 lambda_3$.
– Travis
Oct 28 at 7:47
1
1
The choice of $lambda$ for the variable of the characteristic polynomial is perhaps confusing here. We have $det (t I - B) = cdots - lambda_1 lambda_2lambda_3$, and at $t = 0$, $cdots = 0$, so evaluating both sides there gives $det (-B) = -lambda_1 lambda_2 lambda_3$.
– Travis
Oct 28 at 7:47
The choice of $lambda$ for the variable of the characteristic polynomial is perhaps confusing here. We have $det (t I - B) = cdots - lambda_1 lambda_2lambda_3$, and at $t = 0$, $cdots = 0$, so evaluating both sides there gives $det (-B) = -lambda_1 lambda_2 lambda_3$.
– Travis
Oct 28 at 7:47
add a comment |
1 Answer
1
active
oldest
votes
For any square matrix $A$, we define $$f(x)=|xI-A|$$ also according to definition of eigenvalue$$Av=lambda vto (A-lambda I)v=0$$this means that there is a linear dependence between the rows of $A-lambda I$ (with the coefficients being the entries of $v$) therefore $|A-lambda I|=0$. Conversely, if for some $lambda$ we have $|A-lambda I|=0$ then there exists some linear dependence as described before for which$$(A-lambda I)v=0$$and $lambda$ and $v$ become eigenvalue and eigenvector respectively. Therefore the equation $|A-lambda I|=0$ yields to all eigenvalues. Equivalently (and fortunately!) all the roots of $f(lambda)=0$ are the eigenvalues. This means that we can express $f(lambda)$ as following$$f(lambda)=|lambda I-A|=(lambda-lambda_1)(lambda-lambda_2)cdots (lambda-lambda_n)$$by substituting $0$ we obtain $$|-A|=(-1)^n |A|=(-1)^n lambda_1lambda_2cdots lambda_n$$which leads to $$|A|= lambda_1lambda_2cdots lambda_n$$
answered Nov 24 at 17:59
Mostafa Ayaz
13.7k3836
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%2f2974364%2feigenvalue-of-matrix-determinant-equals-to-the-product-of-all-its-eigenvalue%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
For any square matrix $A$, we define $$f(x)=|xI-A|$$ also according to definition of eigenvalue$$Av=lambda vto (A-lambda I)v=0$$this means that there is a linear dependence between the rows of $A-lambda I$ (with the coefficients being the entries of $v$) therefore $|A-lambda I|=0$. Conversely, if for some $lambda$ we have $|A-lambda I|=0$ then there exists some linear dependence as described before for which$$(A-lambda I)v=0$$and $lambda$ and $v$ become eigenvalue and eigenvector respectively. Therefore the equation $|A-lambda I|=0$ yields to all eigenvalues. Equivalently (and fortunately!) all the roots of $f(lambda)=0$ are the eigenvalues. This means that we can express $f(lambda)$ as following$$f(lambda)=|lambda I-A|=(lambda-lambda_1)(lambda-lambda_2)cdots (lambda-lambda_n)$$by substituting $0$ we obtain $$|-A|=(-1)^n |A|=(-1)^n lambda_1lambda_2cdots lambda_n$$which leads to $$|A|= lambda_1lambda_2cdots lambda_n$$
answered Nov 24 at 17:59
Mostafa Ayaz
13.7k3836
add a comment |
For any square matrix $A$, we define $$f(x)=|xI-A|$$ also according to definition of eigenvalue$$Av=lambda vto (A-lambda I)v=0$$this means that there is a linear dependence between the rows of $A-lambda I$ (with the coefficients being the entries of $v$) therefore $|A-lambda I|=0$. Conversely, if for some $lambda$ we have $|A-lambda I|=0$ then there exists some linear dependence as described before for which$$(A-lambda I)v=0$$and $lambda$ and $v$ become eigenvalue and eigenvector respectively. Therefore the equation $|A-lambda I|=0$ yields to all eigenvalues. Equivalently (and fortunately!) all the roots of $f(lambda)=0$ are the eigenvalues. This means that we can express $f(lambda)$ as following$$f(lambda)=|lambda I-A|=(lambda-lambda_1)(lambda-lambda_2)cdots (lambda-lambda_n)$$by substituting $0$ we obtain $$|-A|=(-1)^n |A|=(-1)^n lambda_1lambda_2cdots lambda_n$$which leads to $$|A|= lambda_1lambda_2cdots lambda_n$$
answered Nov 24 at 17:59
Mostafa Ayaz
13.7k3836
add a comment |
For any square matrix $A$, we define $$f(x)=|xI-A|$$ also according to definition of eigenvalue$$Av=lambda vto (A-lambda I)v=0$$this means that there is a linear dependence between the rows of $A-lambda I$ (with the coefficients being the entries of $v$) therefore $|A-lambda I|=0$. Conversely, if for some $lambda$ we have $|A-lambda I|=0$ then there exists some linear dependence as described before for which$$(A-lambda I)v=0$$and $lambda$ and $v$ become eigenvalue and eigenvector respectively. Therefore the equation $|A-lambda I|=0$ yields to all eigenvalues. Equivalently (and fortunately!) all the roots of $f(lambda)=0$ are the eigenvalues. This means that we can express $f(lambda)$ as following$$f(lambda)=|lambda I-A|=(lambda-lambda_1)(lambda-lambda_2)cdots (lambda-lambda_n)$$by substituting $0$ we obtain $$|-A|=(-1)^n |A|=(-1)^n lambda_1lambda_2cdots lambda_n$$which leads to $$|A|= lambda_1lambda_2cdots lambda_n$$
answered Nov 24 at 17:59
Mostafa Ayaz
13.7k3836
For any square matrix $A$, we define $$f(x)=|xI-A|$$ also according to definition of eigenvalue$$Av=lambda vto (A-lambda I)v=0$$this means that there is a linear dependence between the rows of $A-lambda I$ (with the coefficients being the entries of $v$) therefore $|A-lambda I|=0$. Conversely, if for some $lambda$ we have $|A-lambda I|=0$ then there exists some linear dependence as described before for which$$(A-lambda I)v=0$$and $lambda$ and $v$ become eigenvalue and eigenvector respectively. Therefore the equation $|A-lambda I|=0$ yields to all eigenvalues. Equivalently (and fortunately!) all the roots of $f(lambda)=0$ are the eigenvalues. This means that we can express $f(lambda)$ as following$$f(lambda)=|lambda I-A|=(lambda-lambda_1)(lambda-lambda_2)cdots (lambda-lambda_n)$$by substituting $0$ we obtain $$|-A|=(-1)^n |A|=(-1)^n lambda_1lambda_2cdots lambda_n$$which leads to $$|A|= lambda_1lambda_2cdots lambda_n$$
answered Nov 24 at 17:59
Mostafa Ayaz
13.7k3836
answered Nov 24 at 17:59
Mostafa Ayaz
13.7k3836
answered Nov 24 at 17:59
Mostafa Ayaz
13.7k3836
answered Nov 24 at 17:59
Mostafa Ayaz
13.7k3836
13.7k3836
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.
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%2f2974364%2feigenvalue-of-matrix-determinant-equals-to-the-product-of-all-its-eigenvalue%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
The choice of $lambda$ for the variable of the characteristic polynomial is perhaps confusing here. We have $det (t I - B) = cdots - lambda_1 lambda_2lambda_3$, and at $t = 0$, $cdots = 0$, so evaluating both sides there gives $det (-B) = -lambda_1 lambda_2 lambda_3$.
– Travis
Oct 28 at 7:47