If a diagonalizable matrix is equal to its cube, then its rank is equal to the trace of its square
$begingroup$
Suppose $A in Bbb R ^{n times n}$ is such that $A^3=A$, and $A$ is diagonalizable. Prove that $operatorname{trace}(A^2) = operatorname{rank}(A)$.
I know that the possible eigenvalues of $A$ are $0,1,-1$. But how does that help to prove the required relation between trace and rank?
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked Oct 6 '15 at 7:53
StabiloStabilo
729512
$endgroup$
add a comment |
$begingroup$
Suppose $A in Bbb R ^{n times n}$ is such that $A^3=A$, and $A$ is diagonalizable. Prove that $operatorname{trace}(A^2) = operatorname{rank}(A)$.
I know that the possible eigenvalues of $A$ are $0,1,-1$. But how does that help to prove the required relation between trace and rank?
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked Oct 6 '15 at 7:53
StabiloStabilo
729512
$endgroup$
add a comment |
$begingroup$
Suppose $A in Bbb R ^{n times n}$ is such that $A^3=A$, and $A$ is diagonalizable. Prove that $operatorname{trace}(A^2) = operatorname{rank}(A)$.
I know that the possible eigenvalues of $A$ are $0,1,-1$. But how does that help to prove the required relation between trace and rank?
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked Oct 6 '15 at 7:53
StabiloStabilo
729512
$endgroup$
Suppose $A in Bbb R ^{n times n}$ is such that $A^3=A$, and $A$ is diagonalizable. Prove that $operatorname{trace}(A^2) = operatorname{rank}(A)$.
I know that the possible eigenvalues of $A$ are $0,1,-1$. But how does that help to prove the required relation between trace and rank?
linear-algebra matrices eigenvalues-eigenvectors diagonalization
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked Oct 6 '15 at 7:53
StabiloStabilo
729512
asked Oct 6 '15 at 7:53
StabiloStabilo
729512
edited Oct 11 '15 at 4:55
user147263
asked Oct 6 '15 at 7:53
StabiloStabilo
729512
asked Oct 6 '15 at 7:53
StabiloStabilo
729512
asked Oct 6 '15 at 7:53
StabiloStabilo
729512
729512
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$Tr(A^2)=sum_{i}lambda_i^2$. Let the number of $1$'s be $k$, no. of $-1$'s be $l$, then, $Tr(A^2)=k+l$ and $rank(A)=$ no. of nonzero eigenvalues of $A$, which is again $k+l$.
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.8k2047
$endgroup$
$begingroup$
Two questions 1.Where is the fact that $A$ is diagonalizable used?2.how is rank of A=no. of non-zero eigen values
$endgroup$
– Learnmore
Oct 6 '15 at 10:13
$begingroup$
@learnmore A is similar to diagonal matrix D that its diagonal elements are A's eigenvalues, so tr(A)=tr(D) which is multiplication of diagonal elements = multiplication of eigenvalues. Also rank(A)=rank(D) = no. of non-zero eigenvalues.
$endgroup$
– Stabilo
Oct 6 '15 at 10:40
$begingroup$
$tr(D)$ =sum of diagonal elements ;anyway got your point
$endgroup$
– Learnmore
Oct 6 '15 at 10:44
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%2f1466788%2fif-a-diagonalizable-matrix-is-equal-to-its-cube-then-its-rank-is-equal-to-the-t%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
$begingroup$
$Tr(A^2)=sum_{i}lambda_i^2$. Let the number of $1$'s be $k$, no. of $-1$'s be $l$, then, $Tr(A^2)=k+l$ and $rank(A)=$ no. of nonzero eigenvalues of $A$, which is again $k+l$.
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.8k2047
$endgroup$
$begingroup$
Two questions 1.Where is the fact that $A$ is diagonalizable used?2.how is rank of A=no. of non-zero eigen values
$endgroup$
– Learnmore
Oct 6 '15 at 10:13
$begingroup$
@learnmore A is similar to diagonal matrix D that its diagonal elements are A's eigenvalues, so tr(A)=tr(D) which is multiplication of diagonal elements = multiplication of eigenvalues. Also rank(A)=rank(D) = no. of non-zero eigenvalues.
$endgroup$
– Stabilo
Oct 6 '15 at 10:40
$begingroup$
$tr(D)$ =sum of diagonal elements ;anyway got your point
$endgroup$
– Learnmore
Oct 6 '15 at 10:44
add a comment |
$begingroup$
$Tr(A^2)=sum_{i}lambda_i^2$. Let the number of $1$'s be $k$, no. of $-1$'s be $l$, then, $Tr(A^2)=k+l$ and $rank(A)=$ no. of nonzero eigenvalues of $A$, which is again $k+l$.
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.8k2047
$endgroup$
$begingroup$
Two questions 1.Where is the fact that $A$ is diagonalizable used?2.how is rank of A=no. of non-zero eigen values
$endgroup$
– Learnmore
Oct 6 '15 at 10:13
$begingroup$
@learnmore A is similar to diagonal matrix D that its diagonal elements are A's eigenvalues, so tr(A)=tr(D) which is multiplication of diagonal elements = multiplication of eigenvalues. Also rank(A)=rank(D) = no. of non-zero eigenvalues.
$endgroup$
– Stabilo
Oct 6 '15 at 10:40
$begingroup$
$tr(D)$ =sum of diagonal elements ;anyway got your point
$endgroup$
– Learnmore
Oct 6 '15 at 10:44
add a comment |
$begingroup$
$Tr(A^2)=sum_{i}lambda_i^2$. Let the number of $1$'s be $k$, no. of $-1$'s be $l$, then, $Tr(A^2)=k+l$ and $rank(A)=$ no. of nonzero eigenvalues of $A$, which is again $k+l$.
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.8k2047
$endgroup$
$Tr(A^2)=sum_{i}lambda_i^2$. Let the number of $1$'s be $k$, no. of $-1$'s be $l$, then, $Tr(A^2)=k+l$ and $rank(A)=$ no. of nonzero eigenvalues of $A$, which is again $k+l$.
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.8k2047
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.8k2047
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.8k2047
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.8k2047
13.8k2047
$begingroup$
Two questions 1.Where is the fact that $A$ is diagonalizable used?2.how is rank of A=no. of non-zero eigen values
$endgroup$
– Learnmore
Oct 6 '15 at 10:13
$begingroup$
@learnmore A is similar to diagonal matrix D that its diagonal elements are A's eigenvalues, so tr(A)=tr(D) which is multiplication of diagonal elements = multiplication of eigenvalues. Also rank(A)=rank(D) = no. of non-zero eigenvalues.
$endgroup$
– Stabilo
Oct 6 '15 at 10:40
$begingroup$
$tr(D)$ =sum of diagonal elements ;anyway got your point
$endgroup$
– Learnmore
Oct 6 '15 at 10:44
add a comment |
$begingroup$
Two questions 1.Where is the fact that $A$ is diagonalizable used?2.how is rank of A=no. of non-zero eigen values
$endgroup$
– Learnmore
Oct 6 '15 at 10:13
$begingroup$
@learnmore A is similar to diagonal matrix D that its diagonal elements are A's eigenvalues, so tr(A)=tr(D) which is multiplication of diagonal elements = multiplication of eigenvalues. Also rank(A)=rank(D) = no. of non-zero eigenvalues.
$endgroup$
– Stabilo
Oct 6 '15 at 10:40
$begingroup$
$tr(D)$ =sum of diagonal elements ;anyway got your point
$endgroup$
– Learnmore
Oct 6 '15 at 10:44
$begingroup$
Two questions 1.Where is the fact that $A$ is diagonalizable used?2.how is rank of A=no. of non-zero eigen values
$endgroup$
– Learnmore
Oct 6 '15 at 10:13
$begingroup$
Two questions 1.Where is the fact that $A$ is diagonalizable used?2.how is rank of A=no. of non-zero eigen values
$endgroup$
– Learnmore
Oct 6 '15 at 10:13
$begingroup$
@learnmore A is similar to diagonal matrix D that its diagonal elements are A's eigenvalues, so tr(A)=tr(D) which is multiplication of diagonal elements = multiplication of eigenvalues. Also rank(A)=rank(D) = no. of non-zero eigenvalues.
$endgroup$
– Stabilo
Oct 6 '15 at 10:40
$begingroup$
@learnmore A is similar to diagonal matrix D that its diagonal elements are A's eigenvalues, so tr(A)=tr(D) which is multiplication of diagonal elements = multiplication of eigenvalues. Also rank(A)=rank(D) = no. of non-zero eigenvalues.
$endgroup$
– Stabilo
Oct 6 '15 at 10:40
$begingroup$
$tr(D)$ =sum of diagonal elements ;anyway got your point
$endgroup$
– Learnmore
Oct 6 '15 at 10:44
$begingroup$
$tr(D)$ =sum of diagonal elements ;anyway got your point
$endgroup$
– Learnmore
Oct 6 '15 at 10:44
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%2f1466788%2fif-a-diagonalizable-matrix-is-equal-to-its-cube-then-its-rank-is-equal-to-the-t%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
