Given set M , which of following option is correct?
$begingroup$
let $M = { A = begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z} text{and the eigenvalue of A are in }mathbb{Q}}$ .Then
choose the correct option
$1.$ M is empty
$2.$$M = {begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z}}$
$3.$if $A in M$ ,then the eigenvalue of $A$ are in $mathbb{Z}$
$4.$if$ A,B in M$ are such that $AB=I$ then $det A in {+1,-1}$
MY attempt : option $1)$ is obviously false
option $2)$ is True according to the definition of question
option $3)$ is True take $A=I$
option $4)$ is True take $A= B=I$
is my attempt is True
Any hints/solution
linear-algebra
$endgroup$
add a comment |
$begingroup$
let $M = { A = begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z} text{and the eigenvalue of A are in }mathbb{Q}}$ .Then
choose the correct option
$1.$ M is empty
$2.$$M = {begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z}}$
$3.$if $A in M$ ,then the eigenvalue of $A$ are in $mathbb{Z}$
$4.$if$ A,B in M$ are such that $AB=I$ then $det A in {+1,-1}$
MY attempt : option $1)$ is obviously false
option $2)$ is True according to the definition of question
option $3)$ is True take $A=I$
option $4)$ is True take $A= B=I$
is my attempt is True
Any hints/solution
linear-algebra
$endgroup$
add a comment |
$begingroup$
let $M = { A = begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z} text{and the eigenvalue of A are in }mathbb{Q}}$ .Then
choose the correct option
$1.$ M is empty
$2.$$M = {begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z}}$
$3.$if $A in M$ ,then the eigenvalue of $A$ are in $mathbb{Z}$
$4.$if$ A,B in M$ are such that $AB=I$ then $det A in {+1,-1}$
MY attempt : option $1)$ is obviously false
option $2)$ is True according to the definition of question
option $3)$ is True take $A=I$
option $4)$ is True take $A= B=I$
is my attempt is True
Any hints/solution
linear-algebra
$endgroup$
let $M = { A = begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z} text{and the eigenvalue of A are in }mathbb{Q}}$ .Then
choose the correct option
$1.$ M is empty
$2.$$M = {begin{bmatrix} a & b\c&d end{bmatrix}| a,b ,c, d in mathbb{Z}}$
$3.$if $A in M$ ,then the eigenvalue of $A$ are in $mathbb{Z}$
$4.$if$ A,B in M$ are such that $AB=I$ then $det A in {+1,-1}$
MY attempt : option $1)$ is obviously false
option $2)$ is True according to the definition of question
option $3)$ is True take $A=I$
option $4)$ is True take $A= B=I$
is my attempt is True
Any hints/solution
linear-algebra
linear-algebra
asked Dec 15 '18 at 3:55
jasminejasmine
1,783417
1,783417
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For 1, you should display a matrix that is in $M$. It should be easy to find one.
For 2, they have removed the requirement that the eigenvalues of $A$ are rational. If you can find a matrix where the eigenvalues are not rational you show that this is not true. Can you find one?
For 3, finding one matrix is not enough. You are asked to show that if a matrix has integral entries and rational eigenvalues, the eigenvalues are integral. You could think about the rational root theorem and the characteristic polynomial of matrices in $M$.
4 is the same as 3. In fact it will follow from 3, if 3 is true.
$endgroup$
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%2f3040150%2fgiven-set-m-which-of-following-option-is-correct%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
$begingroup$
For 1, you should display a matrix that is in $M$. It should be easy to find one.
For 2, they have removed the requirement that the eigenvalues of $A$ are rational. If you can find a matrix where the eigenvalues are not rational you show that this is not true. Can you find one?
For 3, finding one matrix is not enough. You are asked to show that if a matrix has integral entries and rational eigenvalues, the eigenvalues are integral. You could think about the rational root theorem and the characteristic polynomial of matrices in $M$.
4 is the same as 3. In fact it will follow from 3, if 3 is true.
$endgroup$
add a comment |
$begingroup$
For 1, you should display a matrix that is in $M$. It should be easy to find one.
For 2, they have removed the requirement that the eigenvalues of $A$ are rational. If you can find a matrix where the eigenvalues are not rational you show that this is not true. Can you find one?
For 3, finding one matrix is not enough. You are asked to show that if a matrix has integral entries and rational eigenvalues, the eigenvalues are integral. You could think about the rational root theorem and the characteristic polynomial of matrices in $M$.
4 is the same as 3. In fact it will follow from 3, if 3 is true.
$endgroup$
add a comment |
$begingroup$
For 1, you should display a matrix that is in $M$. It should be easy to find one.
For 2, they have removed the requirement that the eigenvalues of $A$ are rational. If you can find a matrix where the eigenvalues are not rational you show that this is not true. Can you find one?
For 3, finding one matrix is not enough. You are asked to show that if a matrix has integral entries and rational eigenvalues, the eigenvalues are integral. You could think about the rational root theorem and the characteristic polynomial of matrices in $M$.
4 is the same as 3. In fact it will follow from 3, if 3 is true.
$endgroup$
For 1, you should display a matrix that is in $M$. It should be easy to find one.
For 2, they have removed the requirement that the eigenvalues of $A$ are rational. If you can find a matrix where the eigenvalues are not rational you show that this is not true. Can you find one?
For 3, finding one matrix is not enough. You are asked to show that if a matrix has integral entries and rational eigenvalues, the eigenvalues are integral. You could think about the rational root theorem and the characteristic polynomial of matrices in $M$.
4 is the same as 3. In fact it will follow from 3, if 3 is true.
edited Dec 15 '18 at 4:11
answered Dec 15 '18 at 4:03
Ross MillikanRoss Millikan
297k23198371
297k23198371
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%2f3040150%2fgiven-set-m-which-of-following-option-is-correct%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