$A$ is an invertible $ntimes n$ matrix, where $n$ is an even number. Given that $A^3+A=0$, calculate...












14












$begingroup$


$A$ is an invertible matrix with $n$ columns and $n$ rows, where $n$ is an even number. We are given that $A^3+A=0$ and we need to calculate $det(A^4)$. Here is my solution:
$$A^3+A=0 implies A^{-1}(A^3+A)=0 implies A^2=-I implies A^4=I implies det(A^4)=1.$$ But I did not use the fact that n is even. Am I wrong, or this is not needed? If I'm wrong, please don't tell me the solution yet. Just tell me where I'm wrong. Thanks!






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is correct.
    $endgroup$
    – Aniruddh Agarwal
    Dec 23 '18 at 13:53






  • 1




    $begingroup$
    You're right for me. Probably the author thought of another way, in which $n$ being even is used (one may even easily prove $det A=pm1$).
    $endgroup$
    – Bernard
    Dec 23 '18 at 13:53










  • $begingroup$
    We have $A^4=-A^2$, so, if $n$ is even, $(det A)^4=(det A)^2$, hence $(det A)^2=1$ and $(det A)^4=det(A^4)=1$. Without knowing that $n$ is even, this argument would not work; but yours is definitely better.
    $endgroup$
    – egreg
    Dec 23 '18 at 15:29


















14












$begingroup$


$A$ is an invertible matrix with $n$ columns and $n$ rows, where $n$ is an even number. We are given that $A^3+A=0$ and we need to calculate $det(A^4)$. Here is my solution:
$$A^3+A=0 implies A^{-1}(A^3+A)=0 implies A^2=-I implies A^4=I implies det(A^4)=1.$$ But I did not use the fact that n is even. Am I wrong, or this is not needed? If I'm wrong, please don't tell me the solution yet. Just tell me where I'm wrong. Thanks!






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is correct.
    $endgroup$
    – Aniruddh Agarwal
    Dec 23 '18 at 13:53






  • 1




    $begingroup$
    You're right for me. Probably the author thought of another way, in which $n$ being even is used (one may even easily prove $det A=pm1$).
    $endgroup$
    – Bernard
    Dec 23 '18 at 13:53










  • $begingroup$
    We have $A^4=-A^2$, so, if $n$ is even, $(det A)^4=(det A)^2$, hence $(det A)^2=1$ and $(det A)^4=det(A^4)=1$. Without knowing that $n$ is even, this argument would not work; but yours is definitely better.
    $endgroup$
    – egreg
    Dec 23 '18 at 15:29
















14












14








14


1



$begingroup$


$A$ is an invertible matrix with $n$ columns and $n$ rows, where $n$ is an even number. We are given that $A^3+A=0$ and we need to calculate $det(A^4)$. Here is my solution:
$$A^3+A=0 implies A^{-1}(A^3+A)=0 implies A^2=-I implies A^4=I implies det(A^4)=1.$$ But I did not use the fact that n is even. Am I wrong, or this is not needed? If I'm wrong, please don't tell me the solution yet. Just tell me where I'm wrong. Thanks!






linear-algebra







share|cite|improve this question











$endgroup$




$A$ is an invertible matrix with $n$ columns and $n$ rows, where $n$ is an even number. We are given that $A^3+A=0$ and we need to calculate $det(A^4)$. Here is my solution:
$$A^3+A=0 implies A^{-1}(A^3+A)=0 implies A^2=-I implies A^4=I implies det(A^4)=1.$$ But I did not use the fact that n is even. Am I wrong, or this is not needed? If I'm wrong, please don't tell me the solution yet. Just tell me where I'm wrong. Thanks!






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 18:26









Asaf Karagila♦

303k32429762




303k32429762










asked Dec 23 '18 at 13:42









OmerOmer

3618




3618








  • 1




    $begingroup$
    This is correct.
    $endgroup$
    – Aniruddh Agarwal
    Dec 23 '18 at 13:53






  • 1




    $begingroup$
    You're right for me. Probably the author thought of another way, in which $n$ being even is used (one may even easily prove $det A=pm1$).
    $endgroup$
    – Bernard
    Dec 23 '18 at 13:53










  • $begingroup$
    We have $A^4=-A^2$, so, if $n$ is even, $(det A)^4=(det A)^2$, hence $(det A)^2=1$ and $(det A)^4=det(A^4)=1$. Without knowing that $n$ is even, this argument would not work; but yours is definitely better.
    $endgroup$
    – egreg
    Dec 23 '18 at 15:29
















  • 1




    $begingroup$
    This is correct.
    $endgroup$
    – Aniruddh Agarwal
    Dec 23 '18 at 13:53






  • 1




    $begingroup$
    You're right for me. Probably the author thought of another way, in which $n$ being even is used (one may even easily prove $det A=pm1$).
    $endgroup$
    – Bernard
    Dec 23 '18 at 13:53










  • $begingroup$
    We have $A^4=-A^2$, so, if $n$ is even, $(det A)^4=(det A)^2$, hence $(det A)^2=1$ and $(det A)^4=det(A^4)=1$. Without knowing that $n$ is even, this argument would not work; but yours is definitely better.
    $endgroup$
    – egreg
    Dec 23 '18 at 15:29










1




1




$begingroup$
This is correct.
$endgroup$
– Aniruddh Agarwal
Dec 23 '18 at 13:53




$begingroup$
This is correct.
$endgroup$
– Aniruddh Agarwal
Dec 23 '18 at 13:53




1




1




$begingroup$
You're right for me. Probably the author thought of another way, in which $n$ being even is used (one may even easily prove $det A=pm1$).
$endgroup$
– Bernard
Dec 23 '18 at 13:53




$begingroup$
You're right for me. Probably the author thought of another way, in which $n$ being even is used (one may even easily prove $det A=pm1$).
$endgroup$
– Bernard
Dec 23 '18 at 13:53












$begingroup$
We have $A^4=-A^2$, so, if $n$ is even, $(det A)^4=(det A)^2$, hence $(det A)^2=1$ and $(det A)^4=det(A^4)=1$. Without knowing that $n$ is even, this argument would not work; but yours is definitely better.
$endgroup$
– egreg
Dec 23 '18 at 15:29






$begingroup$
We have $A^4=-A^2$, so, if $n$ is even, $(det A)^4=(det A)^2$, hence $(det A)^2=1$ and $(det A)^4=det(A^4)=1$. Without knowing that $n$ is even, this argument would not work; but yours is definitely better.
$endgroup$
– egreg
Dec 23 '18 at 15:29












1 Answer
1






active

oldest

votes


















18












$begingroup$

It's not possible for $n$ to be odd, which can be extracted from your solution:



$ A^2 = -I_n ; Rightarrow ; det (A^2) = ( det A )^2 = det (-I_n) = (-1)^n$.
But $A$ is invertible, which means that $( det A )^2 > 0$ which means that $(-1)^n > 0$ which only occurs when $n$ is even.



So it was not necessary in your solution, but rather the question breaks down if $n$ is odd.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
    $endgroup$
    – Sambo
    Dec 23 '18 at 14:38








  • 1




    $begingroup$
    @Sambo Sort of; it's impossible for $n$ to be odd, $A$ invertible, and $A^3 + A = 0$ all at once. Remove any one of those conditions and there is no inconsistency.
    $endgroup$
    – Mark Heavey
    Dec 23 '18 at 14:41










  • $begingroup$
    What is In and what why it's determinant is (-1)^n
    $endgroup$
    – user4951
    Dec 23 '18 at 16:48






  • 1




    $begingroup$
    @user4951 : $I_n$ is the $n times n$ identity matrix. No one claims that $det I_n = (-1)^n$. The claim is about $det(-I_n)$, which is a matrix with $0$s in every cell except the diagonal, on which there are $n$ copies of $-1$, so the determinant (by minors, say) simplifies to the product of those $n$ copies of $-1$, i.e., $(-1)^n$.
    $endgroup$
    – Eric Towers
    Dec 23 '18 at 17:32













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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050353%2fa-is-an-invertible-n-times-n-matrix-where-n-is-an-even-number-given-that%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









18












$begingroup$

It's not possible for $n$ to be odd, which can be extracted from your solution:



$ A^2 = -I_n ; Rightarrow ; det (A^2) = ( det A )^2 = det (-I_n) = (-1)^n$.
But $A$ is invertible, which means that $( det A )^2 > 0$ which means that $(-1)^n > 0$ which only occurs when $n$ is even.



So it was not necessary in your solution, but rather the question breaks down if $n$ is odd.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
    $endgroup$
    – Sambo
    Dec 23 '18 at 14:38








  • 1




    $begingroup$
    @Sambo Sort of; it's impossible for $n$ to be odd, $A$ invertible, and $A^3 + A = 0$ all at once. Remove any one of those conditions and there is no inconsistency.
    $endgroup$
    – Mark Heavey
    Dec 23 '18 at 14:41










  • $begingroup$
    What is In and what why it's determinant is (-1)^n
    $endgroup$
    – user4951
    Dec 23 '18 at 16:48






  • 1




    $begingroup$
    @user4951 : $I_n$ is the $n times n$ identity matrix. No one claims that $det I_n = (-1)^n$. The claim is about $det(-I_n)$, which is a matrix with $0$s in every cell except the diagonal, on which there are $n$ copies of $-1$, so the determinant (by minors, say) simplifies to the product of those $n$ copies of $-1$, i.e., $(-1)^n$.
    $endgroup$
    – Eric Towers
    Dec 23 '18 at 17:32


















18












$begingroup$

It's not possible for $n$ to be odd, which can be extracted from your solution:



$ A^2 = -I_n ; Rightarrow ; det (A^2) = ( det A )^2 = det (-I_n) = (-1)^n$.
But $A$ is invertible, which means that $( det A )^2 > 0$ which means that $(-1)^n > 0$ which only occurs when $n$ is even.



So it was not necessary in your solution, but rather the question breaks down if $n$ is odd.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
    $endgroup$
    – Sambo
    Dec 23 '18 at 14:38








  • 1




    $begingroup$
    @Sambo Sort of; it's impossible for $n$ to be odd, $A$ invertible, and $A^3 + A = 0$ all at once. Remove any one of those conditions and there is no inconsistency.
    $endgroup$
    – Mark Heavey
    Dec 23 '18 at 14:41










  • $begingroup$
    What is In and what why it's determinant is (-1)^n
    $endgroup$
    – user4951
    Dec 23 '18 at 16:48






  • 1




    $begingroup$
    @user4951 : $I_n$ is the $n times n$ identity matrix. No one claims that $det I_n = (-1)^n$. The claim is about $det(-I_n)$, which is a matrix with $0$s in every cell except the diagonal, on which there are $n$ copies of $-1$, so the determinant (by minors, say) simplifies to the product of those $n$ copies of $-1$, i.e., $(-1)^n$.
    $endgroup$
    – Eric Towers
    Dec 23 '18 at 17:32
















18












18








18





$begingroup$

It's not possible for $n$ to be odd, which can be extracted from your solution:



$ A^2 = -I_n ; Rightarrow ; det (A^2) = ( det A )^2 = det (-I_n) = (-1)^n$.
But $A$ is invertible, which means that $( det A )^2 > 0$ which means that $(-1)^n > 0$ which only occurs when $n$ is even.



So it was not necessary in your solution, but rather the question breaks down if $n$ is odd.






share|cite|improve this answer









$endgroup$



It's not possible for $n$ to be odd, which can be extracted from your solution:



$ A^2 = -I_n ; Rightarrow ; det (A^2) = ( det A )^2 = det (-I_n) = (-1)^n$.
But $A$ is invertible, which means that $( det A )^2 > 0$ which means that $(-1)^n > 0$ which only occurs when $n$ is even.



So it was not necessary in your solution, but rather the question breaks down if $n$ is odd.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 23 '18 at 14:19









Mark HeaveyMark Heavey

19614




19614












  • $begingroup$
    In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
    $endgroup$
    – Sambo
    Dec 23 '18 at 14:38








  • 1




    $begingroup$
    @Sambo Sort of; it's impossible for $n$ to be odd, $A$ invertible, and $A^3 + A = 0$ all at once. Remove any one of those conditions and there is no inconsistency.
    $endgroup$
    – Mark Heavey
    Dec 23 '18 at 14:41










  • $begingroup$
    What is In and what why it's determinant is (-1)^n
    $endgroup$
    – user4951
    Dec 23 '18 at 16:48






  • 1




    $begingroup$
    @user4951 : $I_n$ is the $n times n$ identity matrix. No one claims that $det I_n = (-1)^n$. The claim is about $det(-I_n)$, which is a matrix with $0$s in every cell except the diagonal, on which there are $n$ copies of $-1$, so the determinant (by minors, say) simplifies to the product of those $n$ copies of $-1$, i.e., $(-1)^n$.
    $endgroup$
    – Eric Towers
    Dec 23 '18 at 17:32




















  • $begingroup$
    In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
    $endgroup$
    – Sambo
    Dec 23 '18 at 14:38








  • 1




    $begingroup$
    @Sambo Sort of; it's impossible for $n$ to be odd, $A$ invertible, and $A^3 + A = 0$ all at once. Remove any one of those conditions and there is no inconsistency.
    $endgroup$
    – Mark Heavey
    Dec 23 '18 at 14:41










  • $begingroup$
    What is In and what why it's determinant is (-1)^n
    $endgroup$
    – user4951
    Dec 23 '18 at 16:48






  • 1




    $begingroup$
    @user4951 : $I_n$ is the $n times n$ identity matrix. No one claims that $det I_n = (-1)^n$. The claim is about $det(-I_n)$, which is a matrix with $0$s in every cell except the diagonal, on which there are $n$ copies of $-1$, so the determinant (by minors, say) simplifies to the product of those $n$ copies of $-1$, i.e., $(-1)^n$.
    $endgroup$
    – Eric Towers
    Dec 23 '18 at 17:32


















$begingroup$
In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
$endgroup$
– Sambo
Dec 23 '18 at 14:38






$begingroup$
In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
$endgroup$
– Sambo
Dec 23 '18 at 14:38






1




1




$begingroup$
@Sambo Sort of; it's impossible for $n$ to be odd, $A$ invertible, and $A^3 + A = 0$ all at once. Remove any one of those conditions and there is no inconsistency.
$endgroup$
– Mark Heavey
Dec 23 '18 at 14:41




$begingroup$
@Sambo Sort of; it's impossible for $n$ to be odd, $A$ invertible, and $A^3 + A = 0$ all at once. Remove any one of those conditions and there is no inconsistency.
$endgroup$
– Mark Heavey
Dec 23 '18 at 14:41












$begingroup$
What is In and what why it's determinant is (-1)^n
$endgroup$
– user4951
Dec 23 '18 at 16:48




$begingroup$
What is In and what why it's determinant is (-1)^n
$endgroup$
– user4951
Dec 23 '18 at 16:48




1




1




$begingroup$
@user4951 : $I_n$ is the $n times n$ identity matrix. No one claims that $det I_n = (-1)^n$. The claim is about $det(-I_n)$, which is a matrix with $0$s in every cell except the diagonal, on which there are $n$ copies of $-1$, so the determinant (by minors, say) simplifies to the product of those $n$ copies of $-1$, i.e., $(-1)^n$.
$endgroup$
– Eric Towers
Dec 23 '18 at 17:32






$begingroup$
@user4951 : $I_n$ is the $n times n$ identity matrix. No one claims that $det I_n = (-1)^n$. The claim is about $det(-I_n)$, which is a matrix with $0$s in every cell except the diagonal, on which there are $n$ copies of $-1$, so the determinant (by minors, say) simplifies to the product of those $n$ copies of $-1$, i.e., $(-1)^n$.
$endgroup$
– Eric Towers
Dec 23 '18 at 17:32




















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050353%2fa-is-an-invertible-n-times-n-matrix-where-n-is-an-even-number-given-that%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more