$A$ 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













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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




















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