$A$ is an invertible $ntimes n$ matrix, where $n$ is an even number. Given that $A^3+A=0$, calculate...
$A$ 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
edited Dec 23 at 18:26
Asaf Karagila♦
301k32423755
asked Dec 23 at 13:42
Omer
2768
add a comment |
$A$ 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
edited Dec 23 at 18:26
Asaf Karagila♦
301k32423755
asked Dec 23 at 13:42
Omer
2768
1
This is correct.
– Aniruddh Agarwal
Dec 23 at 13:53
1
You're right for me. Probably the author thought of another way, in which $n$ being even is used (one may even easily prove $det A=pm1$).
– Bernard
Dec 23 at 13:53
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.
– egreg
Dec 23 at 15:29
add a comment |
$A$ 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
edited Dec 23 at 18:26
Asaf Karagila♦
301k32423755
asked Dec 23 at 13:42
Omer
2768
$A$ 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
edited Dec 23 at 18:26
Asaf Karagila♦
301k32423755
asked Dec 23 at 13:42
Omer
2768
edited Dec 23 at 18:26
Asaf Karagila♦
301k32423755
asked Dec 23 at 13:42
Omer
2768
edited Dec 23 at 18:26
Asaf Karagila♦
301k32423755
edited Dec 23 at 18:26
Asaf Karagila♦
301k32423755
edited Dec 23 at 18:26
Asaf Karagila♦
301k32423755
301k32423755
asked Dec 23 at 13:42
Omer
2768
asked Dec 23 at 13:42
Omer
2768
asked Dec 23 at 13:42
Omer
2768
2768
1
This is correct.
– Aniruddh Agarwal
Dec 23 at 13:53
1
You're right for me. Probably the author thought of another way, in which $n$ being even is used (one may even easily prove $det A=pm1$).
– Bernard
Dec 23 at 13:53
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.
– egreg
Dec 23 at 15:29
add a comment |
1
This is correct.
– Aniruddh Agarwal
Dec 23 at 13:53
1
You're right for me. Probably the author thought of another way, in which $n$ being even is used (one may even easily prove $det A=pm1$).
– Bernard
Dec 23 at 13:53
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.
– egreg
Dec 23 at 15:29
1
1
This is correct.
– Aniruddh Agarwal
Dec 23 at 13:53
This is correct.
– Aniruddh Agarwal
Dec 23 at 13:53
1
1
You're right for me. Probably the author thought of another way, in which $n$ being even is used (one may even easily prove $det A=pm1$).
– Bernard
Dec 23 at 13:53
You're right for me. Probably the author thought of another way, in which $n$ being even is used (one may even easily prove $det A=pm1$).
– Bernard
Dec 23 at 13:53
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.
– egreg
Dec 23 at 15:29
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.
– egreg
Dec 23 at 15:29
add a comment |
1 Answer
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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.
answered Dec 23 at 14:19
Mark Heavey
18614
New contributor
Mark Heavey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
– Sambo
Dec 23 at 14:38
1
@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.
– Mark Heavey
Dec 23 at 14:41
What is In and what why it's determinant is (-1)^n
– user4951
Dec 23 at 16:48
1
@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., $(-1)^n$.
– Eric Towers
Dec 23 at 17:32
add a comment |
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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.
answered Dec 23 at 14:19
Mark Heavey
18614
New contributor
Mark Heavey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
– Sambo
Dec 23 at 14:38
1
@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.
– Mark Heavey
Dec 23 at 14:41
What is In and what why it's determinant is (-1)^n
– user4951
Dec 23 at 16:48
1
@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., $(-1)^n$.
– Eric Towers
Dec 23 at 17:32
add a comment |
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.
answered Dec 23 at 14:19
Mark Heavey
18614
New contributor
Mark Heavey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
– Sambo
Dec 23 at 14:38
1
@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.
– Mark Heavey
Dec 23 at 14:41
What is In and what why it's determinant is (-1)^n
– user4951
Dec 23 at 16:48
1
@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., $(-1)^n$.
– Eric Towers
Dec 23 at 17:32
add a comment |
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.
answered Dec 23 at 14:19
Mark Heavey
18614
New contributor
Mark Heavey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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.
answered Dec 23 at 14:19
Mark Heavey
18614
New contributor
Mark Heavey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 23 at 14:19
Mark Heavey
18614
New contributor
Mark Heavey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 23 at 14:19
Mark Heavey
18614
answered Dec 23 at 14:19
Mark Heavey
18614
18614
New contributor
Mark Heavey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mark Heavey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mark Heavey is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
– Sambo
Dec 23 at 14:38
1
@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.
– Mark Heavey
Dec 23 at 14:41
What is In and what why it's determinant is (-1)^n
– user4951
Dec 23 at 16:48
1
@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., $(-1)^n$.
– Eric Towers
Dec 23 at 17:32
add a comment |
In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
– Sambo
Dec 23 at 14:38
1
@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.
– Mark Heavey
Dec 23 at 14:41
What is In and what why it's determinant is (-1)^n
– user4951
Dec 23 at 16:48
1
@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., $(-1)^n$.
– Eric Towers
Dec 23 at 17:32
In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
– Sambo
Dec 23 at 14:38
In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
– Sambo
Dec 23 at 14:38
1
1
@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.
– Mark Heavey
Dec 23 at 14:41
@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.
– Mark Heavey
Dec 23 at 14:41
What is In and what why it's determinant is (-1)^n
– user4951
Dec 23 at 16:48
What is In and what why it's determinant is (-1)^n
– user4951
Dec 23 at 16:48
1
1
@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., $(-1)^n$.
– Eric Towers
Dec 23 at 17:32
@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., $(-1)^n$.
– Eric Towers
Dec 23 at 17:32
add a comment |
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1
This is correct.
– Aniruddh Agarwal
Dec 23 at 13:53
1
You're right for me. Probably the author thought of another way, in which $n$ being even is used (one may even easily prove $det A=pm1$).
– Bernard
Dec 23 at 13:53
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.
– egreg
Dec 23 at 15:29