Am I wrong, or there is too much information given?












3














$A$ 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




















  • 1




    This is correct.
    – Aniruddh Agarwal
    1 hour ago






  • 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
    1 hour ago
















3














$A$ 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




















  • 1




    This is correct.
    – Aniruddh Agarwal
    1 hour ago






  • 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
    1 hour ago














3












3








3







$A$ 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















$A$ 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 49 mins ago









Aniruddh Agarwal

1098




1098










asked 1 hour ago









Omer

2217




2217








  • 1




    This is correct.
    – Aniruddh Agarwal
    1 hour ago






  • 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
    1 hour ago














  • 1




    This is correct.
    – Aniruddh Agarwal
    1 hour ago






  • 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
    1 hour ago








1




1




This is correct.
– Aniruddh Agarwal
1 hour ago




This is correct.
– Aniruddh Agarwal
1 hour ago




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
1 hour ago




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
1 hour ago










1 Answer
1






active

oldest

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5














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








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
    17 mins ago












  • @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
    14 mins ago











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5














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








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
    17 mins ago












  • @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
    14 mins ago
















5














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








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
    17 mins ago












  • @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
    14 mins ago














5












5








5






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








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.







share|cite|improve this answer








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.









share|cite|improve this answer



share|cite|improve this answer






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 36 mins ago









Mark Heavey

663




663




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
    17 mins ago












  • @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
    14 mins ago


















  • In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
    – Sambo
    17 mins ago












  • @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
    14 mins ago
















In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
– Sambo
17 mins ago






In other words, if $n$ is odd and $A$ is invertible, it's impossible to have $A^3 + A = 0$, correct?
– Sambo
17 mins ago














@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
14 mins ago




@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
14 mins ago


















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