1

This question Prove that there is no 5 × 5 matrix A such that $A^2 = −I$ Raises the questions:

For what values of n $n \times n$ matrix $A$ there is no $\mathrm {A}$ exists s.t. $\mathrm {A}^2=-\mathrm {I}$

or more generally

For what values of $\mathrm {B},p$, $n \times n$ matrix there is no $\mathrm {A}$ exists s.t. $A^p=\mathrm {B}$

jimjim
  • 9,675
  • 3
    You take the determinant and get $\det(A)^2 = (-1)^n$, impossible if $n$ is odd over the reals. – Pedro May 31 '18 at 21:51
  • @Arjang Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Jun 22 '18 at 21:17

2 Answers2

2

Since for A real $$\det(A^2)=[\det(A)]^2\ge0$$ such $A$ doesn't exist for all $n$ odd and $B$ such that $\det(B)<0$.

user
  • 154,566
  • 2
    This is half the proof, it leaves the case of $n$ even undetermined. – Alan May 31 '18 at 23:04
  • Indeed fon n even we can’t conclude anything in general, whereas for n odd we gave a criteria to exclude the existence of a solution . – user Jun 01 '18 at 07:04
2

Whenever $n$ is odd. For example, $det(A)^2=det(A^2)=(-1)^n$, so whenever $n$ is odd, you need a square root of $-1$.

Andres Mejia
  • 20,977