Find A such that $A^2 \neq I$ but $A^4 = I$

Question

Find a $3 \times 3$ matrix A such that $A^2 \neq I$ but $A^4 = I$, where $I$ is the $3 \times 3$ identity matrix.

Is there a simpler way to solve this problem rather than bashing it out by substituting a variable matrix value for A?

Answer 1

If your entries are complex, take a diagonal matrix with all entries equal to the fourth roots of unity.

In general, I guess you want to have $A^4 -I =0$ , but not the factorization:

$(A^2+I)(A^2 -I )=0$ with either factor being zero.

So we could, in theory, have zero-divisors here, i.e., we can have the

product be zero without either factor being zero. But this is not

possible with entries over $\mathbb R$ , since $\mathbb R$ is a field.

Since you want to avoid $A^2=I$ , your

only option is having matrices $A$ with $A^2+I=0$ , which greatly simplifies

the problem.

EDIT : there I something wrong in my answer that I am trying to understand.

Specifically, there are matrices $A$ with $A^4=I$ , but that satisfy neither

$A^2=I$ , nor $A^2 =-I$ . Specifically, the diagonal matrix $D(-1,-1,1)$.

I suspect that there are factorizations for $A^4-I$ other than $(A+I)(A-I)$