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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?

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marked as duplicate by Steven Stadnicki, Community Jul 23 '15 at 0:16

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    $\begingroup$ Rotate by 90 degrees about some fixed axis? $\endgroup$ – lulu Jul 22 '15 at 23:59
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    $\begingroup$ Where are the entries from? Real? Complex? $\endgroup$ – Gary. Jul 23 '15 at 0:03
  • $\begingroup$ @Gary: Real entries. $\endgroup$ – Grace Jul 23 '15 at 0:15
  • $\begingroup$ @Grace: I gave it a try. Please let me know if you liked the answer. $\endgroup$ – Gary. Jul 23 '15 at 0:59
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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 $(A+I)(A-I)$

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  • $\begingroup$ Thanks for the hint. I tried plugging in the value for $I$ and a variable value for $A$, but it became extremely messy, with many different sets of equations. Is there a simpler way to solve this that you would suggest? $\endgroup$ – Grace Jul 23 '15 at 22:32
  • $\begingroup$ Please give me some time and I will look into it. $\endgroup$ – Gary. Jul 24 '15 at 0:50
  • $\begingroup$ Definitely! Thank you. $\endgroup$ – Grace Jul 24 '15 at 1:09
  • $\begingroup$ Actually, I am sorry to have to tell you that there is something wrong in my aswer: there are matrices that satisfy $A^4=I$ that do not satisfy $A^2=I$ nor $A^2 = -I$; specifically, take the diagonal matrix ( -1,-1,1), which represents a rotation by $\pi/2$ about the $x -axis $. I will continue trying to figure it out; you may deselect my answer as the correct one. I suspect the issue is that there are factorizations for $A^4-I$ other than $(A+I)(A-I)$. $\endgroup$ – Gary. Jul 24 '15 at 1:56

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