Can a 3x3 non-invertible matrix $A$ exist such that $A^5-A^3=I_3$?

$I_3$ is the 3x3 identity matrix.

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    $\begingroup$ $A^{-1}$ would simply be $A^4-A^2$ so no such matrix can exist. $\endgroup$
    – user801306
    Commented Jan 10, 2022 at 14:27
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    $\begingroup$ @MatthewH. Oh wow, thanks! I can't believe I didn't see that. Feel free to post as an answer and I will accept. $\endgroup$ Commented Jan 10, 2022 at 14:42
  • $\begingroup$ Please ask a good question, if you plan to answer it. That means abiding by How to ask a good question. Else it can and should be appropriately closed as a Problem Statement/ low-quality question, which it currently is. I say this mostly for any future questions you know the answer to, but post only a problem statement, and proceed to answer the poor posted question. In part, asking only a problem statement question with no context, gives other users the idea that it's okay. $\endgroup$
    – amWhy
    Commented Jan 10, 2022 at 20:21

1 Answer 1


I believe this solution is right, feel free to correct me if I'm wrong.

My solution:

For $A$ to be non-invertible, it must be the case that $det(A) = 0$.

Then, $det(A^5)$ and $det(A^3)$ must also be $0$ since $det(XY) = det(X) \cdot det(Y)$.

The equation $A^5 - A^3 = I_3$ can be factored as $A^3(A^2 - I_3) = I_3$.

Hence, $det(A^3(A^2-I_3)) = det(I_3) = 1$.

Therefore, $det(A^3) \cdot det(A^2 - I_3) = 1$.

This implies that $det(A^3) \neq 0$.

This contradicts the statement that $det(A^3) = 0$.

Hence, such a matrix cannot exist.


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