# Does a non-invertible matrix A exist where some power of A is the identity matrix? [closed]

##### Question:

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

$$I_3$$ is the 3x3 identity matrix.

• $A^{-1}$ would simply be $A^4-A^2$ so no such matrix can exist.
– user801306
Commented Jan 10, 2022 at 14:27
• @MatthewH. Oh wow, thanks! I can't believe I didn't see that. Feel free to post as an answer and I will accept. Commented Jan 10, 2022 at 14:42
• 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. Commented Jan 10, 2022 at 20:21

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.