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For my linear algebra homework, I need to prove that if $T$ is a diagonalizable linear map and $T^3=T^2$, then $T^2=T$. I tried proving that $T$ is invertible (so that I can multiply both sides of $T^3=T^2$ by $T^{-1}$, but apparently diagonalizability does not imply invertibility (see Can a matrix be invertible but not diagonalizable? in the comment section of the first answer).

Could you help me?

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    $\begingroup$ This is quite simple - all you have to do is write $T = SDS^{-1}$ from definition where $D$ is a diagonal matrix. The rest is easy $\endgroup$
    – Jakobian
    Aug 13, 2021 at 18:58
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    $\begingroup$ Does this answer your question? $A$ is diagonalizable and $A^3 = A^2$ – found with Approach0 $\endgroup$
    – Martin R
    Aug 13, 2021 at 19:08
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    $\begingroup$ Another approach is possible if you know the minimal polynomial for a diagonalizable matrix has no repeated roots… $\endgroup$
    – Thomas Andrews
    Aug 13, 2021 at 19:09

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If $T$ is diagonalizable, then you have a diagonal matrix $D$ and an invertible matrix $P$ such that $T=PDP^{-1}$.
Now you have $$T^3=T^2$$ $$PDP^{-1}PDP^{-1}PDP^{-1}=PDP^{-1}PDP^{-1}$$

$$PD^3P^{-1}=PD^2P^{-1}%$$ $$D^3=D^2$$

Now, you can't just apply $D^{-1}$ as that might not exist, but what is the formula entrancewise for $D^n$? from that you should be able to conclude what $D$ has to look like and finish

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