An involutory matrix $\mathbf{A}$ is such that $\mathbf{A}^{2}=\mathbf{I}$.

An idempotent matrix $\mathbf{A}$ is such that $\mathbf{A}^{2}=\mathbf{A}$

So we'd have to satisfy both conditions simultaneously.

I do know the answer to the question is the identity matrix itself. But how can we go about proving it?

One way I thought of is to expose that the diagonal of $\mathbf{A}^{2}$ and condition that all other entries must be zero, but then I would already be biased by knowing the answer beforehand.

It'd be really nice if you guys gave me a hand with this one. In the meantime I'll carry on trying.

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    $\begingroup$ $A^2=A$ and $A^2=I$. Doesn't this naturally mean $A=I=A^2$? $\endgroup$ – Rohit Pandey Mar 14 at 22:50
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    $\begingroup$ $A=A^2=I\implies A= I$ $\endgroup$ – saulspatz Mar 14 at 22:51
  • $\begingroup$ I think you meant involutory $\endgroup$ – J. W. Tanner Mar 14 at 23:05
  • $\begingroup$ Thanks for that @J.W.Tanner. Indeed, Rohit, it would. Yet I am trying to give a more detailed demonstration. Since the $\mathbf{A}^{2}$ yields $a_{ii}^{2}$ for $1 \leq i \leq n$ and that would have to be $1$. Now, as for $a_i * a_j$ for every $i \neq j$ the answer should be zero. Therefore mounting to an identity matrix. Yet it feels like I'd be missing something with such arguments. Hence the post. $\endgroup$ – upStoneLock Mar 14 at 23:20
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    $\begingroup$ @upStoneLock: You know that $A=A^{2}=I,$ so you're done. If you are looking for more than this, then it is unclear what you are asking. Could you clarify what more you want? $\endgroup$ – Will R Mar 15 at 0:15

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