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So the question is that when a square matrix $P$ fulfills the equality that $P^2 = P$

Prove that $P^n = P$ and $(I-P)^n = I -P$

So for first $P^n = P$

I tried to prove by

$P^n = P^{n-2}P^2$

So given $P^2 = P$

$P^n = P^{n-2}P^2 = P^{n-2}P = P^{n-1}$

By same way it we can deduce

$P^{n-1} = P^{n-2} = P^{n-3} = ... = P^2 = P$

or my second attempt is that as $P^2 = P$

$P^{-1}PP = P^{-1}P$

$P = I$

Hence that

$P^n = I^n = I = P$

Which do you think is more suitable as proving???

And for second question now that we know $P = I$

For $(I-P)^n$,

I thought that $I - P = 0$, a zero matrix

and hence that $0^n = 0$

Is my proving, incomplete or not logical?? Please give feedback

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    $\begingroup$ In general, $P$ is a singular matrix, which means that its inverse $P^{-1}$ does not exist. Use a) to prove b) $\endgroup$
    – Jochen
    Commented Apr 23, 2022 at 8:52
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    $\begingroup$ For the second eqaulity, note that $(I-P)^2=I^2-2P+P^2=I-P$. $\endgroup$ Commented Apr 23, 2022 at 8:54
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    $\begingroup$ Do you really think $P=I$ is correct? Note that the zer0 matrix also satisfies the hypothesis. $\endgroup$ Commented Apr 23, 2022 at 8:56
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    $\begingroup$ hints : For the first one, use induction on $n$. For the second one, use the binomial theorem on $(I-P)^n$. $\endgroup$
    – David Lui
    Commented Apr 23, 2022 at 8:58
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    $\begingroup$ Yes, that's a good idea. You're right, if $P$ is regular, then $P=I$. $\endgroup$
    – Jochen
    Commented Apr 23, 2022 at 9:00

1 Answer 1

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Given that $P^2 = P$.

To prove that $$ P^n = P, \ \ \ (n \in \mathbf{N}) $$ we can use the principle of induction.

Clearly, $P^n = P$ for $n = 1$.

Assume that $P^m = P$ for some positive integer $m$.

Then $$ P^{m + 1} = P^m P = P P = P^2 = P $$

This completes the induction.

Hence, $$ P^n = P, \ \ \mbox{for all} \ \ n \in \mathbf{N} $$

Next, when $n = 1$, $$ (I - P)^n = (I - P)^1 = I - P $$

Assume that $$ (I - P)^m = I - P $$ for some positive integer $m$.

Then we find that $$ (I - P)^{m + 1} = (I - P)^m (I - P) = (I - P) (I - P) = I - 2 P + P^2 $$

Since $P^2 = P$, we get $$ (I - P)^{m + 1} = I - 2 P + P = I - P $$

This completes the induction.

Hence, we conclude that $$ (I - P)^n = (I - P) \ \ \mbox{for all} \ \ n \in \mathbf{N} $$

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  • $\begingroup$ There's no need to repeat the same inductive proof for the second case. Instead. simply verify that $\,(I-P)^2 = I-P,\,$ so the claim follows from the first proof, by replacing $\,P\,$ by $\,I-P.\ \ $ $\endgroup$ Commented May 12, 2022 at 16:18

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