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What we know about projection matrices:

$A^T=A $

$A=A^2 $

I want to prove that $(I-A)^T $is also projection matrix :

My solution:

$(I-A)^T =I-A^T=I-A=I-A^2$

we can see that $ I-A=I-A^2 => 0=A-A^2$, $A^2=A$ then $A-A=0$

Does it look correct?

Thanks for help!

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  • $\begingroup$ You haven't proven that $(I-A)^2=I-A$. $\endgroup$ – WoolierThanThou Nov 21 at 15:01
  • $\begingroup$ Note that $(I-A)^T = I^T-A^T = I - A$. $\endgroup$ – Dustan Levenstein Nov 21 at 15:08
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Hint:

You need to prove that $$\left((I-A)^T\right)^T = (I-A)^T, \quad \left((I-A)^T\right)^2 = (I-A)^T.$$

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  • $\begingroup$ $((I-A)^T)^T =(I^T-A^T)^T=(I-A)^T$ Maybe it is ok $\endgroup$ – BlackSwan_22 Nov 21 at 15:14
  • $\begingroup$ @BlackSwan_22 That's it. Now the other one. $\endgroup$ – mechanodroid Nov 21 at 15:17
  • $\begingroup$ $(I−A^T)^2=(I−A)^T(I−A)^T=((I−A)(I−A))^T=(I−2A+A^2)^T=(I−A)^T$ $\endgroup$ – BlackSwan_22 Nov 21 at 15:20
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    $\begingroup$ @BlackSwan_22 The first one should be $\left((I-A)^T\right)^2$ but yes. $\endgroup$ – mechanodroid Nov 21 at 15:22
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I don't understand what you mean by "we can see that $ I-A=I-A^2 => 0=A-A^2$ $A^2=A$ then $A-A=0$".

However

$$\begin{aligned} \left(I - A^T\right)^2 & = (I-A)^T(I-A)^T\\ &=\left((I-A)(I-A)\right)^T\\ &= (I-2A + A^2)^T\\ &=(I-A)^T \end{aligned}$$

So $(I-A)^T$ which satisfies the two required identities to be a projection is indeed a projection.

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