# Is this projection matrix?

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!

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

Hint:

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

• $((I-A)^T)^T =(I^T-A^T)^T=(I-A)^T$ Maybe it is ok – BlackSwan_22 Nov 21 at 15:14
• @BlackSwan_22 That's it. Now the other one. – mechanodroid Nov 21 at 15:17
• $(I−A^T)^2=(I−A)^T(I−A)^T=((I−A)(I−A))^T=(I−2A+A^2)^T=(I−A)^T$ – BlackSwan_22 Nov 21 at 15:20
• @BlackSwan_22 The first one should be $\left((I-A)^T\right)^2$ but yes. – mechanodroid Nov 21 at 15:22

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.