Can a Direct Proof Show That 1-A is its Own Inverse?

[torquerotates]

So my teacher said that when proving something, I can't start out with what I'm trying to prove. But what if it is an "if this than that proof"

For example,

If A(squared)=A, then I-A=(I-A)inverse

Well, I started using what I'm trying to prove by multiplying both sides by I-A

I get (I-A)squared=I
implies I-4A+4AA=I
implies I-4A+4A=I b/c A(squared)=A using the hypothesis
implies I+0=I
implies I=I both sides equal

The thing is that I have proven that if AA=A, then I-A=(I-A)inverse by using the hypothesis somewhere in the solution. Would this be a logical conclusion?

 

[LukeD]

No, everything that you have just written is complete nonsense. First of all, (I-A)^2 = I - 2A + A^2, so your proof that I=I is flawed anyway.

What you have tried to do is to take some statement and to use it to derive a true statement. However, I can easily derive a true statement from a false one. For instance, you have assumed that (I-A)^2 = I-4A+4A^2 and that A^2 = A. These two statements are not in general true (This is in fact only true if A = 0). Then you have used these statements to derive that I=I, which is true, but meaningless.

 

[mathwonk]

if every step in your proof is correct and reversable, then after geting to a true stTEMENT, just reverse field and reason bCKWrds to the desired statement.

 

[mathwonk]

here i would try proving the statement directly, i.e. ask whether indeed 1-A is its own inverse by squaring it and seeing if you get 1. along the way you get to replace A^2 by A.

i.e. (1-A)^2 = 1-2A + A^2 = 1-2A+A = 1-A. this does not seem to prove what you asked for.

this seems to prove that if you start from a projection operator and subtract it from 1, you get another projection operator.