Playing with projection matrices: How did A^3 become the identity matrix?

[Lifprasir]

*new question, playing with projection matrix.

Homework Statement

This is the first problem of our practice exam.
http://puu.sh/1jReL

And here's the solution.
http://puu.sh/1jRg3

Homework Equations

The standard matrix has to be 2x2 to be compatible u v w, so this must be the reason why he limited the matrix to A (u v) = (v w) instead of A (u v w) = (v w u).

And then he used matrix properties to solve for A, and got it.

However what I don't understand is the part where it says "Observe that A^3u = u, A^3 u = v.

The Attempt at a Solution

I've tried doing,

(u v) = D
(v w) = X

so, AD = X
A = XD^-1

Then I tried to reverse the observation by doing,

A^3u = u
A^3 = I
(XD^-1)(XD^-1)(XD^-1)=I, but that's as far as I'm able to go.

I really don't understand how he was able to see how A^3 was I..

Well I have a new question about the A(A^TA)^-1A^T matrix.

http://puu.sh/1jXT6

I was able to show that B^T was idempotent, but my manipulation was a bit different from the teacher for B²

Let B² = (A(A^TA)^-1A^T)(A(A^TA)^-1A^T)

I did (AA^-1(A^T)^-1A^T)A(A^TA)^-1A^T = IA(A^TA)^-1A^T=A(A^TA)^-1A^T=B

The teacher did it in another method and I don't think my method is correct because it doesn't make sense that B = I.

 

[tiny-tim]

Hi Lifprasir!

(try using the X² button just above the Reply box )

However what I don't understand is the part where it says "Observe that A^3u = u, A^3 u = v.

Because S³(u) = u, S³(v) = v

 

[Lifprasir]

Ooooooooooooooooooooh. Thank you so much!