(adsbygoogle = window.adsbygoogle || []).push({}); *new question, playing with projection matrix.

Well I have a new question about the A(A1. The problem statement, all variables and given/known data

This is the first problem of our practice exam.

http://puu.sh/1jReL [Broken]

And here's the solution.

http://puu.sh/1jRg3 [Broken]

2. Relevant 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.

3. 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..^{T}A)^{-1}A^{T}matrix.

http://puu.sh/1jXT6 [Broken]

I was able to show that B^{T}was idempotent, but my manipulation was a bit different from the teacher for B^{2}

Let B^{2}= (A(A^{T}A)^{-1}A^{T})(A(A^{T}A)^{-1}A^{T})

I did (AA^{-1}(A^{T})^{-1}A^{T})A(A^{T}A)^{-1}A^{T}= IA(A^{T}A)^{-1}A^{T}=A(A^{T}A)^{-1}A^{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.

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# Homework Help: Linear Algebra (Find A^43)

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