What is $\|\Phi x\|_2^2$ given that $\Phi\in R^{d\times d}$ is an $m$-dimensional orthogonal projection matrix uniformly drawn at random?

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The problem is extracted from Page 3 and Section 6.2 of Page 15 in the paper http://arxiv.org/pdf/1506.00898v2.pdf , and here I have listed all the information I can get through the whole paper.

The paper has shown an answer and related proof: To characterize $\|\Phi x\|_2^2$, by rotational invariance, we can think of $\Phi$ as fixed, and $x$ as a uniformly distributed unit vector. By setting $\Phi$ to project onto the first $m$ standard basis elements and using the given fact (Let $a\sim \chi_{m}^2$, $b\sim \chi_{d-m}^2$ be Chi-squared distributed random variables. Then, $\frac{a}{a+b}\sim \text{Beta}(\frac{m}{2}, \frac{d-m}{2})$ ), it is easy to see that $\|\Phi x\|_2^2\sim \omega$ where $\omega \sim \text{Beta}(\frac{m}{2}, \frac{d-m}{2})$.

Questions: can someone show some detailed procedure for the above proof? Since my difficult results from that the paper only offers a very brief proof and gives no detailed definition for $\Phi\in R^{d\times d}$, which is an $m$-dimensional orthogonal projection matrix uniformly drawn at random. I google http://www.mathworks.com/matlabcentral/newsreader/view_thread/298500 and find we can uniformly generate $d$-dimensional orthogonal matrix by Matlab code: [Q, R] = qr(randn(d)); Q = Q*diag(sign(diag(R))); where randn(d) means $d\times d$-dimensional standard Gaussian matrix with mean 0 variance 1.