Orthogonal projection operator - squared of cosine

Question

I am reading a paper making use of orthogonal projection and I came across an expression $ \frac{z^TPz}{z^Tz} $ where P = $ S(S^TS)^{-1}S^T $ which is basically the orthogonal proejction operator. z is just a vector of variables. Then, since $P^2 = P$ for projection matrices, the equation is then rewritten as $ \frac{\lVert Pz \rVert ^2}{\lVert z \rVert ^2} $ which is $ cos^2\theta $.

Can someone explain to me why ? I am only able to do this.

$ z^TPz \\= z^TP^2 z \\ = z^TPPz$

So I don't get how that becomes $\lVert Pz \rVert ^2$. I know the dot product of 2 vectors is the cosine of the 2 vectors. I am not able to manipulate the expression to reach the squared of the cosine. However, incase someone still insists that he or she needs to know the problem before coming up with the solution, here is the paper. The expression can be found on page 100. Help thanks.

Answer 1

You also have $P=P^\top$, so $z^\top PPz=z^\top P^\top Pz=(Pz)^\top(Pz)$. Then apply the definition of the norm.

Answer 2

The dot product satisfies $$y^Tx=\|x\|\,\|y\|\,\cos\theta, $$ where $\theta $ is the angle between $x $ and $y $.

So $$\frac {\|Pz\|^2}{\|z\|^2}=\frac {z^TPz}{z^Tz}=\frac {\|z\|\,\|Pz\|\,\cos\theta}{\|z\|^2} $$ Simplifying, $$ \frac {\|Pz\|}{\|z\|}=\cos\theta. $$