How to calculate Hat matrix for penalized spline regressions? The book "Semiparametric Regression" by Ruppert et al. (2003) provided a computationally fast algorithm for Penalized Spline Regression. I put a part of the algorithm here. Does anybody can do algebra to prove equation that is mentioned on step 4? Any help would be much appreciated.

 A: I get an ever so slightly different formula, but it only differs in where a transpose is placed.
Let's start with the expression $$C^t C + \alpha D $$ and use the hypothesized decomposition $$C^t C = R^t R$$ to write:
$$\begin{align*}
  C^t C + \alpha D &= R^t R + \alpha D \\
  &=  R^t R + \alpha R^{t} R^{-t} D R^{-1} R \\ 
  &= R^t ( I + \alpha R^{-t} D R^{-1} ) R \\
  &= R^t ( I + \alpha U \text{diag}(s) U^t ) R
\end{align*}$$
In the last step we've used $R^{-t} D R^{-1} =  U \text{diag}(s) U^t$, another hypothesis.  Note that $U^t U = I$, since $U$ is an orthogonal matrix.
Breaking down the inverse:
$$\begin{align*}
  ( I + \alpha U \text{diag}(s) U^t )^{-1} &= ( U U^t + \alpha U \text{diag}(s) U^t ) \\
  &= ( U (I + \alpha \ \text{diag}(s)) U^t)^{-1} \\
  &= U^{-t} ( I + \alpha \ \text{diag}(s) )^{-1} U^{-1}
\end{align*}$$
Now the matrix $I + \alpha \ \text{diag}(s)$ is itself diagonal, so its inverse is as well.  We can write its inverse conveniently as
$$ \frac{1}{1 + \alpha \ \text{diag}(s)} $$
Now put it all together to get:
$$\begin{align*}
  C ( C^t C + \alpha D )^{-1} C^t y &= C R^{-1} U^{-t} \frac{1}{1 + \alpha \ \text{diag}(s)} U^{-1} R^{-t} C^{t} y \\
  &= C R^{-1} U^{-t} \frac{1}{1 + \alpha \ \text{diag}(s)} ( C R^{-1} U^{-t} )^t y \\
  &= A \frac{1}{1 + \alpha \ \text{diag}(s)} A^{t} y
\end{align*}$$
which, up to a substitution of $U^t$ with $U$, is the formula quoted.
