Here is my question
in the equation 2.11 A is N by N matrix, so there is not feasible if N is large the textbook say in the euqation 2.12, we only need to invert size n by n.
But I think $K$ is 1 by 1 scalar, since $K=\Phi^T\Sigma_{p}\Phi$ and $\Phi^T$ is $1 \times N$ and $\Sigma_{p}$ is $N \times N$.
Am I right?
It looks like $\Phi$ is an $N\times n$ matrix with columns $\phi(x_i)$ for $i=1\cdots n$. I guess that $n$ is the number of training samples from which you want to learn the function $f$. Thus inverting $A$ which has the same size as $\Phi \Phi^\top$ ($N\times N$) is more computationally expensive than inverting $K$ which has the same size as $\Phi^\top \Phi$ ($n\times n$) when $n<N$.