I'm trying to obtain the Ridge Regression solution from the mean of predictive distribution of a Gaussian Process with a linear kernel.

The mean of the predictive distribution of a GP is $$\boldsymbol k^T(\boldsymbol K+\sigma^2I)^{-1}y$$

where $\boldsymbol K=k(\boldsymbol X^T, \boldsymbol X)$ is the Kernel Gram Matrix (Symetric Positive definite) and $\boldsymbol k = k(x_{n+1},\boldsymbol X)$. (Note that $x_{n+1}$ is a vector while $\boldsymbol X$ is a matrix

Now I'm assuming a linear kernel, thus $k(\boldsymbol X^T, \boldsymbol X) = \boldsymbol X^T \boldsymbol X$ and $\boldsymbol k = k(x_{n+1},\boldsymbol X) = x_{n+1}^T* \boldsymbol X$. So I can write $$\boldsymbol k^T(\boldsymbol K+\sigma^2*I)^{-1}y = (x_{n+1} \boldsymbol X^T) (\boldsymbol X^T \boldsymbol X+\sigma^2*I)^{-1}y $$

Now the issue: The Ridge Regression should be $$x_{n_1}(X^TX+\sigma ^2 I)^{-1}\boldsymbol X^T y$$ while I have $$(x_{n+1} \boldsymbol X^T) (\boldsymbol X^T \boldsymbol X+\sigma^2*I)^{-1}y $$

So I need to commute $\boldsymbol X^T$ and $(\boldsymbol K+\sigma^2I)^{-1}$ but I know matrix cannot commute, so I'm stucked.

What am I missing? Can you help me?

Note that $(\boldsymbol K+\sigma^2I)^{-1}$ is a symmetric matrix.


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