Inverse of sum of matrices (SVD, ridge regression) Looking at these slides, I've found the following:
$X=UDV^T$, where $U$ and $V$ are orthogonal matrices, $V$ is a square matrix, and $D$ contains the singular values of $X$.
The author then writes
$$X(X^TX + \lambda I)^{−1}X^Ty$$
$$=UD(D^2 + \lambda I)^{−1}DU^Ty.$$
Why does that equality hold? I've tried substituting $X=UDV^T$, have got
$$UDV^T(VD^2V^T+\lambda I)^{-1}VDU^Ty.$$
If there wasn't the $+\lambda I$ term, then, as $V$ is a square matrix, we would have
$$UDV^T V^{-T}D^{-1}D^{-1}V^{-1} VDU^Ty$$
$$=UU^Ty.$$
However, it's not the case that there isn't the $+\lambda I$ term, so I don't know how to proceed.

Here is the point :
In: $\ UDV^T(VD^2V^T+\lambda I)^{-1}VDU^Ty$
you should write $I=VV^T$ giving the factorization:
$UDV^T(V(D^2+\lambda I)V^T)^{-1}VDU^Ty$
$=UDV^T(V^T)^{-1}(D^2+\lambda I)^{-1}V^{-1}VDU^Ty$
yielding the looked-for formula.
