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Let $X$ be an $D \times N$ matrix. Let $I$ be a $D \times D$ identity matrix. Also let $y$ be a $N \times 1$ column vector. Suppose we are trying to solve $(X X ^T + k I) w = Xy$ for a $D$ dimensional column vector $w$. Assuming that $X X ^T + kI$ is invertible, the solution is given by $w_1 = (XX ^ T + k I w) ^ {-1} Xy$.

Now suppose we want to find a solution for the equation $(X X ^T + (k + 1) I) w = Xy$. Assuming that $X X ^T + (k + 1)I$ is invertible, the solution is given by $w_2 = (XX ^ T + (k + 1) I w) ^ {-1} Xy$. Is it possible to express $w_2$ recursively using $w_1$? I tried using Woodbury Matrix Identity, but I couldn't continue much.

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