sircelj

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Check your derivation again, it's easy to mix up $w_j$ and $w_i$ from the original error function and the regularization part. In matrix form the answer should be

$$(A + \lambda I) w = T,$$

where $A$ and $T$ are the same as before. Going from this form back, you can see that the new expression for $A$ (say $A'$) is

$$A'_{ij} = A_{ij} + \lambda \delta_{ij},$$

where $\delta_{ij}$ is the Kronecker delta, meaning 1 if $i=j$ and 0 if $i\neq j$.

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