Quoting Prof. Bengio from his Deep Learning text (http://www.iro.umontreal.ca/~bengioy/dlbook/regularization.html),

$ w = (X^{T}X + \alpha I)^{-1}X^{T}y $

We can see L2 regularization causes the learning algorithm to “perceive” the input X as having higher variance, which makes it shrink the weights on features whose covariance with the output target is low compared to this added variance

After spending an hour, I can't understand how to approach the proof of this. Can anybody help me get an intuition for this?

  • $\begingroup$ $X^Ty$ is your cross product matrix (akin to $E[XY]$). $X^TX+\alpha I$ is the variability of your $x$ values, with added ridge weights. $\endgroup$
    – user75138
    Jun 29, 2015 at 16:48
  • 1
    $\begingroup$ Shouldn't we consider the entire $ (X^{T}X + \alpha I )^{-1} X^{T} $ for the proof? Can you please elaborate? $\endgroup$ Jun 29, 2015 at 16:55
  • $\begingroup$ Yes, you should, but I was offering pointers on what the compoents of the results are. $\endgroup$
    – user75138
    Jun 29, 2015 at 16:56
  • $\begingroup$ Basically $ w_{new} - w = (( X^{T}X + \alpha I)^{-1}X^{T} - ( X^{T}X)^{-1}X^{T})y $ and due to the addition of $ \alpha \delta_{ij} $, the values should be lesser in $ ( X^{T}X + \alpha I)^{-1}X^{T} $, since $ \alpha > 0 $ and the resulting matrix has greater determinant, hence lesser value. Though this indicates towards a more stronger result. Am i correct in my reasoning? $\endgroup$ Jun 29, 2015 at 18:31

1 Answer 1


The solution was along the similar lines of previous approach taken for the hessian approximation in the book. For getting exact relationship between $ w^{\sim} $ and $ w^{*} $, one has to do some kind of decomposition. Here, instead of eigenvalue decomposition (which is not possible, since $ X $,is not a square matrix) we do an SVD decomposition.

For unregularized version,

\begin{align} \newcommand{\new}{{\rm new}} X &= UDV^{T} \\ w &= (X^{T}X)^{-1}X^{T}y \end{align}

Putting the values of the decomposition into $w$, we get:

$$ w = D^{-1}U^{T}y $$

However if you consider the least square regularized version,

$$ w_{\new} = (X^{T}X + \alpha I)^{-1}X^{T}y $$

We get:

$$ w_{\new} = (D^2 + \lambda I)^{-1}D^{2}U^{T}y $$

We can see that $ w_{\new} $ is nothing but a scaled version of $ w $, namely $ w_{\new,j} = d_{j}^{2}/(d_{j}^{2} + \lambda_{j})w_{j} $. Hence proved.

  • $\begingroup$ @user3303, do not edit someone else's answer to 'correct' it. If you think it is wrong, leave a comment, downvote, or post your own (correct) answer to set the record straight. $\endgroup$ Apr 27, 2017 at 15:37
  • $\begingroup$ shouldn't it be $w_{new, j} = d_{j}/(d_{j}^{2} + \lambda_{j})w_{j} $ because $w$ contains $D^{-1}$ so it cancels out the square of $d_{j}$? $\endgroup$
    – Agile Bean
    May 11, 2021 at 7:26

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