Derivation of Closed Form solution of Regualrized Linear Regression I can follow the derivation of the closed form solution for the regualarized linear regression like shown here up to a specific point:

Where I get stuck is the part on the bottom of the slide:
$\mathbf{X}^{T}\mathbf{X}\mathbf{w}+\lambda N\mathbf{w}=\mathbf{X}^{T}\mathbf{y}$
$\mathbf{X}^{T}\mathbf{X}\mathbf{w}+\lambda N\mathbf{w}=\mathbf{X}^{T}\mathbf{y}$
$\mathbf{w}(\mathbf{X}^{T}\mathbf{X}+\lambda N\mathbf{I})=\mathbf{X}^{T}\mathbf{y}$
I understand that $\mathbf{X}^{T}\mathbf{X}$ is a $mxm$ matrix and $\lambda N$ is a scalar. So to add the two parts we must make the dimensions match. This is the point where I begin to struggle. I (rarely) understand that we can multiply $\lambda N$ with a $mxm$ identity ($\mathbf{I}$) to get the same dimensions as $\mathbf{X}^{T}\mathbf{X}$. At this point I miss some intuition why we can do this since: $\lambda N\mathbf{I}$ has the values $\lambda N$ on its diagonal and zeros everywhere else. If we add this to $\mathbf{X}^{T}\mathbf{X}$, the diagonal values of $\mathbf{X}^{T}\mathbf{X}$ are increased/decreased by $\lambda N$. But why do we add this value only to the diagonal elements and why not to all elements of $\mathbf{X}^{T}\mathbf{X}$. So you see I miss some intuition in the application of the identity. Can anybody please explain me why this is useful and give me some intuition.
 A: This depends on the form of your "regularization". Note that $\| w \|^2 \le r$ is an $m$  dimensional closed ball. Namely, if $r$ is not too large, the solution will be on the boundary of this ball. Now, recall that the variance of the estimators of $w$ in the unregularized problem are proportional to the diagonal terms of $(X'X)^{-1}$. Thus, the greater the diagonal terms of $X'X$, the lower the variance. $L_2$ regularization shrinks the estimators themselves and their variance. If you will to add something to the non-diagonal entries of $X'X$, just use more complex form of penalty, i.e., $w'Aw = \|  A^{1/2}w \|^2\le r$.  Where the exact geometric form depends on the entries of the matrix $A$. Sometimes this is called a generalized ridge regression. 
EDIT:
Let us consider a simple case where $L_2$ regularization may be helpful. Assume a complete multicolinearity between $x_1$ and $x_2$, in such a case the inverse of $X'X$ does not exist at all as $X'X$ is semi positive definite matrix. That is, exists a non-zero vector $v$ such that $v'X'Xv = \| Xv \|^2 = 0$. For the linear regression it means that you have an infinite number of possible estimates $\hat{w}$ that solve the unregularized minimization problem. Consider now the ridge solution, i.e., the inverse of $X'X + \lambda I$, note that for every positive lambda $v'(X'X + \lambda I)v = \|Xv\|^2 + \lambda\|v\|^2 >0$. Namely, a unique solution exists. Therefore, it was enough to penalize the main diagonal in order to solve the problem. For centered matrix $X$, $X'X$ is the covariance matrix of $X$ and its main diagonal is the variance of each variable $x_j$. Intuitively speaking each $\hat{w}_j$ is more-or-less the ratio between the covariance of corresponding $x_j$ and $y$ and the variance of $x_j$. Adding $\lambda$ to the main diagonal of $X'X$ increases the variance of $x_j$ without effecting its covariance with $y$, hence you are shrinking the estimate of $w_j$ by only dealing with the diagonal terms of $X'X$.  
