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Jan 10, 2022 at 21:20 answer added Aksakal timeline score: 1
Jan 10, 2022 at 14:33 answer added Sycorax timeline score: 6
Jan 10, 2022 at 9:36 vote accept Mas A
Jan 10, 2022 at 7:55 answer added Eric Johnson timeline score: 3
Jan 7, 2022 at 20:48 comment added Mas A Thank you @whuber for your comments and the url that you shared!
Jan 7, 2022 at 20:46 comment added Mas A How could I not think of that, of course, it makes sense! Thank you so much for your explanations @Sycorax
Jan 7, 2022 at 18:59 comment added whuber stats.stackexchange.com/questions/69205 is closely related: it discusses Ridge Regression, which is an example of this form of regularization.
Jan 7, 2022 at 18:31 comment added Sycorax @whuber That’s a much better demonstration!
Jan 7, 2022 at 18:26 comment added whuber @Sycorax That machinery is overkill. In the context $A=\Phi^\prime T\Phi$ must be square and positive semi-definite. (Otherwise the conclusion is incorrect.) Now, $\lambda+A$ is singular for a positive real number $\lambda$ if and only if there exists nonzero $x$ for which $(\lambda + A)x=0$ (that's the definition of "singular"). That's obviously equivalent to the eigenvalue equation $Ax=-\lambda x,$ yet we know (by supposition) that $A$ has no negative eigenvalues, QED.
Jan 7, 2022 at 15:40 comment added Sycorax Here's a sketch: Any $M=\Phi^T \Phi$ (for real $\Phi$) must be positive semi-definite (sums of squares must be non-negative). Adding $\lambda I$ (for $\lambda > 0$) guarantees that the singular values must be positive (sum of positive number and non-negative number must be positive). You can demonstrate this by writing down the SVD of $\Phi$ and working through the algebra of $M + \lambda I$.
Jan 7, 2022 at 15:27 history asked Mas A CC BY-SA 4.0