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Nov 28, 2022 at 1:41 history edited statmerkur CC BY-SA 4.0
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Oct 29, 2022 at 23:59 history edited statmerkur CC BY-SA 4.0
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Oct 15, 2022 at 17:01 comment added usεr11852 Thank you for the edits.
Oct 15, 2022 at 16:38 comment added statmerkur @User1865345 Thanks for the feedback.
Oct 15, 2022 at 16:32 comment added User1865345 The post looks complete and concise, @statmerkur.
Oct 15, 2022 at 16:30 comment added User1865345 @zzzhhh yes. They are linearly independent almost surely.
Oct 15, 2022 at 16:21 comment added statmerkur @Sycorax made an edit. Hope it's clearer now.
Oct 15, 2022 at 16:18 history edited statmerkur CC BY-SA 4.0
added 95 characters in body
Oct 15, 2022 at 16:13 comment added zzzhhh Thank you for pointing out the condition that the columns of $\math\Phi$ should be linearly independent. This is true almost surely because the elements of the design matrix are (functions of) random variables.
Oct 15, 2022 at 16:08 vote accept zzzhhh
Oct 15, 2022 at 15:44 comment added Sycorax This answer might be clearer if you state up front that the claim in the question is not true in general, unless we make an additional assumption about $\Phi$.
Oct 15, 2022 at 15:42 comment added statmerkur @usεr11852 I do say that in my answer: "is positive definite if the columns of $\Phi$ are linearly independent"
Oct 15, 2022 at 15:38 comment added usεr11852 You don't say that about $\Phi$ in your answer nor is mentioned in the OP's question.
Oct 15, 2022 at 15:26 comment added statmerkur @usεr11852 No, positive definite. If $\Phi$ has full column rank then $\Phi v \neq 0_{ \mathbb{R}^N}$ and thus $v^\top \left(\Phi^\top\Phi\right) v = \|\Phi v \|_2^2 > 0$ for all $v \in \mathbb{R}^M \setminus \left\{ 0_{ \mathbb{R}^M} \right\}$
Oct 15, 2022 at 15:16 comment added usεr11852 semi-definite...
Oct 15, 2022 at 14:56 history answered statmerkur CC BY-SA 4.0