Timeline for Why does the condition number of the covariance matrix explode as number of variables increases?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 17, 2020 at 19:24 | comment | added | proof_by_accident | Any matrix $A$ (of any dimension, regardless of whether its square or not) can be decomposed as $A = U \Sigma V^T$ where $U$ and $V$ are orthogonal matrices and $\Sigma$ is diagonal. This is called the singular value decomposition. If $A$ is symmetric (as the covariance matrix $M$ is) then it's not hard to convince yourself that $U=V$. | |
| Sep 17, 2020 at 19:11 | comment | added | Matthew Drury | It's part of the definition of the singular value decomposition of $M$. I believe if the singular values of $M$ are all distinct (which is almost certainly true of any set of data obtained in nature), then the orthogonal matrix $V$ is uniquely determined by $M$. | |
| Sep 17, 2020 at 18:56 | comment | added | develarist | $V$ is an orthogonal matrix, but where did it come from | |
| Sep 17, 2020 at 18:54 | vote | accept | develarist | ||
| Sep 17, 2020 at 18:46 | history | answered | proof_by_accident | CC BY-SA 4.0 |