Timeline for Is a sample covariance matrix always symmetric and positive definite?
Current License: CC BY-SA 3.0
13 events
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| S Apr 10, 2014 at 15:32 | history | suggested | Akseli Palén | CC BY-SA 3.0 |
Corrected a typo: positve -> positive. Also added link from abbreviation pdf to Wolfram's page of probability density function to make the abbreviation absolutely clear.
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| Apr 10, 2014 at 15:28 | review | Suggested edits | |||
| S Apr 10, 2014 at 15:32 | |||||
| May 7, 2013 at 14:07 | history | edited | Konstantin Schubert | CC BY-SA 3.0 |
spelling
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| Mar 23, 2013 at 22:46 | comment | added | Konstantin Schubert | Unfortunately, I only know German literature for these things. | |
| Mar 23, 2013 at 22:43 | history | edited | Konstantin Schubert | CC BY-SA 3.0 |
Added remarks about what holds for the sample covariance.
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| Mar 23, 2013 at 22:28 | history | edited | Konstantin Schubert | CC BY-SA 3.0 |
added 4 characters in body
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| Mar 23, 2013 at 22:17 | comment | added | Konstantin Schubert | @Morten When you think in coordinates, the argument goes like this: When $A$ is your transformation matrix then: $v'=Av$ with $v'$ as the transformed coordinate-vector, $M'=AMA^T$, so when you transform each element in the equation $v^T M v > 0$, you get $ v'^T M' v' = (Av)^T A M A^T A v > 0$, which equals $v^T A^T A M A^T A v > 0$, and, because A is orthogonal, $A^T A$ is the unit matrix and we again get $v^T M v > 0$, which means that the transformed and the untransformed equation have the same scalar as result, so their are either both or both not greater zero. | |
| Mar 23, 2013 at 22:09 | comment | added | Konstantin Schubert | @Morten The transformation-invariance is pretty clear if you understand a matrix multiplication geometrically. Think of your vector as an arrow. The numbers that describe your vector change with the coordinate system, but the direction and length of your vector doesnt. Now, a multiplication with a matrix means that you change length and direction of that arrow, but again the effect is geometrically the same in each coordinate system. The same is with a scalar product: It is defined geometrically and Geometriy is transformation-invariant. So your equation has the same result in all systems. | |
| Mar 23, 2013 at 16:21 | comment | added | whuber♦ | Morten, the symmetry is immediate from the formula. To show semi-definiteness, you need to establish that $uQ_nu'\ge 0$ for any vector $u$. But $Q_n$ is $1/n$ times a sum of $v_iv_i'$ (where $v_i=x_i-\bar{x})$, whence $n uQ_nu'$ is a sum of $u(v_iv_i')u'$ = $(uv_i)(uv_i)'$, which is the squared length of the vector $uv_i$. Because $n\gt 0$ and a sum of squares cannot ever be negative, $uQ_nu'\ge 0$, QED. This also shows that $uQ_nu'=0$ precisely for those vectors $u$ which are orthogonal to all the $v_i$ (i.e., $uv_i=0$ for all $i$). When the $v_i$ span, then $u=0$ and $Q_n$ is definite. | |
| Mar 23, 2013 at 15:39 | comment | added | Morten | No Konstantine you got it right:-) I did not know that the definition of definity is transformation-invariant. But that I would like to look in to, do you know you litteratur or links that could be usefull? | |
| Mar 22, 2013 at 15:19 | comment | added | Konstantin Schubert | But if you want to know whether your sampling algorithm guarantees it, you will have to state how you are sampling. | |
| Mar 22, 2013 at 13:53 | comment | added | Konstantin Schubert | PS: I am starting to think that this wasn't your question... | |
| Mar 22, 2013 at 13:53 | history | answered | Konstantin Schubert | CC BY-SA 3.0 |