Author Topic: TT's Maths Thread (Read 119437 times) Tweet Share

TrueTears · « **Reply #1335 on:** July 14, 2013, 04:53:42 pm »

Let $\mathbf{M}$ be a n by n symmetric, idempotent matrix with rank (hence its trace as well) equal to n-k < n (hence $\mathbf{M}$ is singular). Since $\mathbf{M}$ can be written as $\mathbf{M} = \mathbf{MM'}$ then $\mathbf{M}$ is positive semi definite and hence can be decomposed as $\mathbf{M} = \mathbf{P}\mathbf{P'}$ where $\mathbf{P}$ is a n by n-k matrix with full column rank (that is with rank n-k). Show that $\mathbf{PP'} = \mathbf{M} \iff \mathbf{P'P} = \mathbf{I_{n-k}}$ where $\mathbf{I_{n-k}}$ denotes the (n-k) by (n-k) identity matrix.

kamil9876 · « **Reply #1336 on:** July 14, 2013, 05:32:01 pm »

If you are familiar with Jordan Normal Form, then it's possible to solve it with that. (I mean how else did you know that the trace=rank for idempotent matrices?)

TrueTears · « **Reply #1337 on:** July 14, 2013, 05:34:10 pm »

lol I was given the latter fact.

TrueTears · « **Reply #1338 on:** July 18, 2013, 12:23:54 am »

Anyone wanna do a bit of algebra soup? nothing difficult, just annoying lol

Inverted Gamma pdf is: $f(x; \alpha, \beta) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{-\alpha - 1}\exp\left(-\frac{\beta}{x}\right)$

My equation:

$f(\sigma; ?, ?) = \frac{2}{\Gamma\left(\frac{v}{2}\right)}\left(\frac{v\hat{\sigma}^2}{2}\right)^{\frac{v}{2}} \frac{1}{\sigma^{v+1}}\exp\left[\frac{-v\hat{\sigma}^2}{2\sigma^2}\right]$

Clearly $\alpha = \frac{v}{2}$ , what's $\beta$ ?

nvm finally got it lol

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