User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)


 * $u \ v \ \mathsf{u} \ \mathsf{v} \ \nu \ \upsilon$

Anyone else have a hard t--GFauxPas 14:32, 19 February 2012 (EST)ime distinguishing between $u$ and $v$? I would like it to look more like this, does it confuse anyone else? It seems PW doesn't have the upgreek package. --GFauxPas 07:49, 27 January 2012 (EST)


 * Nope. Multiple years of extensive TeX writing and reading have trained my eye. I agree that referenced $v$ looks more distinguished, but imagine it is hard to implement. --Lord_Farin 08:08, 27 January 2012 (EST)

Exponential Definitions
I am discussing the equivalence of the definitions of exponential here:

http://forums.xkcd.com/viewtopic.php?f=17&t=80256

For anyone who has been following my progress or lack thereof on exponent combination laws/log laws etc, feel free to look on. --GFauxPas 16:59, 6 February 2012 (EST)


 * Okay, it looks like $e^{xy} = e^xe^y$ was the hardest one to prove! I was expecting a walk uphill the whole way. Oh, my Linear Algebra book came in the mail, so I guess I'll work on vectors next. And one of these days I'll have to tie up loose ends with Tarski. --GFauxPas 16:57, 10 February 2012 (EST)

Matrix Multiplication
Let $\mathbf {A,B}$ be square matrices of order $n$

Let $\mathbf I$ be the $n \times n$ Identity Matrix.

Let $\mathbf{A}$ and $\mathbf{B}$ be invertible.

Then the matrix product $\mathbf {AB}$ is also invertible, and:


 * $\left({\mathbf{AB}}\right)^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}$

Proof
By the definition of inverse and the given that $\mathbf{A}$ and $\mathbf{B}$ are invertible:


 * $\mathbf{AA}^{-1} = \mathbf{A}^{-1}\mathbf{A} = \mathbf{I}$

and


 * $\mathbf{BB}^{-1} = \mathbf{B}^{-1}\mathbf{B} = \mathbf{I}$

Now, observe that:

Similarly,

The result follows from the definition of inverse.

I'm letting this sit in the sandbox for a bit while I decide what's redundant on PW. Perhaps I should just say "follows from Inverse of Product" but see what I wrote on the talk page there. --GFauxPas 15:03, 23 February 2012 (EST)