$$(I-A)^{-1}$$ trick


The geometric series equals if and only if .

By squinting at the notation, we can notice (not prove) a “well-known” matrix identity. (I keep forgetting it, so it’s not well-known to me).

If we let and make the following replacements:

(where is some matrix and is the identity matrix), we can get the following “identity”:

Now we interpret “1 over something” as its multiplicative inverse, so the right hand side becomes .

Therefore, by our “identity”,

which is actually true.

Related Posts

How to Disable Disqus Ads on your Blog

Derivation of Reservoir Sampling

Fun with Python Iterators: Linked Lists Made Easy

Notes for November 11, 2018

Underrated Vim Option: undofile and undodir

Hot Take on Solo Travel: Starve

Alan Perlis

Book Notes: The Map of My Life by Goro Shimura

Prague

Way to remember the definition of local finiteness