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

Handy command line benchmarking tool

Stan Rogers

Ultimate Hot Couch Guy

Quote on Java Generics

The Programmer Tendency

Figure out undocumented JSON with gron

Mental Model of Dental Hygiene

Book Review: Swastika Night

Is there a name for this construction?

Fun with negation and idioms