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

The geometric series \(\sum_{k=0}^{\infty}{ar^k}\) equals \(\frac{a}{1-r}\) if and only if \(|r| < 1\).

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 \(a = 1\) and make the following replacements:

\[r \leftarrow A \\ 1 \leftarrow I \\\]

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

\[\sum_{k=0}^{\infty}{A^k} = \frac{1}{I-A}\]

Now we interpret “1 over something” as its multiplicative inverse, so the right hand side becomes \((I-A)^{-1}\).

Therefore, by our “identity”,

\[(I-A)^{-1}= A + A^2 + A^3 + A^4...\]

which is actually true.

Related Posts

Use of emphasis in speech

Generating a lot of language data with a theorem prover

"Litany Against Fear" in Present Tense

When it's time to party we will party hard

these are people who died

divine carrot

the frog

what it’s like to get nail phenolization

Why 0 to the power of 0 is 1

Lines and Points are Circles