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

Random Thought: LC Theorem

I finally have an answer to "who's your favorite singer?"

My Top Tip for Helping People Get Started Programming


Random paper on angles

An Image is Worth 16x16 Words

Random stuff

Lossless Data Compression with Neural Networks by Fabrice Bellard

Downscaling Numerical Weather Models With GANs (My CI 2019 Paper)

Learning Differential Forms and Questions