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

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