$K_m(A, v) \equiv span \{ v, Av, A^2v, \dots, A^m v\}$

Clearly, $K_m \subseteq K_{m+1}$, and there is a maximum $K_N$ that we can span
(the full vector space). We are interested in the smallest index $M$
such that $K_M = K_{M+1}$.
We notice that $K_M$ is invariant under the action of $A$.
Now, let's consider:
$\begin{aligned}
K_m(A, x) &\equiv span \{x, Ax, A^2x, \dots A^m x \} \\
&= span \{ A^{-1} b, b, Ab, \dots A^{m-1} x \} \qquad \text{(substitute $x = A^{-1}b$)} \\
&= A span \{ A^{-1} b, b, Ab, \dots A^{m-1} b\} \qquad \text{(Invariance of Krylov subspace)} \\
&= span \{b, Ab, \dots A^m b\} \\
&= K_m(A, b)
\end{aligned}$

We learnt that $Ax = b$ has a solution in $K_m(A, b)$. Using this, we can build
solvers that exploit the Krylov subspace. We will describe GMRES and CG.