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  "body": "\n## Definition\n\nLet $F$ be $\\R$ or $\\C$, let $A$ be an $n \\times n$ [[matrix|0VD9ZSWWJG]] over $F$, let $b \\in F^n$, and let $r \\ge 1$ be an integer.\nThe *Krylov subspace* of order $r$ generated by $A$ and $b$ is the [[linear subspace|00YC1D652K]]\n$$\\mathcal{K}_r(A, b) = \\operatorname{span} \\set{b, A b, A^2 b, \\dots, A^{r-1} b} \\subseteq F^n.$$\n\n## Remarks\n\nEquivalently, $\\mathcal{K}_r(A, b) = \\set{p(A) b : p \\in F[t], \\ \\deg p \u003c r}$.\nThe subspaces grow, $\\mathcal{K}_1 \\subseteq \\mathcal{K}_2 \\subseteq \\cdots$, and their dimensions increase by one until they stop growing.\nOnce $\\mathcal{K}_{r+1} = \\mathcal{K}_r$, the subspace is invariant under $A$ and all later ones are equal to it.\nThis happens at the least degree $d$ of a monic polynomial $p$ with $p(A) b = 0$.\nThat degree is at most the degree of the [[minimal polynomial|03NDVCSHSB]] of $A$, hence at most $n$.\nIf $A$ is [[invertible|0EB9N0S33C]], then $A^{-1} b$ lies in $\\mathcal{K}_n(A, b)$, since the minimal polynomial of $A$ then has nonzero constant term.\nExample: for $A = \\begin{pmatrix} 0 \u0026 1 \\\\ 0 \u0026 0 \\end{pmatrix}$ and $b = e_2$, one has $A b = e_1$, so $\\mathcal{K}_2(A, b) = F^2$.\nNon-example: for $A = I$, every Krylov subspace $\\mathcal{K}_r(I, b)$ is the line through $b$ or $\\set{0}$.\nSo for $n \\ge 2$ the whole space $F^n$ is not a Krylov subspace of the identity.\n",
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