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  "body": "\n## Statement\n\nLet $n \\ge 1$, let $A$ be an $n \\times n$ complex matrix, and let $\\norm{\\cdot}$ be a [[matrix norm|54T6Y6DSY4]] on $\\C^{n \\times n}$.\nThen the [[spectral radius|9V7YQA85PC]] of $A$ is the [[limit|05M5SWSZZD]]\n$$\\rho(A) = \\lim_{k \\to \\infty} \\norm{A^k}^{1/k}.$$\n\n## Remarks\n\nFor a submultiplicative norm, $\\rho(A)^k = \\rho(A^k) \\le \\norm{A^k}$, which gives $\\rho(A) \\le \\norm{A^k}^{1/k}$ for every $k$.\nFor $\\varepsilon \u003e 0$, the matrix $A / (\\rho(A) + \\varepsilon)$ has spectral radius less than $1$, so its powers tend to $0$ by the Jordan normal form; this gives $\\norm{A^k} \\le (\\rho(A) + \\varepsilon)^k$ for large $k$.\nAny two norms are equivalent, and $c^{1/k} \\to 1$ for $c \u003e 0$, so the limit does not depend on the norm.\nExample: $A = \\begin{pmatrix} 1 \u0026 1 \\\\ 0 \u0026 1 \\end{pmatrix}$ has $A^k = \\begin{pmatrix} 1 \u0026 k \\\\ 0 \u0026 1 \\end{pmatrix}$, and with the maximum row sum norm $\\norm{A^k}^{1/k} = (1 + k)^{1/k} \\to 1 = \\rho(A)$, although $\\norm{A} = 2$.\nThe same formula holds for elements of a complex Banach algebra; see the [[spectral radius formula|KX2XY7S3X6]].\n\n## Explanations\n\n### L1 human\n\nThe spectral radius is the largest absolute value of an eigenvalue.\nA norm of $A$ can overestimate it, but the norms of the powers $A^k$ grow like $\\rho(A)^k$.\nTaking the $k$-th root and letting $k$ grow recovers $\\rho(A)$ exactly, whatever norm is used.\n\n### L2 human\n\nThe formula explains why $A^k \\to 0$ if and only if $\\rho(A) \u003c 1$, which is the convergence criterion for iterative methods such as Jacobi and Gauss–Seidel.\nIt also shows that $\\rho(A) \\le \\norm{A}$ for every submultiplicative norm, and that for every $\\varepsilon \u003e 0$ there is an operator norm with $\\norm{A} \\le \\rho(A) + \\varepsilon$.\nIn the Banach algebra setting, the same statement is proved with the resolvent and complex analysis instead of the Jordan form.\n",
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