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  "body": "\n## Definition\n\nLet $\\mathbb{K}$ be $\\R$ or $\\C$, let $m, n \\ge 1$, and let $\\mathbb{K}^{m \\times n}$ be the vector space of $m \\times n$ [[matrices|0VD9ZSWWJG]] over $\\mathbb{K}$.\nA *matrix norm* on $\\mathbb{K}^{m \\times n}$ is a map $\\norm{\\cdot} \\colon \\mathbb{K}^{m \\times n} \\to \\R$ such that, for all $A, B \\in \\mathbb{K}^{m \\times n}$ and $c \\in \\mathbb{K}$:\n\n1. $\\norm{A} \\ge 0$, and $\\norm{A} = 0$ only if $A = 0$.\n2. $\\norm{cA} = \\abs{c} \\norm{A}$.\n3. $\\norm{A + B} \\le \\norm{A} + \\norm{B}$.\n\nA matrix norm on $\\mathbb{K}^{n \\times n}$ is *submultiplicative* if moreover $\\norm{AB} \\le \\norm{A} \\norm{B}$ for all $A, B \\in \\mathbb{K}^{n \\times n}$.\n\n## Remarks\n\nClauses 1 to 3 say that $\\norm{\\cdot}$ is a [[norm|0V2WP9B1MF]] on $\\mathbb{K}^{m \\times n}$; many authors include submultiplicativity in the word \"matrix norm\".\nExample: for a vector norm $\\abs{\\cdot}$ on $\\mathbb{K}^n$, the [[operator norm|06EKW1AKXF]] $\\norm{A} = \\max_{\\abs{x} = 1} \\abs{Ax}$ is a submultiplicative matrix norm, since $\\abs{ABx} \\le \\norm{A} \\abs{Bx} \\le \\norm{A}\\norm{B}\\abs{x}$.\nNon-example: $\\norm{A}_{\\max} = \\max_{i,j} \\abs{a_{ij}}$ is a norm but not submultiplicative.\nFor $A = \\begin{pmatrix} 1 \u0026 1 \\\\ 1 \u0026 1 \\end{pmatrix}$ one has $A^2 = 2A$, so $\\norm{A^2}_{\\max} = 2 \u003e 1 = \\norm{A}_{\\max}^2$.\nAll norms on $\\mathbb{K}^{m \\times n}$ are equivalent, because the space has finite dimension.\n",
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