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  "body": "\n## Statement\n\nLet $A$ be an $n \\times n$ complex matrix that is [[diagonalisable|07SQCNV233]], with $A = V \\Lambda V^{-1}$ for an invertible matrix $V$ and a diagonal matrix $\\Lambda$.\nLet $1 \\le p \\le \\infty$, let $\\norm{\\cdot}_p$ be the [[operator norm|06EKW1AKXF]] induced by the $p$-norm on $\\C^n$, and let $\\kappa_p(V) = \\norm{V}_p \\norm{V^{-1}}_p$ be the [[condition number|AB3S6MHJ8X]] of $V$.\nLet $E$ be an $n \\times n$ complex matrix, and let $\\mu$ be an [[eigenvalue|0BQAA2BK3Q]] of $A + E$.\nThen there is an eigenvalue $\\lambda$ of $A$ with\n$$\\abs{\\lambda - \\mu} \\le \\kappa_p(V) \\norm{E}_p.$$\n\n## Remarks\n\nIf $\\mu$ is an eigenvalue of $A$, there is nothing to prove.\nOtherwise $\\Lambda - \\mu I$ is invertible, and $A + E - \\mu I = V(\\Lambda - \\mu I)\\big(I + (\\Lambda - \\mu I)^{-1} V^{-1} E V\\big) V^{-1}$ is singular.\nSo $1 \\le \\norm{(\\Lambda - \\mu I)^{-1} V^{-1} E V}_p \\le \\kappa_p(V) \\norm{E}_p / \\min_\\lambda \\abs{\\lambda - \\mu}$, which is the claim.\n\nIf $A$ is [[normal|0MR0BYF0K2]], $V$ can be chosen unitary, so $\\kappa_2(V) = 1$ and each eigenvalue of $A + E$ is within $\\norm{E}_2$ of an eigenvalue of $A$.\n\nThe diagonalisability hypothesis is needed for a bound linear in $\\norm{E}$.\nFor $A = \\begin{pmatrix} 0 \u0026 1 \\\\ 0 \u0026 0 \\end{pmatrix}$ and $E = \\begin{pmatrix} 0 \u0026 0 \\\\ \\varepsilon \u0026 0 \\end{pmatrix}$ with $\\varepsilon \u003e 0$, the eigenvalues of $A + E$ are $\\pm \\sqrt{\\varepsilon}$, much larger than $\\varepsilon$ when $\\varepsilon$ is small.\n\n## Explanations\n\n### L1 human\n\nIf a diagonalisable matrix is changed by a small amount, its eigenvalues move by a small amount too.\nThe theorem gives the size of the move: the size of the change times a factor that measures how far the eigenvectors are from being orthogonal.\n\n### L2 human\n\nThe theorem is a basic tool of [[eigenvalue perturbation|VR89YHC0TX]] theory and of the error analysis of eigenvalue algorithms.\nIt shows that the eigenvalue problem of a normal matrix is well conditioned.\nWhen the eigenvectors are nearly parallel, $\\kappa_p(V)$ is large and the eigenvalues can be very sensitive.\n",
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