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  "body": "\n## Definition\n\nLet $A$ and $B$ be $n \\times n$ complex matrices, and let $A(\\varepsilon) = A + \\varepsilon B$ for $\\varepsilon \\in \\C$.\nLet $\\lambda_0$ be a simple [[eigenvalue|0BQAA2BK3Q]] of $A$, that is, a root of [[multiplicity|03XVP8Q4GY]] $1$ of the characteristic polynomial.\nLet $x$ be a right [[eigenvector|0N6960W0BV]] and $y$ a left eigenvector of $A$ for $\\lambda_0$: $Ax = \\lambda_0 x$ and $y^* A = \\lambda_0 y^*$, with $x, y \\ne 0$.\nThen $y^* x \\ne 0$.\nThere are a neighbourhood $D$ of $0$ in $\\C$ and a unique analytic function $\\lambda \\colon D \\to \\C$ with $\\lambda(0) = \\lambda_0$ and the following property.\nFor each $\\varepsilon \\in D$, the number $\\lambda(\\varepsilon)$ is a simple eigenvalue of $A(\\varepsilon)$.\nThe *eigenvalue perturbation* of $\\lambda_0$ in the direction $B$ is this function $\\lambda$.\nIts *first-order coefficient* is\n$$\\lambda'(0) = \\frac{y^* B x}{y^* x}.$$\n\n## Remarks\n\nSo $\\lambda(\\varepsilon) = \\lambda_0 + \\varepsilon \\, \\dfrac{y^* B x}{y^* x} + O(\\varepsilon^2)$.\nTo obtain the coefficient, differentiate $A(\\varepsilon) x(\\varepsilon) = \\lambda(\\varepsilon) x(\\varepsilon)$ at $0$ and multiply on the left by $y^*$; the terms with $x'(0)$ cancel.\n\nExample: for a Hermitian $A$ one may take $y = x$, so $\\lambda'(0) = x^* B x / x^* x$.\nFor $A = \\operatorname{diag}(1, 2)$ and $B = \\begin{pmatrix} 0 \u0026 1 \\\\ 1 \u0026 0 \\end{pmatrix}$, with $\\lambda_0 = 1$ and $x = y = e_1$, this gives $\\lambda'(0) = 0$.\nIndeed the eigenvalues of $A + \\varepsilon B$ are $\\frac{3}{2} \\pm \\sqrt{\\frac{1}{4} + \\varepsilon^2}$, and the smaller one is $1 - \\varepsilon^2 + O(\\varepsilon^4)$.\n\nNon-example: for $A = \\begin{pmatrix} 0 \u0026 1 \\\\ 0 \u0026 0 \\end{pmatrix}$, the eigenvalue $0$ is not simple.\nWith $B = \\begin{pmatrix} 0 \u0026 0 \\\\ 1 \u0026 0 \\end{pmatrix}$, the eigenvalues of $A + \\varepsilon B$ are $\\pm \\sqrt{\\varepsilon}$, which are not analytic in $\\varepsilon$ at $0$.\nHere the right eigenvector $x = e_1$ and the left eigenvector $y = e_2$ satisfy $y^* x = 0$, so the formula for $\\lambda'(0)$ has no meaning.\n\n## Explanations\n\n### L1 human\n\nWhen a matrix depends smoothly on a small parameter, a simple eigenvalue moves smoothly too.\nIts speed at the start is given by a short formula with the left and right eigenvectors.\nA small value of $y^* x$ means that the eigenvalue is sensitive to changes.\n",
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