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  "body": "\n## Statement\n\nLet $H$ be a complex [[Hilbert space|HM2HHE3J51]] and let $T \\colon H \\to H$ be a bounded linear operator.\nThen the [[numerical range|XWXZ0DGW3Z]]\n$$W(T) = \\set{\\ip{Tx}{x} : x \\in H, \\ \\norm{x} = 1}$$\nis a [[convex subset|02AS45T3H9]] of $\\C$.\n\n## Remarks\n\nProof idea: take two points $\\ip{Tx}{x}$ and $\\ip{Ty}{y}$ of $W(T)$ and let $M$ be the span of the unit vectors $x$ and $y$.\nThe numerical range of the compression $P_M T|_M$ of $T$ to $M$ is contained in $W(T)$ and contains both points, so it is enough to treat a $2 \\times 2$ matrix.\nFor a $2 \\times 2$ matrix, a direct computation shows that the numerical range is a closed elliptical disc with foci at the two eigenvalues.\nThe disc may degenerate to a segment or a point.\n\nExample: for $T = \\begin{pmatrix} 1 \u0026 0 \\\\ 0 \u0026 -1 \\end{pmatrix}$ on $\\C^2$ and a unit vector $x = (x_1, x_2)$, one gets $\\ip{Tx}{x} = \\abs{x_1}^2 - \\abs{x_2}^2$, so $W(T) = [-1, 1]$, a convex set.\nThe set $\\set{1, -1}$ of eigenvalues is not convex, but $W(T)$ fills in the segment between them.\n\n## Explanations\n\n### L1 human\n\nFeed a unit vector $x$ into $T$ and measure how much of $Tx$ points back along $x$; that complex number is $\\ip{Tx}{x}$.\nThe set of all these numbers is the numerical range.\nThe theorem says that this set is convex: the segment between any two of its points lies inside it.\n\n### L2 human\n\nThe numerical range contains the eigenvalues of $T$, and its closure contains the [[spectrum|BMKATJ802A]].\nWith convexity, the closure of $W(T)$ contains the convex hull of the spectrum, with equality for normal operators.\nThe numerical radius $w(T) = \\sup \\set{\\abs{z} : z \\in W(T)}$ satisfies $\\tfrac12 \\norm{T} \\le w(T) \\le \\norm{T}$, so it is a norm on bounded operators equivalent to the operator norm.\n",
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  "name": "Toeplitz–Hausdorff theorem",
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