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  "body": "\n## Definition\n\nLet $\\mathbb{K}$ be a field, let $A$ be an associative algebra over $\\mathbb{K}$ with unit $1$, and let $a \\in A$.\nThe *spectrum* of $a$ is the set\n$$\\sigma(a) = \\set{\\lambda \\in \\mathbb{K} : a - \\lambda 1 \\text{ is not invertible in } A}.$$\nHere invertible means that $a - \\lambda 1$ is a [[unit|0XVEZDE9J8]] of $A$.\nIf $A$ has no unit, the spectrum of $a$ is its spectrum in the [[unitization|7SB03M850M]] $A^+$.\n\n## Remarks\n\nThe spectrum depends on the algebra: if $a$ lies in a subalgebra $B$ of $A$ that contains $1$, then $\\sigma_A(a) \\subseteq \\sigma_B(a)$, and the inclusion can be strict.\nIn a non-unital algebra, $0$ always lies in the spectrum, because no $(0, a)$ is invertible in $A^+$.\n\nExample: for $A = M_n(\\C)$, $a - \\lambda 1$ is invertible exactly when $\\det(a - \\lambda 1) \\ne 0$.\nSo $\\sigma(a)$ is the set of eigenvalues of $a$, the [[spectrum of the matrix|0XPX5NC67E]].\n\nExample: for $A = C([0,1])$ with complex values and $f \\in A$, $\\sigma(f) = f([0,1])$.\nIf $\\lambda \\notin f([0,1])$, then $1/(f - \\lambda)$ is continuous and inverts $f - \\lambda$; if $\\lambda = f(t_0)$, then $f - \\lambda$ vanishes at $t_0$ and has no inverse.\n\nFor bounded operators on $\\ell^2$, the right shift $S$ has no eigenvalues, but $0 \\in \\sigma(S)$ because $S$ is not surjective.\nSo in infinite dimension the spectrum can contain points that are not eigenvalues.\n",
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