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  "body": "\n## Proof\n\nWe argue by induction on $n = \\dim V$. Steps 2 to 11 show that $T$ has an eigenvalue in the scalar field.\n\n1. Let $\\mathbb{K}$ be $\\R$ or $\\C$. For $n \\ge 0$, let $P(n)$ be the following claim. Let $V$ be an $n$-dimensional inner product space over $\\mathbb{K}$, and let $T \\colon V \\to V$ be [[self-adjoint|0MB8HQW495]]. Then $V$ has an [[orthonormal basis|0FWJ0VW561]] of [[eigenvectors|0B5H4208Z9]] of $T$.\n2. Base case $n = 0$. Then $V = \\set{0}$, and the empty family is an orthonormal basis. It vacuously consists of eigenvectors. So $P(0)$ holds.\n3. Let $n \\ge 1$, and let $V$ and $T$ be as in $P(n)$, with [[inner product|0VGSYBWHV2]] $\\ip{\\cdot}{\\cdot}$. If $\\mathbb{K} = \\C$, then $\\C$ is [[algebraically closed|02SJG1HC32]] by the [[fundamental theorem of algebra|0JP0W07Y9A]]. So $T$ has an [[eigenvalue|08CJB2NQ92]].\n4. Now let $\\mathbb{K} = \\R$. Take an [[orthonormal basis|0Y85GYEGSF]] $e_1, \\dots, e_n$ of $V$, and set $a_{ij} = \\ip{T(e_j)}{e_i} \\in \\R$. By the [[formula for coordinates|09DZ8CD1AK]], $T(e_j) = \\sum_{i} a_{ij} e_i$.\n5. Over $\\R$, conjugate symmetry is symmetry. So self-adjointness gives $a_{ji} = \\ip{T(e_i)}{e_j} = \\ip{e_i}{T(e_j)} = \\ip{T(e_j)}{e_i} = a_{ij}$. Let $A = (a_{ij})$, a real symmetric matrix.\n6. View $A$ as a complex matrix. Since $\\C^n \\ne \\set{0}$, the [[existence of an eigenvalue|08CJB2NQ92]] over $\\C$, applied to $z \\mapsto Az$, gives $\\lambda \\in \\C$ and $z \\in \\C^n$ with $z \\ne 0$ and $Az = \\lambda z$.\n7. Set $s = \\sum_{i,j} \\overline{z_i}\\, a_{ij} z_j$. Since $\\sum_j a_{ij} z_j = \\lambda z_i$, we get $s = \\lambda \\sum_i \\abs{z_i}^2$.\n8. Since the $a_{ij}$ are real, $\\overline{s} = \\sum_{i,j} z_i a_{ij} \\overline{z_j}$. Exchanging the names $i$ and $j$ and using $a_{ji} = a_{ij}$ gives $\\overline{s} = s$. So $s$ is real.\n9. The number $\\sum_i \\abs{z_i}^2$ is real and positive, as $z \\ne 0$. By step 7, $\\lambda$ is real.\n10. Write $z = x + i y$ with $x, y \\in \\R^n$. Since $A$ and $\\lambda$ are real, comparing real and imaginary parts of $Az = \\lambda z$ gives $Ax = \\lambda x$ and $Ay = \\lambda y$.\n11. As $z \\ne 0$, one of $x, y$ is nonzero; call it $c$. Set $v = \\sum_j c_j e_j$, which is nonzero. By step 4, $T(v) = \\sum_i \\big(\\sum_j a_{ij} c_j\\big) e_i = \\sum_i \\lambda c_i e_i = \\lambda v$. So $\\lambda \\in \\R$ is an eigenvalue of $T$.\n12. In both cases, let $\\lambda$ be an eigenvalue of $T$ and $v$ an eigenvector for it. By the [[theorem on eigenvalues|0C4NT4GT3X]], $\\lambda$ is real.\n13. Set $e = v / \\norm{v}$. Then $\\ip{e}{e} = 1$ and $T(e) = \\lambda e$.\n14. Let $W$ be the [[orthogonal complement|079D68S3AW]] of the line $L = \\set{a e : a \\in \\mathbb{K}}$. A vector $w$ lies in $W$ if and only if $\\ip{w}{e} = 0$, by linearity in the first argument and conjugate symmetry.\n15. By the [[dimension formula|0E2NYPVSP9]], $\\dim W = n - \\dim L = n - 1$.\n16. Let $w \\in W$. Then $\\ip{T(w)}{e} = \\ip{w}{T(e)} = \\ip{w}{\\lambda e} = \\overline{\\lambda \\ip{e}{w}} = \\lambda \\overline{\\ip{e}{w}} = \\lambda \\ip{w}{e} = 0$, as $\\lambda$ is real. So $T(w) \\in W$.\n17. The restriction of $\\ip{\\cdot}{\\cdot}$ to $W$ is an inner product. Let $S \\colon W \\to W$ be $S(w) = T(w)$. It maps into $W$ by step 16, and it is linear since $T$ is.\n18. For $w, w' \\in W$, $\\ip{S(w)}{w'} = \\ip{T(w)}{w'} = \\ip{w}{T(w')} = \\ip{w}{S(w')}$. So $S$ is self-adjoint on $W$.\n19. By $P(n - 1)$, $W$ has an orthonormal basis $f_2, \\dots, f_n$ of eigenvectors of $S$. Each $f_k$ is nonzero and satisfies $T(f_k) = S(f_k) = \\mu_k f_k$ for some $\\mu_k$, so it is an eigenvector of $T$.\n20. Put $f_1 = e$. Each $f_k$ with $k \\ge 2$ lies in $W$, so $\\ip{f_k}{f_1} = 0$, and $\\ip{f_1}{f_k} = \\overline{0} = 0$. With steps 13 and 19, $f_1, \\dots, f_n$ is orthonormal.\n21. An orthonormal family is [[linearly independent|0JWM6GJNQW]]. These $n$ vectors in the $n$-dimensional space $V$ therefore form [[a basis|096V7H13H9]].\n22. So $f_1, \\dots, f_n$ is an orthonormal basis of eigenvectors of $T$. This proves $P(n)$, and by induction $P(n)$ holds for all $n$.\n23. The eigenvalues of these eigenvectors are real, by the [[theorem on eigenvalues|0C4NT4GT3X]].\n\n## Idea\n\nFind one unit eigenvector $e$, then look at the hyperplane orthogonal to it. Self-adjointness makes this hyperplane invariant, and $T$ stays self-adjoint there, so induction gives the rest of the basis. Over $\\R$, the eigenvalue comes from viewing the symmetric matrix over $\\C$: its complex eigenvalues are real, so a real eigenvector exists.\n",
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