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  "body": "\n## Proof\n\nWe compute $\\ip{T(v)}{v}$ in two ways for an eigenvector $v$.\n\n1. Let $V$ be a finite-dimensional inner product space over $\\mathbb{K} = \\R$ or $\\C$, with [[inner product|0VGSYBWHV2]] $\\ip{\\cdot}{\\cdot}$. Let $T \\colon V \\to V$ be [[self-adjoint|0MB8HQW495]], so $\\ip{T(v)}{w} = \\ip{v}{T(w)}$ for all $v, w \\in V$.\n2. If $\\mathbb{K} = \\R$, every scalar is real, so there is nothing to prove. Let $\\mathbb{K} = \\C$.\n3. Let $\\lambda$ be an [[eigenvalue|0ZND7NQ970]] of $T$. So there is $v \\in V$ with $v \\ne 0$ and $T(v) = \\lambda v$.\n4. Linearity in the first argument gives $\\ip{T(v)}{v} = \\ip{\\lambda v}{v} = \\lambda \\ip{v}{v}$.\n5. Conjugate symmetry and linearity in the first argument give $\\ip{v}{\\lambda v} = \\overline{\\lambda \\ip{v}{v}} = \\overline{\\lambda}\\, \\ip{v}{v}$, because $\\ip{v}{v}$ is real.\n6. By step 1, $\\ip{T(v)}{v} = \\ip{v}{T(v)} = \\ip{v}{\\lambda v}$. So steps 4 and 5 give $\\lambda \\ip{v}{v} = \\overline{\\lambda} \\ip{v}{v}$.\n7. Since $v \\ne 0$, positivity gives $\\ip{v}{v} \u003e 0$. Dividing by it gives $\\lambda = \\overline{\\lambda}$.\n8. A complex number equal to its [[complex conjugate|0BJV24C4SY]] has imaginary part $0$. So $\\lambda$ is real.\n\n## Idea\n\nSelf-adjointness lets $T$ move from one side of the inner product to the other. On an eigenvector, moving $T$ to the left gives the factor $\\lambda$, and moving it to the right gives $\\overline{\\lambda}$. So $\\lambda = \\overline{\\lambda}$.\n",
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