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  "body": "\n## Proof\n\nWe compute $\\ip{T(v)}{w}$ in two ways.\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]].\n2. Let $v$ and $w$ be [[eigenvectors|0B5H4208Z9]] with $T(v) = \\lambda v$ and $T(w) = \\mu w$, where $\\lambda \\ne \\mu$.\n3. Linearity in the first argument gives $\\ip{T(v)}{w} = \\lambda \\ip{v}{w}$.\n4. Conjugate symmetry and linearity in the first argument give $\\ip{v}{T(w)} = \\ip{v}{\\mu w} = \\overline{\\mu} \\ip{v}{w}$.\n5. The eigenvalue $\\mu$ is [[real|0C4NT4GT3X]], so $\\overline{\\mu} = \\mu$.\n6. By self-adjointness, $\\ip{T(v)}{w} = \\ip{v}{T(w)}$. Steps 3 to 5 give $\\lambda \\ip{v}{w} = \\mu \\ip{v}{w}$.\n7. So $(\\lambda - \\mu) \\ip{v}{w} = 0$. Since $\\lambda - \\mu \\ne 0$, we get $\\ip{v}{w} = 0$. So $v$ and $w$ are [[orthogonal|0TS7VZJBWD]].\n8. Now let $E_\\lambda$ and $E_\\mu$ be the [[eigenspaces|0AYEPBQMAW]] for $\\lambda \\ne \\mu$. Let $v \\in E_\\lambda$ and $w \\in E_\\mu$.\n9. If $v = 0$, then $\\ip{v}{w} = \\ip{0 \\cdot v}{w} = 0 \\cdot \\ip{v}{w} = 0$. If $w = 0$, then $\\ip{v}{w} = \\overline{\\ip{w}{v}} = 0$ in the same way.\n10. Otherwise $v$ and $w$ are eigenvectors for $\\lambda$ and $\\mu$, and step 7 gives $\\ip{v}{w} = 0$. So $E_\\lambda$ and $E_\\mu$ are orthogonal to each other.\n\n## Idea\n\nMoving $T$ across the inner product gives the factor $\\lambda$ on one side and $\\overline{\\mu} = \\mu$ on the other. Two different numbers times the same inner product can agree only when the inner product is zero.\n",
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        "locator": "Theorem 7.22",
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