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  "body": "\n## Proof\n\nWe compute both matrices entry by entry with the formula for coordinates in an orthonormal basis.\n\n1. Let $V$ and $W$ be finite-dimensional inner product spaces over $\\mathbb{K} = \\R$ or $\\C$, with [[inner products|0VGSYBWHV2]] written $\\ip{\\cdot}{\\cdot}$. Let $B = (b_1, \\dots, b_n)$ and $C = (c_1, \\dots, c_m)$ be [[orthonormal bases|0FWJ0VW561]] of $V$ and $W$.\n2. Let $T \\colon V \\to W$ be linear, with [[adjoint|0ST8QF9T3T]] $T^* \\colon W \\to V$. So $\\ip{T(v)}{w} = \\ip{v}{T^*(w)}$ for all $v \\in V$ and $w \\in W$.\n3. By the definition of the [[matrix of a linear map|0YYMY95FWS]], $[T]_{C,B} = (a_{ij})$ is the $m \\times n$ matrix with $T(b_j) = \\sum_{i=1}^{m} a_{ij} c_i$.\n4. Likewise, $[T^*]_{B,C} = (d_{ji})$ is the $n \\times m$ matrix with $T^*(c_i) = \\sum_{j=1}^{n} d_{ji} b_j$, for $1 \\le i \\le m$.\n5. By the [[formula for coordinates|09DZ8CD1AK]] in the orthonormal basis $C$, the coordinates of $T(b_j)$ are $a_{ij} = \\ip{T(b_j)}{c_i}$.\n6. By the same formula in the orthonormal basis $B$, the coordinates of $T^*(c_i)$ are $d_{ji} = \\ip{T^*(c_i)}{b_j}$.\n7. By conjugate symmetry and step 2, $d_{ji} = \\overline{\\ip{b_j}{T^*(c_i)}} = \\overline{\\ip{T(b_j)}{c_i}} = \\overline{a_{ij}}$.\n8. By the definition of the [[conjugate transpose|0EX045E2BP]], the $(j, i)$ entry of $([T]_{C,B})^*$ is $\\overline{a_{ij}}$.\n9. So $[T^*]_{B,C}$ and $([T]_{C,B})^*$ have the same size $n \\times m$ and the same entries. Hence $[T^*]_{B,C} = ([T]_{C,B})^*$.\n\n## Idea\n\nIn an orthonormal basis, each matrix entry is an inner product: the $(i, j)$ entry of $[T]$ is $\\ip{T(b_j)}{c_i}$. The defining property of the adjoint moves $T$ across the inner product. Conjugate symmetry then turns the entry of $[T^*]$ into the conjugate of the mirrored entry of $[T]$.\n",
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