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  "body": "\n## Proof\n\nWe apply the spectral theorem to the map $x \\mapsto Ax$ on $\\C^n$ with the standard inner product.\n\n1. Let $A$ be a [[Hermitian|0EKZCX4K3P]] $n \\times n$ complex matrix, so $A^* = A$, where $^*$ is the [[conjugate transpose|0EX045E2BP]]. Write vectors of $\\C^n$ as columns.\n2. For $x, y \\in \\C^n$, set $\\ip{x}{y} = y^* x = \\sum_{i=1}^{n} x_i \\overline{y_i}$. It is linear in $x$, and $\\ip{y}{x} = \\overline{\\ip{x}{y}}$. Also $\\ip{x}{x} = \\sum_i \\abs{x_i}^2 \u003e 0$ for $x \\ne 0$.\n3. So $\\ip{\\cdot}{\\cdot}$ is an [[inner product|0VGSYBWHV2]] on $\\C^n$. Let $T \\colon \\C^n \\to \\C^n$ be $T(x) = Ax$.\n4. For a product of matrices, $(MN)^* = N^* M^*$, since $\\overline{\\sum_k m_{jk} n_{ki}} = \\sum_k \\overline{n_{ki}}\\, \\overline{m_{jk}}$. Also $(M^*)^* = M$. Hence $\\ip{Ax}{y} = y^* A x = (A^* y)^* x = \\ip{x}{A^* y} = \\ip{x}{Ay}$.\n5. So $T$ is [[self-adjoint|0MB8HQW495]]. By the [[spectral theorem|06J4PXRG22]], $\\C^n$ has an orthonormal basis $u_1, \\dots, u_n$ with $A u_j = \\lambda_j u_j$ and $\\lambda_j \\in \\R$.\n6. Let $U$ be the $n \\times n$ matrix with columns $u_1, \\dots, u_n$. Let $D$ be the [[diagonal matrix|0Q9PGXD9E4]] with diagonal entries $\\lambda_1, \\dots, \\lambda_n$, which is real.\n7. The $(i, j)$ entry of $U^* U$ is $u_i^* u_j = \\ip{u_j}{u_i} = \\delta_{ij}$. So $U^* U = I_n$, and $U$ is [[unitary|0D6RMJ10M9]].\n8. By the [[one-sided inverse theorem|0753Y2X5K6]], also $U U^* = I_n$.\n9. By the definition of [[matrix multiplication|076JGP6R3T]], column $j$ of $AU$ is $A u_j = \\lambda_j u_j$, and column $j$ of $UD$ is $\\lambda_j u_j$. So $AU = UD$.\n10. Multiplying on the right by $U^*$ and using step 8 gives $A = A U U^* = U D U^*$. This proves the first claim.\n11. Now let $A$ be a real [[symmetric matrix|0N28QN4CYA]], so $A^T = A$. Use $\\R^n$ with $\\ip{x}{y} = y^T x$, which is an inner product by the real case of step 2.\n12. As in step 4, $\\ip{Ax}{y} = y^T A x = (A^T y)^T x = \\ip{x}{Ay}$. So $x \\mapsto Ax$ is self-adjoint on $\\R^n$.\n13. The spectral theorem gives an orthonormal basis $q_1, \\dots, q_n$ of $\\R^n$ with $A q_j = \\lambda_j q_j$ and $\\lambda_j \\in \\R$.\n14. Let $Q$ have columns $q_1, \\dots, q_n$, and $D = \\operatorname{diag}(\\lambda_1, \\dots, \\lambda_n)$. As in step 7, $Q^T Q = I_n$, so $Q$ is real and [[orthogonal|0QRBVFJ50X]].\n15. As in steps 8 to 10, $Q Q^T = I_n$ and $AQ = QD$. So $A = Q D Q^T$, and $Q = Q^*$ shows that $U = Q$ can be chosen real.\n\n## Idea\n\nA Hermitian matrix acts on $\\C^n$ as a self-adjoint map for the standard inner product. The spectral theorem gives an orthonormal basis of eigenvectors. Put these as the columns of $U$: orthonormal columns make $U$ unitary, and $AU = UD$ becomes $A = UDU^*$.\n",
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