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  "body": "\n## Proof\n\n1. By the [[spectral theorem|0DGZBQWZWN]] for the real [[symmetric matrix|0N28QN4CYA]] $A$, there are an [[orthogonal matrix|0QRBVFJ50X]] $P$ and a real diagonal matrix $D$ with $A = P D P^T$.\n2. Let $\\lambda_1, \\dots, \\lambda_n$ be the diagonal entries of $D$. By definition $P^T P = I$, and the [[one-sided inverse theorem|0753Y2X5K6]] gives $P P^T = I$.\n3. By step 2, $P^T A P = P^T P D P^T P = D$.\n4. Let $y \\in \\R^n$ and $x = Py$. By the [[transpose of a product|09GK5PS824]], $Q(x) = y^T P^T A P y = y^T D y$.\n5. By step 3, $y^T D y = \\lambda_1 y_1^2 + \\cdots + \\lambda_n y_n^2$. This is the formula of the statement.\n6. Let $p_1, \\dots, p_n$ be the columns of $P$. The $(i, j)$ entry of $P^T P = I$ is $p_i^T p_j$. So the columns form an [[orthonormal family|09VYHEF2VT]] for the dot product, and each $p_i \\ne 0$.\n7. From $A = P D P^T$ and $P^T P = I$ we get $AP = PD$. Comparing column $i$ gives $A p_i = \\lambda_i p_i$.\n8. By steps 6 and 7, each $p_i$ is an [[eigenvector|0B5H4208Z9]] of $A$, and each $\\lambda_i$ is an [[eigenvalue|0ZND7NQ970]] of $A$.\n9. Since $D = P^{-1} A P$, the [[characteristic polynomials agree|0TYMASG33Y]]. So $\\det(tI - A) = (t - \\lambda_1) \\cdots (t - \\lambda_n)$.\n10. By [[the roots of the characteristic polynomial|053QVHF6N7]], the eigenvalues of $A$ are exactly $\\lambda_1, \\dots, \\lambda_n$, each with its multiplicity.\n11. So $Q(Py) = \\sum_i \\lambda_i y_i^2$, where the $\\lambda_i$ are the eigenvalues of $A$ and the columns of $P$ are orthonormal eigenvectors of $A$.\n\n## Idea\n\nThe spectral theorem rotates coordinates so that $A$ becomes diagonal. In the new coordinates $y = P^T x$, the cross terms of $Q$ vanish, and only the eigenvalues remain as coefficients of the squares.\n",
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