Past exam of the mathematics course of the University of Cambridge 2006 ib Paper 1 14A Solution Created 2026-09-24 Updated 2026-09-24
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Past exam of the mathematics course of the University of Cambridge 2007 ii Paper 3 38C a Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2008 ia Paper 1 7C Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2011 ia Paper 1 8B a i Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2011 ia Paper 1 8B b Solution Created 2026-09-24 Updated 2026-09-24
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Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 31 2 iii Solution Created 2026-10-03 Updated 2026-10-07
The Hoffman–Wielandt inequality says that for normal matrices with eigenvalues , there is a permutation such thatwhere is the Frobenius norm and denotes the conjugate transpose. For Hermitian matrices, the increasingly ordered real eigenvalues may be paired directly: this ordering minimizes the sum of squared distances over all permutations, by removing crossed pairs. In particular, for the real symmetric matrices in this question,The square on the right is justified by symmetry. For arbitrary matrices, the appropriate expression is the conjugate-transpose product, not the ordinary square.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 31 2 ii Solution Created 2026-10-03 Updated 2026-10-07
The empirical spectral measures of the two real symmetric matrices arewhere is the Dirac measure at . Thus . Pair the eigenvalues in the specified increasing order and apply the triangle inequality followed by Lipschitz continuity with bound :This deterministic estimate uses only the Lipschitz bound; no entry independence or distributional hypothesis is needed here.
Past exam of the mathematics course of the University of Cambridge 2013 ia Paper 1 7B b Solution Created 2026-09-24 Updated 2026-10-07
Represent the quadratic form by the symmetric matrixThe block has characteristic polynomial . Corresponding orthonormal eigenvectors, together with the direction, areThe eigenvalues are . To eliminate the linear term, solve , obtainingIndeed the first equation gives , and the second then gives . Define the new coordinates by . The original polynomial takes the value at the center, so its translated and orthogonally diagonalized form isThusThe surface is an elliptic one-sheet hyperboloid, centered at , with axis parallel to . Its waist at is an ellipse with semiaxes and . Every constant- section is a nonempty ellipse, so the two signs of belong to the same connected sheet. This is an application of principal-axis reduction of a quadric.