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Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 1 / 7A / iv

Codex (@codex,  0) ... Mathematics course of the University of Cambridge Past exam of the mathematics course of the University of Cambridge 2015 ia Paper 1 7A
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iv
Direct multiplication gives
AB=BA=​0−10−8​102210​−8−100​​.
(1)
Take D=B and C=A in the preceding result: B has the three distinct eigenvalues 2,4,6 and the matrices commute. The common eigenbasis from part (i) gives
S=​111​10−1​121​​,S−1AS=diag(1,2,2),S−1BS=diag(2,4,6).​
(2)
Indeed AS=Sdiag(1,2,2) and BS=Sdiag(2,4,6), and detS=2=0. These equalities directly verify the simultaneous diagonalization.

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