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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 331 / 3 / c / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 331 3 c
Created 2026-10-03 Updated 2026-10-05  0 By others on same topic  0 Discussions Create my own version
For real λ1​,λ2​, compute
LLT−LTL=(−1λ1​−λ2​​λ1​−λ2​1​)=0.
(1)
Thus L is a non-normal matrix. For distinct eigenvalues, take
v1​=(λ1​−λ2​1​),v2​=(01​).
(2)
These satisfy Lvj​=λj​vj​ but have inner product v1T​v2​=1. The eigenvectors are therefore nonorthogonal. When the eigenvalues coincide, only the direction v2​ remains as an eigenvector; L is a defective Jordan block.

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