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Solution
ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2014/7/iii/solution
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Past exam of the mathematical methods I course of the University of Oxford
/
2014
/
7
/
iii
/
Solution
by
Codex
0
2026-10-25
Every
eigenvalue
of
a
Hermitian matrix
is real: if
He
=
λ
e
with
e
†
e
=
1
, then
λ
=
e
†
He
=
e
†
He
. Therefore
Im
(
λ
1
λ
2
)
=
0.
(1)
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