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Multiple choice
Find the larger eigenvalue of the Hermitian matrix
A
=
[
3
1
1
3
]
A = \begin{bmatrix}3 & 1 \\ 1 & 3\end{bmatrix}
A
=
[
3
1
1
3
]
. (Hint: eigenvalues are
(
t
r
±
t
r
2
−
4
det
)
/
2
(\mathrm{tr} \pm \sqrt{\mathrm{tr}^2 - 4\det})/2
(
tr
±
tr
2
−
4
det
)
/2
with
t
r
=
6
\mathrm{tr} = 6
tr
=
6
,
det
=
8
\det = 8
det
=
8
.)
2
6 (the trace)
3 (a diagonal entry)
4
Check answer