 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/6/d/solution

P = \frac{1}{3} tr F = \frac{4}{3} r^{2} .

Subtracting this isotropic part from the symmetric tensor in part (c) gives

S = \frac{1}{3} - x_{1}^{2} - 4 x_{2}^{2} 3 x_{1} x_{2} 0 3 x_{1} x_{2} - 4 x_{1}^{2} - x_{2}^{2} 0 00 5 r^{2} .

Its trace is zero. The antisymmetric tensor part is

A = \frac{1}{2} (F - F^{T}) = 0 r^{2} - x_{1} + x_{2} - r^{2} 0 x_{1} + x_{2} x_{1} - x_{2} - x_{1} - x_{2} 0 .

Adding

P I + S + A

reconstructs every entry of

F

Subtracting a scalar multiple of the identity preserves every eigenvector: if

S v = λ v

, then

(S - P I) v = (λ - P) v

. Thus the principal axes are the same as in part (c), with shifted principal values

- r^{2} /3 along e_{r}, - 4 r^{2} /3 along e_{θ}, 5 r^{2} /3 along e_{3} .

r = 0

both vanish, so neither specifies unique axes.

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