 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/b/solution

S = (T + T^{T}) /2

and

A = (T - T^{T}) /2

. The transformed transpose satisfies

(T^{'})^{T} = (MT M^{T})^{T} = M T^{T} M^{T}

. Therefore

S^{'} = \frac{1}{2} (T^{'} + (T^{'})^{T}) = M \frac{1}{2} (T + T^{T}) M^{T} = MS M^{T},

and similarly

A^{'} = \frac{1}{2} (T^{'} - (T^{'})^{T}) = M \frac{1}{2} (T - T^{T}) M^{T} = M A M^{T} .

Both obey the tensor component transformation law. Thus the symmetric and antisymmetric parts are themselves second-order tensors. Taking a transpose and forming linear combinations commute with this orthogonal change of frame.

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