Past exam of the mathematical methods I course of the University of Oxford / 2015 / 13 / b / ii / Solution

The second gradient is ${\nabla }_{v}b=[A+B-(\alpha +\beta )I]v$. Combining its vanishing with $(A-\alpha I)v=0$ givesThe zero vector always works. A nonzero common stationary vector exists exactly when $A,B$ have a common eigenvector with respective eigenvalues $\alpha ,\beta$; both stationary values are then zero. Commuting symmetric matrices have a common orthonormal eigenbasis, so commutation is sufficient to find such vectors for the paired eigenvalues in that basis. It does not guarantee an arbitrary prescribed pair.

Commutation is also not necessary for one common eigenvector. For example,are real symmetric matrices sharing $e_1$ with eigenvalues $1,\beta$, but $AB\ne BA$. This separates the question's common-vector condition from the stronger simultaneous diagonalization of the entire space.