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They argue that if A and B are complex square matrices of order n which commute, and if all eigenvalues of B are distinct, then given any eigenvector of B, say x, corresponding to an eigenvalue u, Ax is also an eigenvector of B. This apparently simply follows from the fact that B(Ax)=A(Bx) = u (Ax).

However they skip mentioning why Ax cannot be the zero vector. Is this obvious? I am clearly missing something.