Let S and T denote linear transformations of a real finite dimensional vector space V. Suppose S(T(v))=T(S(v)) for all v that is in V.

(a) Prove that T maps each eigensapce of S onto itself.

(b) Suppose that each S and T may be represented by diagonal matrices. Prove that there exists a basis for V with respect to which both S and T are given by diagonal matrices simultaneously.

i can do part a) but i need a help on b). i dont even know if a) could be useful to prove b). please help.

The key to (b) is (a). If S is diagonalizable, then there exist a basis for the vector space in which S is diagonal. But apply S to each of those basis vectors multiplies each of the basis vectors by the corresponding diagonal value: A matriix is diagonalizable if and only if there exist a basis consisting entirely of eigenvectors of that matrix. Each eigenspace of S is the space spanned by a single basis vector. Now use (a): T maps each eigenspace of S onto itself: that is it maps each basis vector to multiple of itself.

i have a question. If S has the complete set of eigenvectors, does it mean that each eigenvenctor corresponds to distinct eigen value? what if an eigenvalue has the multiplicity of 2 and two eigenvectors correspond to it? then the eigenspace will be spanned by these two vectors. in that case does T still take each basis vector to a multiple of itself?