a) Let \(v\) be an eigenvector of \(A\) with eigenvalue $\lambda$, so \(A v=\lambda v\). Then,

\[B A v = B (\lambda v) = \lambda B v\].

Since \(A B=B A\), we have

\[\mathrm{ABv}=\mathrm{BAv}\],

which implies that \(B v\) is also an eigenvector of \( A \) with eigenvalue \(\lambda \).

b) Consider the \(2 \times 2\) matrices

\[A = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}\]

\[B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\].

Then, \(A\) and \(B\) commute, and both have eigenvalue 1 with eigenvector \((1,0)\) in the subspace \( \mid\) mathcal \(\{V\}_{-} 1 \). However, this vector is not an eigenvector of \(B\).

c) The statement is true by the spectral theorem, which states that any normal matrix (a matrix that commutes with its conjugate transpose) can be diagonalized by a unitary matrix. If all the eigenvalues of \(A\) are distinct, then the eigenvectors form a linearly independent set, and can be used to form a basis for the vector space. In this basis, both \( A \) and \( B \) will be diagonal.

However, if some eigenvalue of \( A \) has multiplicity greater than one, then there are linearly dependent eigenvectors, and it may not be possible to find a basis in which both \( \mathrm{~A} \) and \(\mathrm{~B} \) are diagonal. For example, consider the \(2 \times 2\) matrices

\[A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\]

\[B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\].

Both \(A\) and \(B\) commute, but there is no basis in which both are diagonal.