Question

Let \(A\) and \(B\) be \(n \times n\) complex matrices that commute: \(A B=B A\). If \(\lambda\) is an eigenvalue of \(A\), let \(\mathcal{V}_{\lambda}\) be the subspace of all eigenvectors having this eigenvalue.
a) Show there is an vector \(v \in \mathcal{V}_{\lambda}\) that is also an eigenvector of \(B\), possibly with a different eigenvalue.
b) Give an example showing that some vectors in \(\mathcal{V}_{\lambda}\) may not be an eigenvectors of B.
c) If all the eigenvalues of \(A\) are distinct (so each has algebraic multiplicity one), show that there is a basis in which both \(A\) and \(B\) are diagonal. Also, give an example showing this may be false if some eigenvalue of \(A\) has multiplicity greater than one.

Answer 1

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.