1 like 0 dislike
357 views
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.
| 357 views

0 like 0 dislike
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.
by Diamond (50,339 points)

1 like 0 dislike
1 like 0 dislike
1 like 0 dislike
1 like 0 dislike
1 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike