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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.
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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 (64.2k points)

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