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Let $Dx=\lambda x$ with $x\ne 0$. Commutation givesThus $Cx$ belongs to the eigenspace $E_{\lambda }(D)$, with the zero vector allowed. Since $D$ has $n$ distinct eigenvalues, each eigenspace is one-dimensional. Hence $Cx=\mu x$ for some scalar $\mu$. This is the fact that commuting maps preserve eigenspaces.

The standard linear independence theorem for eigenvectors belonging to distinct eigenvalues supplies a basis $x_1,\dots ,x_n$ of eigenvectors of $D$. The preceding argument makes every one an eigenvector of $C$ as well. Their column matrix therefore simultaneously diagonalizes $C$ and $D$. This is simple spectrum gives simultaneous diagonalization.

For real matrices, this conclusion always holds over $\mathbb{C}$; a real common basis follows when the distinct eigenvalues of $D$ are real. Real matrix entries alone do not ensure a real eigenbasis: a quarter-turn of ${\mathbb{R}}^2$, with $C=D$, has eigenvalues $\pm i$. The matrices in this question have real spectra, so their common basis is real.

Let $A{v}_j=a_j{v}_j$. Commutation gives $A(B{v}_j)=B(A{v}_j)=a_{j}B{v}_j$, so $B$ preserves each eigenspace of $A$. Since the eigenvalues of $A$ are nondegenerate, each eigenspace is one-dimensional and $B{v}_j=b_j{v}_j$. The spectral theorem for a Hermitian matrix lets us choose these vectors as an orthonormal eigenbasis. With $U=(v_1,\dots ,v_n)$,This is simple spectrum gives simultaneous diagonalization. The common eigenvectors can be chosen up to their phases and ordering. In fact degeneracy does not prevent a common eigenbasis for commuting Hermitian matrices, but the one-dimensional argument already suffices here.

The convergent matrix exponential series respects similarity, so $e^A=U{e}^{{\Lambda }_A}{U}^{†}$ and likewise for $B$. Scalar exponentials multiply on each diagonal entry, givingThe essential hypothesis is commutation; Hermiticity supplies this convenient diagonalization proof.

Suppose first that $AB=BA$, and take an eigenvector $v_j$ of $A$, with $A{v}_j=a_j{v}_j$. Commutation givesSince the eigenvalue $a_j$ is simple, its eigenspace is one-dimensional, so $B{v}_j=b_j{v}_j$ for some scalar $b_j$. Thus every vector in an eigenbasis of $A$ is also an eigenvector of $B$. The simple spectrum of $B$ ensures these account for its eigenvector directions as well. Eigenvectors are understood up to nonzero scaling and permutation, rather than as literally identical numerical vectors.

Conversely, if $v_1,\dots ,v_n$ are a common eigenbasis with $A{v}_j=a_j{v}_j$ and $B{v}_j=b_j{v}_j$, thenfor every basis vector. Hence $AB-BA=0$. ThereforeThis is simple spectrum gives simultaneous diagonalization. Without simple spectra, commuting Hermitian matrices still possess a common orthonormal eigenbasis, but not every arbitrary eigenvector of one matrix must be an eigenvector of the other.

Work over the field containing the stated $n$ distinct eigenvalues of $B$; in a real formulation these must be real to obtain real eigenvectors. Choose one eigenvector for each distinct eigenvalue. By independence of eigenvectors for distinct eigenvalues, these $n$ vectors are a basis. Consequently each eigenspace $E_{\lambda }(B)$ has dimension one: in an expansion in this eigenbasis, $Bw=\lambda w$ forces every coefficient with a different eigenvalue to vanish.

Now let $v\ne 0$ be any eigenvector of $B$. By part (d), $Av$ belongs to its one-dimensional eigenspace, so $Av=\mu v$ for some scalar $\mu$. This proves that $v$ is an eigenvector of $A$, including the case $Av=0$, where $\mu =0$. This is the crucial one-dimensional step in simple spectrum gives simultaneous diagonalization.