= Solution
Let $Dx=\lambda x$ with $x\ne0$. Commutation gives
$$
D(Cx)=C(Dx)=\lambda Cx.
$$
Thus $Cx$ belongs to the <eigenspace> $E_\lambda(D)$, with the zero vector allowed. Since $D$ has $n$ distinct <eigenvalues>, each <eigenspace> is one-dimensional. Hence \b[$Cx=\mu x$ 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> $x_1,\ldots,x_n$ 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.
Back to article page
