= Solution
Let $I$ denote the identity endomorphism. The two-root <minimal polynomial> gives $(\alpha-\lambda_1I)(\alpha-\lambda_2I)=0$. Every <vector> has the decomposition
$$
v=\frac{(\alpha-\lambda_2I)v}{\lambda_1-\lambda_2}+\frac{(\alpha-\lambda_1I)v}{\lambda_2-\lambda_1}.
$$
The first summand lies in the <eigenspace> $E_1=\ker(\alpha-\lambda_1I)$, and the second in $E_2$. If a <vector> belongs to both, subtracting its two <eigenvalue> equations gives $(\lambda_1-\lambda_2)v=0$, hence $v=0$. Thus $V=E_1\oplus E_2$.
For distinct real roots $\lambda_1,\ldots,\lambda_m$, define <polynomial> projectors
$$
P_i=\prod_{j\ne i}\frac{\alpha-\lambda_jI}{\lambda_i-\lambda_j}.
$$
The <Lagrange interpolation> <polynomials> in this expression sum to one: their sum has degree at most $m-1$ and equals one at the $m$ distinct roots. Hence $\sum_iP_i=I$. Also $(\alpha-\lambda_iI)P_i=0$, because its numerator is the <minimal polynomial> evaluated at $\alpha$. Thus every <vector> is a sum of <eigenvectors>, $v=\sum_iP_iv$. On $E_j$, $P_i$ acts as $\delta_{ij}I$, so applying $P_i$ to any zero sum of <vectors> from the <eigenspaces> shows that every summand is zero. Therefore
$$
\boxed{V=\bigoplus_{i=1}^m\ker(\alpha-\lambda_iI)}.
$$
Choosing a basis in each <eigenspace> proves <diagonalizability> over the reals. Conversely, if a real <matrix> is similar over the reals to a diagonal <matrix>, a <polynomial> annihilates it exactly when it vanishes at each distinct real diagonal entry. Its <minimal polynomial> is therefore the product of those distinct linear factors. We have proved \b[real diagonalizability is equivalent to a <minimal polynomial> with distinct real roots].
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