= Solution
The printed word “eigenvalues” must read “eigenvectors”: <eigenvalues> are real numbers, whereas a <basis> of $\mathbb R^n$ consists of <vectors>. The intended assertion is
$$
\boxed{A\text{ is diagonalizable over }\mathbb R\iff\mathbb R^n\text{ has a basis of eigenvectors of }A.}
$$
This wording error is present in the original scan.<Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation examination paper by the University of Oxford (2005)>[Printed p. 2, Q2(a), “basis consisting of eigenvalues”]{ref}
If $X^{-1}AX=D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$ for an <invertible matrix> $X$, write $x_i$ for its columns. The <matrix> identity $AX=XD$ says $Ax_i=\lambda_ix_i$. The columns of an <invertible matrix> are a <basis>, hence they give the required <basis> of <eigenvectors>.
Conversely, put a <basis> of real <eigenvectors> $x_1,\ldots,x_n$ into the columns of $X$, with corresponding <eigenvalues> $\lambda_i$. The <basis> property makes $X$ an <invertible matrix>. The equations $Ax_i=\lambda_ix_i$ give $AX=XD$, so $X^{-1}AX=D$. This is precisely <diagonalizability> over the real numbers.
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