= Solution
\b[The <all-ones matrix> is diagonalizable for every $n\geq1$.] The vector $u$ together with the $n-1$ vectors from part (ii) is an <eigenbasis>. Independence follows because a multiple of $u$ has coordinate sum $n$ times that multiple, whereas every other <basis> vector has coordinate sum zero. In this <basis> the <diagonal matrix> is $\operatorname{diag}(n,0,\ldots,0)$. At $n=1$ the one-vector <basis> suffices. Alternatively, the <spectral theorem for real symmetric matrices> applies. If dimension zero is allowed by the convention for $\mathbb N$, the empty $0\times0$ <matrix> has no <eigenvalues>, an empty <basis> and a vacuous <diagonalization>.
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