= Solution
The given <equicorrelation covariance matrix> can be written
$$
\Sigma=(1-r)I_4+r\mathbf1\mathbf1^T.
$$
On $\mathbf1=(1,1,1,1)^T$, the rank-one matrix $\mathbf1\mathbf1^T$ acts as multiplication by four. On the three-dimensional orthogonal complement $\{v:\sum_jv_j=0\}$ it acts as zero. Hence the <eigenvalues> are \b[$1+3r$ and $1-r$ with multiplicity three].
For $0<r<1$, the largest <eigenvalue> is $1+3r$ and its normalized <eigenvector> is $\mathbf1/2$. Therefore the first <population principal component> is
$$
\boxed{Y_1=\tfrac12\sum_{j=1}^4(X_j-\mu_j),\qquad
\operatorname{Var}(Y_1)=1+3r.}
$$
Omitting the centering merely changes the component by a constant and has no effect on its <variance>. Since $\operatorname{tr}\Sigma=4$, its explained fraction is
$$
\boxed{\frac{1+3r}{4}.}
$$
It rises from one quarter towards one as the positive common correlation increases. Outside the specified range, a valid covariance requires $-1/3\le r\le1$; for negative $r$ the sum direction is no longer the leading one.
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