Proof. By the spectral decomposition of Section 1.9, \(\Sigma = P\Lambda P'\) with \(P = (\underline {e}_1,\dots ,\underline {e}_p)\) orthogonal and \(\Lambda = \operatorname {diag}(\lambda _1,\dots ,\lambda _p)\). Put \(\underline {b} = P'\underline {a}\); then \(\underline {b}'\underline {b} = \underline {a}'PP'\underline {a} = \underline {a}'\underline {a} = 1\), and \[\underline {a}'\Sigma \underline {a} = \underline {a}'P\Lambda P'\underline {a} = \underline {b}'\Lambda \underline {b} = \sum ^{p}_{i=1}\lambda _i b_i^{2}.\] This is a weighted average of the \(\lambda _i\) with weights \(b_i^{2}\) summing to one, so it is at most \(\lambda _1\), with equality when \(b_1=1\) and all other \(b_i=0\) — that is, when \(\underline {a}=\underline {e}_1\). Hence \(\operatorname {var}(Y_1)=\lambda _1\) is attained at \(\underline {e}_1\).

For the \(i\)th component, the constraint \(\operatorname {cov}(Y_i,Y_k)=0\) for \(k<i\) forces \(b_1=\cdots =b_{i-1}=0\), and the same argument on the remaining coordinates gives the maximum \(\lambda _i\) at \(\underline {a}=\underline {e}_i\). Finally \(\operatorname {cov}(Y_i,Y_k)=\underline {e}_i'\Sigma \underline {e}_k =\lambda _k\underline {e}_i'\underline {e}_k = 0\) by orthonormality. □