= Solution
Let $v_2,v_3,v_4$ be the three supplied coefficient vectors, in their displayed order. Multiplication by the <sample covariance matrix> gives
$$
\begin{aligned}
\Sigma v_2&=(408,0,-408,4080)^T=408v_2,\\
\Sigma v_3&=(300,-600,300,0)^T=300v_3,\\
\Sigma v_4&=(2040,0,-2040,-408)^T=204v_4.
\end{aligned}
$$
The four vectors are pairwise orthogonal, with norms $\sqrt3$, $\sqrt{102}$, $\sqrt6$ and $\sqrt{204}$. Hence $u_j=v_j/\|v_j\|$ form an <orthonormal basis> of <eigenvectors>. The <variance> of the <principal component> with unit coefficient vector $u_j$ is its <eigenvalue> $\lambda_j$. Using unnormalized coefficient vectors would instead give variances $\lambda_j\|v_j\|^2$ and would not produce the desired proportions.
The total <variance> is
$$
\operatorname{tr}\Sigma=904+950+904+404=3162=2250+408+300+204.
$$
Therefore the <explained variance of a principal component> is
$$
\boxed{\begin{array}{c|r|r}
\text{component}&\lambda_j&100\lambda_j/3162\\\hline
1&2250&71.16\%\\
2&408&12.90\%\\
3&300&9.49\%\\
4&204&6.45\%
\end{array}}
$$
The first two <principal components> together account for $2658/3162\simeq84.06\%$, and the first three account for $2958/3162\simeq93.55\%$ of the total <variance>.
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