Past exam of the mathematical and Computational Biology course of the University of Cambridge / 2024 / Paper 1 / E8 / b / Solution

Place observations in rows of the data matrix $X$. Since the coordinate sample means are zero, the Gram matrix of its columns isThe sample covariance matrix is $S=X^{T}X/(3-1)$; using divisor three instead changes eigenvalues by a common factor and leaves principal component loadings unchanged. For $X^{T}X$, the characteristic polynomial isFor the larger eigenvalue $150$, the eigenvector equation gives $-36\alpha -48\beta =0$, so $3\alpha +4\beta =0$. Normalize to unit Euclidean norm and choose $v_1=(4/5,-3/5{)}^T$. An orthogonal unit eigenvector for $50$ is $v_2=(3/5,4/5{)}^T$. ThereforeThe sample variances of these principal components are $75$ and $25$, respectively. Their principal component scores for observations $A,B,C$ areEither eigenvector may be multiplied by $-1$ without changing its principal component direction or its variance. No coordinate standardization is applied here: that would instead be principal component analysis on a correlation matrix.