Question

 Determine the principal components Y1, Y2 and Y3 for the covariance matrix:



equation

Also calculate the proportion of total population variance for the first principal component.

10 Jan 2026
Answer :
Word Count : 893
We are asked to find the principal components of the covariance matrix [ \Sigma = \begin{pmatrix} 1 & -2 & 0 \ -2 & 5 & 0 \ 0 & 0 & 1 \end{pmatrix} ] and the proportion of total variance for the first principal component. --- Step 1: Find the eigenvalues of Σ Eigenvalues satisfy (\det(\Sigma - \lambda I) = 0). [ \Sigma - \lambda I = \begin{pmatrix} 1-\lambda & -2 & 0 \ -2 & 5-\lambda & 0 \ 0 & 0 & 1-\lambda \end{pmatrix} ] Determinant: [ \det(\Sigma - \lambda I) = (1-\lambda)\begin{vmatrix}5-\lambda & 0 \ 0 & 1-\lambda\end{vmatrix} - (-2)\begin{vmatrix}-2 & 0 \ 0 & 1-\lambda\end{vmatrix} + 0 ] Step by step: [ (1-\lambda)[(5-\lambda)(1-\lambda) - 0] - (-2)[(-2)(1-\lambda) - 0] = (1-\lambda)(5-\lambda)(1-\lambda) + 4(1-\lambda) ] [ = (1-\lambda)[(5-\lambda)(1-\lambda) + 4] ] Expand ((5-\lambda)(1-\lambda) = 5 - 5\lambda - \lambda + \lambda^2 = \lambda^2 - 6\lambda +5) Add 4: (\lambda^2 - 6\lambda + 9 = (\lambda - 3)^2) So determinant: ((1-\lambda)(\lambda-3)^2 = 0) Thus, eigenvalues: [ \lambda_1 = 1, \quad \lambda_2 = 3, \quad \lambda_3 = 3 ] --- Step 2: Find eigenvectors for each eigenvalue (a) For (\lambda = 1): ((\Sigma - I)v = 0) [ \begin{pmatrix} 0 & -2 & 0 \ -2 & 4 & 0 \ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix}x\y\z\end{pmatrix} = 0 ] Equations: 1. (-2y = 0 \implies y=0) 2. (-2x + 4y = -2x = 0 \implies x=0) 3. (0 = 0 \implies z) free So eigenvector: (v_1 = (0,0,1)^T) --- (b) For (\lambda = 3): ((\Sigma - 3I)v = 0) [ \begin{pmatrix}-2 & -2 & 0 \ -2 & 2 & 0 \ 0 & 0 & -2 \end{pmatrix} \begin{pmatrix}x\y\z\end{pmatrix} = 0 ] Equation from third row: (-2 z = 0 \implies z=0) First two rows: 1. (-2x - 2y =0 \implies x + y =0 \implies y=-x) 2. (-2x + 2y=0 \implies -2x + 2(-x)= -4x=0 \implies x=0) Wait, this seems inconsistent. Let's carefully solve: Matrix: [ \begin{pmatrix}-2 & -2 \ -2 & 2\end{pmatrix} ]
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