a.

• \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0.25 & -0.25 \\ 0 & -0.25 & 0.25 \end{bmatrix}\)

b.

• \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0.25 & -0.25 \\ 0 & -0.25 & 0.25 \end{bmatrix} \begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix} \)
• normalized is \(\frac{ \begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}} {1} \)
• first eigenvector is \(\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}\)
• first eigenvalue is 1

c.

• \(B = A - \lambda vv^T = \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0.25 & -0.25 \\ 0 & -0.25 & 0.25 \end{bmatrix} - \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \(\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0.25 & -0.25 \\ 0 & -0.25 & 0.25 \end{bmatrix}\)

d.

e and f.

2.

a.

\(\begin{bmatrix} 4.26 & 0.323 & 0.323 \\ 3.14 & 0.238 & 0.561 \\ 2.0 & 0.152 & 0.713 \\ 1.92 & 0.146 & 0.859 \\ 1.03 & 0.078 & 0.936 \\ 0.84 & 0.064 & 1.0 \end{bmatrix} \)

b.

The first two principal directions correspond to the first two eigenvalues (largest 2). The variance from these two directions is the sum of these eigenvalues \(4.26 + 3.14 = 7.4\)

c.

The number of eigenvectors needed to explain at least 80% of the variance is 4. This is shown in the table above by the cumulative proportion of variance captured, which is 85.9% at 4 eigenvalues.