Data: $X =\begin{bmatrix}2&0\\0&1\\0&2\end{bmatrix}$, $n=3$, $d=2$

Step 1: Mean center. $\bar x = [(2+0+0)/3, (0+1+2)/3]=[0.667,1]$

Centered $X_c =\begin{bmatrix}1.333&-1\\-0.667&0\\-0.667&1\end{bmatrix}$

Step 2: Covariance $C = \frac1{n-1} X_c^T X_c$ (unbiased; if $1/n$ divide by 3, mention both).

$X_c^T X_c =\begin{bmatrix}1.333^2+0.667^2+0.667^2 &1.333*(-1)+(-0.667)*0+(-0.667)*1\\ sym &1+0+1\end{bmatrix}$
$= \begin{bmatrix}2.667 & -2\\ -2 &2\end{bmatrix}$

Hence:

• Unbiased ($1/(n-1)=1/2$): $C =\begin{bmatrix}1.333 & -1\\ -1 &1\end{bmatrix}$
• Biased ($1/n$): $C_{biased}=\begin{bmatrix}0.889 & -0.667\\ -0.667 &0.667\end{bmatrix}$ (scale by $2/3$)

Step 3: Eigenvalues solve $|C-\lambda I|=0$:

$(1.333-\lambda)(1-\lambda)-1=0 \Rightarrow \lambda^2 -2.333\lambda+0.333=0$
$\lambda = \frac{2.333\pm\sqrt{5.444-1.333}}{2}= \frac{2.333\pm2.028}{2}$
$\lambda_1 \approx 2.18$, $\lambda_2 \approx 0.152$ (Unbiased)
For biased version: $\lambda_1\approx1.454$, $\lambda_2\approx0.102$ (2/3 times).

Explained variance: $93.5\%$ by PC1, $6.5\%$ by PC2.

Step 4: Eigenvectors (PCs):
For $\lambda_1=2.18$: $(C-\lambda I)v=0\Rightarrow -0.847 v_1 -1 v_2=0\Rightarrow v\propto[1,-0.847]$ normalized $v_1\approx[0.763,-0.647]$
For $\lambda_2=0.152$: $v_2\approx[0.647,0.763]$ orthogonal.

Normalized PCs (unit): $PC1\approx [-0.76,0.65]$ direction of max variance (roughly along $x-y$ diagonal), $PC2$ orthogonal.

Projection $Z=X_c V$ gives reduced 1D representation capturing $>93\%$ variance. First PC is the principal component.