Let's break down the steps for **Principal Component Analysis (PCA)**: --- ### **Step 1: Write the Data as a Matrix** Let the data matrix \( X \) be: \[ X = \begin{bmatrix} 4 & 0 & -2 & 4 & 2 \\ 3 & 1 & 2 & -1 & 0 \end{bmatrix} \] --- ### **Step 2: Center the Data** Find the mean of each row (dimension): \[ \text{Mean} = \begin{bmatrix} \frac{4 + 0 + (-2) + 4 + 2}{5} \\ \frac{3 + 1 + 2 + (-1) + 0}{5} \end{bmatrix} = \begin{bmatrix} \frac{8}{5} \\ \frac{5}{5} \end{bmatrix} = \begin{bmatrix} 1.6 \\ 1 \end{bmatrix} \] Subtract the mean from each column (center the data): \[ X_{\text{centered}} = \begin{bmatrix} 4-1.6 & 0-1.6 & -2-1.6 & 4-1.6 & 2-1.6 \\ 3-1 & 1-1 & 2-1 & -1-1 & 0-1 \end{bmatrix} = \begin{bmatrix} 2.4 & -1.6 & -3.6 & 2.4 & 0.4 \\ 2 & 0 & 1 & -2 & -1 \end{bmatrix} \] --- ### **Step 3: Compute the Covariance Matrix** \[ \text{Cov}(X) = \frac{1}{n-1} X_{\text{centered}} X_{\text{centered}}^T \] where \( n = 5 \). \[ X_{\text{centered}} X_{\text{centered}}^T = \begin{bmatrix} 2.4 & -1.6 & -3.6 & 2.4 & 0.4 \\ 2 & 0 & 1 & -2 & -1 \end{bmatrix} \begin{bmatrix} 2.4 & 2 \\ -1.6 & 0 \\ -3.6 & 1 \\ 2.4 & -2 \\ 0.4 & -1 \end{bmatrix} \] Calculate each entry: - \( a_{11} = 2.4^2 + (-1.6)^2 + (-3.6)^2 + 2.4^2 + 0.4^2 = 5.76 + 2.56 + 12.96 + 5.76 + 0.16 = 27.2 \) - \( a_{12} = 2.4 \cdot 2 + (-1.6)\cdot 0 + (-3.6)\cdot 1 + 2.4 \cdot (-2) + 0.4 \cdot (-1) = 4.8 + 0 - 3.6 - 4.8 - 0.4 = -4 \) - \( a_{22} = 2^2 + 0^2 + 1^2 + (-2)^2 + (-1)^2 = 4 + 0 + 1 + 4 + 1 = 10 \) Symmetric, so \( a_{21} = a_{12} \). Therefore, \[ X_{\text{centered}} X_{\text{centered}}^T = \begin{bmatrix} 27.2 & -4 \\ -4 & 10 \end{bmatrix} \] Divide by \( n-1 = 4 \): \[ \text{Cov}(X) = \frac{1}{4} \begin{bmatrix} 27.2 & -4 \\ -4 & 10 \end{bmatrix} = \begin{bmatrix} 6.8 & -1 \\ -1 & 2.5 \end{bmatrix} \] --- ### **Step 4: Find the First Principal Component (Eigenvector with Largest Eigenvalue)** Solve \( \det(\text{Cov}(X) - \lambda I) = 0 \): \[ \begin{vmatrix} 6.8-\lambda & -1 \\ -1 & 2.5-\lambda \end{vmatrix} = (6.8-\lambda)(2.5-\lambda) - 1 = 0 \] \[ (6.8-\lambda)(2.5-\lambda) - 1 = 0 \] \[ 17 - 6.8\lambda - 2.5\lambda + \lambda^2 - 1 = 0 \] \[ \lambda^2 - 9.3\lambda + 16 = 0 \] Solve for \( \lambda \): \[ \lambda = \frac{9.3 \pm \sqrt{9.3^2 - 4 \cdot 1 \cdot 16}}{2} \] \[ = \frac{9.3 \pm \sqrt{86.49 - 64}}{2} = \frac{9.3 \pm \sqrt{22.49}}{2} = \frac{9.3 \pm 4.75}{2} \] So, - \( \lambda_1 = \frac{9.3 + 4.75}{2} = 7.025 \) - \( \lambda_2 = \frac{9.3 - 4.75}{2} = 2.275 \) Take the eigenvector for \( \lambda_1 = 7.025 \): \[ (\text{Cov}(X) - 7.025 I)\vec{v} = 0 \] \[ \begin{bmatrix} -0.225 & -1 \\ -1 & -4.525 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = 0 \] First row: \( -0.225 v_1 - v_2 = 0 \implies v_2 = -0.225 v_1 \) So eigenvector is proportional to \( \begin{bmatrix} 1 \\ -0.225 \end{bmatrix} \). Normalize it: \[ \|\vec{v}\| = \sqrt{1^2 + (-0.225)^2} = \sqrt{1 + 0.050625} = \sqrt{1.050625} \approx 1.025 \] So normalized: \[ \mathbf{u}_1 = \frac{1}{1.025} \begin{bmatrix} 1 \\ -0.225 \end{bmatrix} \approx \begin{bmatrix} 0.9756 \\ -0.2196 \end{bmatrix} \] --- ### **Step 5: Project Data onto First Principal Component** Let \( \mathbf{u}_1 \) be the first principal component. The 1D representation of each centered data point \( \mathbf{x}_j \) is: \[ y_j = \mathbf{u}_1^T (\mathbf{x}_j - \text{mean}) \] For each \( j \): - \( \mathbf{x}_1 - \text{mean} = \begin{bmatrix} 2.4 \\ 2 \end{bmatrix} \) - \( \mathbf{x}_2 - \text{mean} = \begin{bmatrix} -1.6 \\ 0 \end{bmatrix} \) - \( \mathbf{x}_3 - \text{mean} = \begin{bmatrix} -3.6 \\ 1 \end{bmatrix} \) - \( \mathbf{x}_4 - \text{mean} = \begin{bmatrix} 2.4 \\ -2 \end{bmatrix} \) - \( \mathbf{x}_5 - \text{mean} = \begin{bmatrix} 0.4 \\ -1 \end{bmatrix} \) So: - \( y_1 = 0.9756 \cdot 2.4 + (-0.2196) \cdot 2 = 2.3414 - 0.4392 = 1.9022 \) - \( y_2 = 0.9756 \cdot (-1.6) + (-0.2196)\cdot 0 = -1.561 \) - \( y_3 = 0.9756 \cdot (-3.6) + (-0.2196)\cdot 1 = -3.512 - 0.2196 = -3.7316 \) - \( y_4 = 0.9756 \cdot 2.4 + (-0.2196)\cdot (-2) = 2.3414 + 0.4392 = 2.7806 \) - \( y_5 = 0.9756 \cdot 0.4 + (-0.2196)\cdot (-1) = 0.39024 + 0.2196 = 0.6098 \) --- ## **Final Answers:** ### (a) **First Principal Component:** \[ \boxed{ \mathbf{u}_1 \approx \begin{bmatrix} 0.9756 \\ -0.2196 \end{bmatrix} } \] ### (b) **Lower-dimensional 1d representations:** \[ \boxed{ \begin{align*} y_1 &\approx 1.90 \\ y_2 &\approx -1.56 \\ y_3 &\approx -3.73 \\ y_4 &\approx 2.78 \\ y_5 &\approx 0.61 \\ \end{align*} } \] Let me know if you’d like to see a plot or further steps!