DSC 140B
Problems tagged with centering

Problems tagged with "centering"

Problem #061

Tags: pca, centering, covariance, quiz-04, lecture-06

Center the data set:

$$\begin{align*}\vec x^{(1)}&= (2, 4)^T \\\vec x^{(2)}&= (4, 2)^T \\\vec x^{(3)}&= (6, 6)^T \end{align*}$$
Solution

First, compute the mean:

\[\bar{\vec x} = \frac{1}{3}\left[(2, 4)^T + (4, 2)^T + (6, 6)^T\right]= \frac{1}{3}(12, 12)^T = (4, 4)^T \]

Then subtract the mean from each data point:

$$\begin{align*}\tilde{\vec x}^{(1)}&= \vec x^{(1)} - \bar{\vec x} = (2, 4)^T - (4, 4)^T = (-2, 0)^T \\\tilde{\vec x}^{(2)}&= \vec x^{(2)} - \bar{\vec x} = (4, 2)^T - (4, 4)^T = (0, -2)^T \\\tilde{\vec x}^{(3)}&= \vec x^{(3)} - \bar{\vec x} = (6, 6)^T - (4, 4)^T = (2, 2)^T \end{align*}$$

Problem #062

Tags: pca, centering, covariance, quiz-04, lecture-06

Center the data set:

$$\begin{align*}\vec x^{(1)}&= (4, -2, 2)^T \\\vec x^{(2)}&= (2, 3, 0)^T \\\vec x^{(3)}&= (3, -1, 1)^T \end{align*}$$
Solution

First, compute the mean:

\[\bar{\vec x} = \frac{1}{3}\left[(4, -2, 2)^T + (2, 3, 0)^T + (3, -1, 1)^T\right]= \frac{1}{3}(9, 0, 3)^T = (3, 0, 1)^T \]

Then subtract the mean from each data point:

$$\begin{align*}\tilde{\vec x}^{(1)}&= \vec x^{(1)} - \bar{\vec x} = (4, -2, 2)^T - (3, 0, 1)^T = (1, -2, 1)^T \\\tilde{\vec x}^{(2)}&= \vec x^{(2)} - \bar{\vec x} = (2, 3, 0)^T - (3, 0, 1)^T = (-1, 3, -1)^T \\\tilde{\vec x}^{(3)}&= \vec x^{(3)} - \bar{\vec x} = (3, -1, 1)^T - (3, 0, 1)^T = (0, -1, 0)^T \end{align*}$$