DSC 140B

Problem #058

Suppose the direction of maximum variance in a centered data set is

\[\vec u = \frac{1}{\sqrt 2}(1, 1)^T \]

Let \(\vec x = (2, 4)^T\) be a centered data point.

Reduce \(\vec x\) to one dimension by projecting onto the direction of maximum variance. What is the new feature \(z\) obtained from this projection?

Solution

\(z = 3\sqrt 2\).

The projection onto the direction of maximum variance is given by the dot product with \(\vec u\):

\[ z = \vec x \cdot\vec u = (2, 4) \cdot\frac{1}{\sqrt 2}(1, 1) = \frac{2 + 4}{\sqrt 2} = \frac{6}{\sqrt 2} = 3\sqrt 2 \]

Suppose the direction of maximum variance in a centered data set is

\[\vec u = \left(\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\right)^T \]

Let \(\vec x = (2, 1, 2)^T\) be a centered data point.

Reduce \(\vec x\) to one dimension by projecting onto the direction of maximum variance. What is the new feature \(z\) obtained from this projection?

Solution

\(z = 3\).

The projection onto the direction of maximum variance is given by the dot product with \(\vec u\):

\[ z = \vec x \cdot\vec u = (2, 1, 2) \cdot\left(\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\right)= \frac{4 + 1 + 4}{3} = 3 \]

Suppose the direction of maximum variance in a centered data set is

\[\vec u = \frac{1}{2}(1, 1, 1, 1)^T \]

Let \(\vec x = (3, 1, -1, 5)^T\) be a centered data point.

Reduce \(\vec x\) to one dimension by projecting onto the direction of maximum variance. What is the new feature \(z\) obtained from this projection?

Solution

\(z = 4\).

The projection onto the direction of maximum variance is given by the dot product with \(\vec u\):

\[ z = \vec x \cdot\vec u = (3, 1, -1, 5) \cdot\frac{1}{2}(1, 1, 1, 1) = \frac{3 + 1 - 1 + 5}{2} = 4 \]