Question

Then, which one of the following is TRUE?

• A. \( S \) is compact and connected
• B. \( S \) is neither compact nor connected
• C. \( S \) is compact but not connected
• D. \( S \) is connected but not compact

Hint:

When dealing with transformations, especially diagonalizable ones, the resulting set might not retain properties such as compactness and connectedness, even if the original set is confined to a bounded space like the unit sphere.

Answer 1

The Correct Option is B

We are given that \( \mathbf{w} \in \mathbb{R}^3 \) lies on the unit sphere, i.e., \( \| \mathbf{w} \| = 1 \). This implies that the set \( S \) represents all unit vectors in \( \mathbb{R}^3 \), which form the surface of a sphere.

However, the diagonalizability condition changes the context of the problem. This condition implies that the transformation described by the matrix is not necessarily related to a closed, bounded space like the unit sphere, but instead represents a linear transformation that may result in a set that is neither compact nor connected. This typically happens when the transformation leads to vectors that can "stretch" or "move apart," breaking compactness and connectedness.

Step 1: Compactness
A set is compact if it is both closed and bounded.
The condition that \( \| \mathbf{w} \| = 1 \) suggests that \( S \) should be bounded.
However, the diagonalizability condition implies that this transformation leads to non-compactness, as the transformation might extend beyond the unit sphere's boundary.

Step 2: Connectedness
A set is connected if any two points in the set can be joined by a continuous path.
The transformation described by diagonalizability may separate points in the set, leading to a situation where the set is no longer connected.

Thus, combining these observations, we conclude that the set \( S \) is neither compact nor connected. Hence, the correct answer is:

\[ \boxed{S \text{ is neither compact nor connected.}} \]