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

To determine which statement about the set \( S \) is true, we need to analyze the provided conditions.

The set \( S \) is defined as:

S = \left\{ \mathbf{w} = \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix} \in \mathbb{R}^3 : \begin{bmatrix} w_1 & w_2 & w_3 \end{bmatrix} \begin{bmatrix} -1 \\ 3 \\ 2 \end{bmatrix} \text{ is diagonalizable and } \|\mathbf{w}\| = 1 \right\}

where:

\|\mathbf{w}\| = \left(w_1^2 + w_2^2 + w_3^2\right)^{\frac{1}{2}}

1. Analyzing Diagonalizability:
   • A matrix is diagonalizable if there exist enough linearly independent eigenvectors.
   • The product \begin{bmatrix} w_1 & w_2 & w_3 \end{bmatrix} \begin{bmatrix} -1 \\ 3 \\ 2 \end{bmatrix} is a scalar, which inherently implies the matrix is always diagonalizable as it is effectively 1-dimensional.
2. Condition on Norm:
   • The condition \|\mathbf{w}\| = 1 describes a unit sphere in \(\mathbb{R}^3\).
3. Compactness:
   • A set is compact in \(\mathbb{R}^n\) if it is closed and bounded.
   • The condition \|\mathbf{w}\| = 1 describes a bounded and closed surface (sphere), so this condition alone would be compact.
4. Connectedness:
   • A set is connected if there are no separate parts. The condition given does not necessarily ensure connectivity if parts are omitted or spliced due to the 'diagonalizable' condition in terms of selection of \(\mathbf{w}\).

Given these insights, the main issue is the 'diagonalizability' of a product of vectors which inherently affects connectivity aspect without influencing compactness, but results in breakdown of connectivity.

The correct answer is: \( S \) is neither compact nor connected, because the additional constraint, even though affecting the connectivity, may not strictly influence the defined bounded characteristic, given the specific matrix condition for specific \(\mathbf{w}\) choices.