Question

If U and W are distinct 4-dimensional subspaces of a vector space V of dimension 6, then the possible dimensions of \( U \cap W \) is:

• A. 1 or 2
• B. exactly 4
• C. 3 or 4
• D. 2 or 3

Hint:

Remember the dimension formula for subspaces: \( \dim(U+W) = \dim(U) + \dim(W) - \dim(U \cap W) \). This formula is fundamental for solving problems involving intersections and sums of subspaces. Always use the fact that the dimension of a subspace cannot exceed the dimension of the containing space.

Answer 1

The Correct Option is D

Step 1: Introduction
This problem deals with subspace dimensions, their sum, and their intersection. We will use the formula for the dimension of the sum of two subspaces.

Step 2: Core Formula and Given Information
The formula for the sum and intersection of subspaces U and W is: \[ \dim(U+W) = \dim(U) + \dim(W) - \dim(U \cap W) \]We are given: - \(\dim(V) = 6\) - \(\dim(U) = 4\) - \(\dim(W) = 4\) - U and W are distinct subspaces.

Step 3: Solution
Rearrange the formula to find the dimension of the intersection: \[ \dim(U \cap W) = \dim(U) + \dim(W) - \dim(U+W) \]Substitute the known values: \[ \dim(U \cap W) = 4 + 4 - \dim(U+W) = 8 - \dim(U+W) \]Determine possible values for \(\dim(U+W)\). - The sum \(U+W\) is a subspace of V, so \(\dim(U+W) \le \dim(V) = 6\). - Since U and W are subspaces of \(U+W\), \(\dim(U+W) \ge \dim(U) = 4\) and \(\dim(U+W) \ge \dim(W) = 4\). - U and W are distinct, meaning \(\dim(U+W)>\dim(U)\). Thus, possible integer values for \(\dim(U+W)\) are 5 and 6. Find the corresponding values for \(\dim(U \cap W)\): - If \(\dim(U+W) = 5\), then \(\dim(U \cap W) = 8 - 5 = 3\). - If \(\dim(U+W) = 6\), then \(\dim(U \cap W) = 8 - 6 = 2\). Therefore, the possible dimensions of \(U \cap W\) are 2 or 3.

Step 4: Answer
The possible dimensions of the intersection are 2 or 3.