Question:
Let V be a 7-dimensional vector space. If U, W are subspaces of V with dimensions 4 and 5 respectively, then which of the following is NOT a possible value of dimension of \(U \cap W\)?
Let V be a 7-dimensional vector space. If U, W are subspaces of V with dimensions 4 and 5 respectively, then which of the following is NOT a possible value of dimension of \(U \cap W\)?
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Use dim(U+W) = dim U + dim W - dim(U cap W), which must be at most dim V = 7.
Updated On: Jul 3, 2026
- 4
- 3
- 2
- 1
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The Correct Option is D
Solution and Explanation
Step 1: Use the dimension formula for subspaces of a finite-dimensional vector space: \[\dim(U+W) = \dim U + \dim W - \dim(U \cap W)\] Here \(\dim U = 4\) and \(\dim W = 5\), so \[\dim(U+W) = 9 - \dim(U\cap W)\]
Step 2: Since \(U+W\) is a subspace of V, which has dimension 7, we need \(\dim(U+W) \le 7\), that is \[9 - \dim(U\cap W) \le 7 \implies \dim(U\cap W) \ge 2\]
Step 3: Also, \(U \cap W\) is a subspace of both U and W, so \(\dim(U\cap W) \le \min(\dim U, \dim W) = \min(4,5) = 4\).
Step 4: Combining both bounds, \(2 \le \dim(U\cap W) \le 4\), so the possible values are exactly 2, 3, and 4, and each is achievable by choosing U and W to overlap in a subspace of that dimension within the 7-dimensional space.
Step 5: The value 1 is below the lower bound of 2, so \(\dim(U\cap W)=1\) is impossible. \[\boxed{1}\]
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