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If V1 and V2 are 3-dimensional subspaces of a 4-dimensional vector space V, then the smallest possible dimension of V1 ∩ V2 is _______.
  • a)
    1
  • b)
    2
  • c)
    3
  • d)
    4
Correct answer is option 'C'. Can you explain this answer?
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If V1and V2are 3-dimensional subspaces of a 4-dimensional vector space...
Explanation:

Dimension of the Intersection of Two Subspaces:
- The dimension of the intersection of two subspaces V1 and V2 is given by dim(V1∩V2) = dim(V1) + dim(V2) - dim(V1+V2).
- Here, V1 and V2 are both 3-dimensional subspaces of a 4-dimensional vector space V.
- Therefore, dim(V1) = 3 and dim(V2) = 3.

Calculating Dimension of V1 + V2:
- Since V1 and V2 are subspaces of V, their sum V1 + V2 is also a subspace of V.
- By the dimension theorem, dim(V1 + V2) = dim(V1) + dim(V2) - dim(V1∩V2).
- Substituting the known values, dim(V1 + V2) = 3 + 3 - dim(V1∩V2) = 6 - dim(V1∩V2).

Conclusion:
- We know that the dimension of V1 + V2 cannot exceed 4 since it is a subspace of a 4-dimensional vector space V.
- Therefore, 6 - dim(V1∩V2) ≤ 4.
- Solving for dim(V1∩V2), we get dim(V1∩V2) ≥ 2.
- Hence, the smallest possible dimension of V1∩V2 is 2.
Therefore, the correct answer is option b) 2.
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If V1and V2are 3-dimensional subspaces of a 4-dimensional vector space V, then the smallest possible dimension of V1∩ V2is _______.a)1b)2c)3d)4Correct answer is option 'C'. Can you explain this answer?
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