GATE CS 2014 Set 3 Q.15 (CS Q.5): Linear Algebra

If $V_1$ and $V_2$ are 4-dimensional subspaces of a 6-dimensional vector space $V$ , then the smallest possible dimension of $V_1 \cap V_2$ is _____.

Your answer

Official answer and solution

Solution

Apply Grassmann dimension theorem for vector subspaces to minimize the dimension of the intersection.

By Grassmann's dimension formula for any two subspaces $V_1$ and $V_2$ of a vector space $V$ :
$\dim(V_1 + V_2) = \dim(V_1) + \dim(V_2) - \dim(V_1 \cap V_2).$
Rearranging for the dimension of the intersection:
$\dim(V_1 \cap V_2) = \dim(V_1) + \dim(V_2) - \dim(V_1 + V_2).$
Given:
- $\dim(V_1) = 4$
- $\dim(V_2) = 4$
- $V_1 + V_2$ is a subspace of $V$ , so $\dim(V_1 + V_2) \le \dim(V) = 6$ .
To minimize $\dim(V_1 \cap V_2)$ , we maximize $\dim(V_1 + V_2)$ :
$\dim(V_1 \cap V_2) = 4 + 4 - \dim(V_1 + V_2) = 8 - \dim(V_1 + V_2) \ge 8 - 6 = 2.$

Therefore, the smallest possible dimension of $V_1 \cap V_2$ is $2$ .

Final answer

Common trap

Assuming the subspaces can be disjoint (dimension 0), forgetting that their dimensions sum to 8 inside a 6-dimensional space.

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