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35 votes
35 votes
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 _____.

2 Answers

Best answer
70 votes
70 votes
A $6$-dimensional vector space $\{ a_{1},a_{2},a_{3},a_{4},a_{5},a_{6}\}$

Let $V_1$ be $\{a_{1},a_{2},a_{3},a_{4}\}$ and $V_2$ be $\{a_{3},a_{4},a_{5},a_{6}\}$

$V_{1}\cap V_{2} = \{a_{3},a_{4}\}$   
This is the smallest possible dimension, which is $2.$

The largest possible dimension will be $4,$ when $V_{1} =  V_{2}$
edited by
26 votes
26 votes
First, note that V1+V2 is still in V,

so dim(V1+V2)≤ 6.

We know that dim(V1+V2)=dimV1+dimV2−dim(V1∩V2).

So 6≥dim(V1+V2)=dimV1+dimV2−dim(V1∩V2) dim(V1∩V2)≥4+4−6=2. The answer is B.
Answer:

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