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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 _____.
asked in Linear Algebra by Veteran (112k points) | 1.7k views
+1
Application of pigeon hole principle. :)
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@ Chhotu is this in syllabus ??
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Yes.
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@Warrior can u suggest a place to study this from unless nptel and reference books :p
+2

Gilbert strang videos are really helpful for Vectorspace,subspace ,basis ,dimension,spanning.Just watch lect 5 to 10.

reference:https://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/ 

0

reference

Let K be a field (such as the real numbers), V be avector space over K, and let W be a subset of V. Then W is a subspace if: The zero vector, 0, is in W. If u and v are elements of W, then the sum u + v is an element of W. If u is an element of W and c is a scalar from K, then the scalar product cu is an element of W

https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/subspace-basis/v/linear-subspaces

https://math.stackexchange.com/questions/1165726/prove-that-a-subspace-of-dimension-n-of-a-vector-space-of-dimension-n-is-the

http://math.mit.edu/~gs/linearalgebra/ila0306.pdf

2 Answers

+33 votes
Best answer
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 $V_{1} =  V_{2}$
answered by Active (3.7k points)
edited by
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From which topic, this question is?
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Topic name : Dimension for Vector space
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this type of question is part of GATE 2018 syllabus?
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Yes.
+1
This is a beautiful question
+9 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.
answered by Loyal (8.7k points)
Answer:

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