FORM GR1268 Question 4

Question:
Let \(V\) and \(W\) be 4-dimensional subspaces of a 7-dimensional vector space \(X\). Which of the following CANNOT be the dimension of the subspace \(V\cap W\)?

(A) 0     (B) 1     (C) 2      (D) 3     (E) 4

Answer:
(A)

Answer Key:
A basis of \(V\) has four vectors, and these four vectors are linearly independent.
A basis of \(W\) has four vectors, and these four vectors are linearly independent.
Because there cannot be more than seven linearly independent vectors in a 7-dimensional vector space, these eight vectors are not linearly independent.
That is, \(V\) and \(W\) has common dimension.

Here's the intuition:
We need four vectors for a basis of \(V\) and four vectors for a basis of \(W\).
But there are only seven vectors in a basis of \(X\).
So, at least one of them has to be used for both \(V\) and \(W\).

For visual learners:
Consider the usual 3-dimensional Euclidean space.
Can you make non-overlapping two 2-dimensional subspaces?
That is, can you choose two planes that go through the origin and do not share a line?
The answer is no.