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Wednesday, 24 August 2011

Math GRE - #23

Suppose B is a basis for a real vector space V of dimension greater than 1. Which of the following statements could be true?

The zero vector of V is an element of B.

B has a proper subset that spans V.

B is a proper subset of a linearly independent subset of V.

There is a basis for V that is disjoint from B.

One of the vectors in B is a linear combination of the other vectors in B.

Solution :

Let us first recall the definition of a basis:

"A basis for a vector space is a linearly independent set of vectors that spans the vector space".

B is linearly independent because it is a basis. This removes choice 5.

Because a basis is linearly independent, the zero vector cannot be part of it.
This removes choice 1.

For B to have a proper subset that spans V, there must exist vectors in B which are not linearly independent (since we can remove vectors and still span V). However, this is impossible since B is a basis (i.e. we cannot remove any vectors and still hope to span V). This removes choice 2.

Similarly, we cannot add any more linearly independent vectors to B because it already spans all of V. Thus B cannot be a proper subset of a linearly independent subset of V. This removes choice 3.

The only choice left is choice 4. Thus choice 4 is the answer (there are in fact an infinite number of bases that are disjoint from B).

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