Thread: Subspace: why?

I'm sure you seen this theorem in one wording or the other that W is a subspace of V iff: 1) W is nonempty, 2) W is closed under multiplication by scalars, and 3) W is closed under vector addition.

One of the questions in the textbook asks "Explain why a subset of a vector space must be nonempty in order for it to be a subspace."

I was guessing that if W was empty, then #2 and 3 in that theorem would be disrupted since there's nothing to multiply the scalars with in #2.

What's your take on this question?

Originally Posted by seadog

I'm sure you seen this theorem in one wording or the other that W is a subspace of V iff: 1) W is nonempty, 2) W is closed under multiplication by scalars, and 3) W is closed under vector addition.

One of the questions in the textbook asks "Explain why a subset of a vector space must be nonempty in order for it to be a subspace."

I was guessing that if W was empty, then #2 and 3 in that theorem would be disrupted since there's nothing to multiply the scalars with in #2.

What's your take on this question?

Strictly speaking, if W has no members, the requirements regarding addition of vectors and scalar product are "vacuously" true: they are of the form "if A then B" and "A" is true. Such a statement is true whether or not B is.

You must show that a set satisfying #2 and #3 does not satisfy at least one of the requirements for a subspace. The standard definition of "subspace" of a vector space is that it is a subset of the vector space that forms a vector space itself with the addition and scalar multiplication "inherited" from the vecto space. And one of the requirements for a vector space is that it have a "0" vector.