Proving Subset and Subspace Properties | V is a Subspace of Rn

[negation]

Homework Statement

Show that if V is a subspace of R n, then V must contain the zero vector.

The Attempt at a Solution

If a set V of vectors is a subspace of Rn, then, V must contain the zero vector, must be closed under addition, and, closed under scalar multiplication.

Let u = (u1,u2,u3...un), Let w = (w1,w2,w3...wn) and k scalar is a real number.

for ku = (0,0,0,...0n), k = 0 (is this a valid?)

for u+w = 0, (u1+w1, u2+w2,u3+w3...un+wn) = (0,0,0,...0n)

i.e., u1+w1 = 0, then u1=-w1
if u1 = 1,then w1=-1 (is this valid?)

 

[Dick]

Homework Statement

Show that if V is a subspace of R n, then V must contain the zero vector.

The Attempt at a Solution

If a set V of vectors is a subspace of Rn, then, V must contain the zero vector, must be closed under addition, and, closed under scalar multiplication.

Let u = (u1,u2,u3...un), Let w = (w1,w2,w3...wn) and k scalar is a real number.

for ku = (0,0,0,...0n), k = 0 (is this a valid?)

for u+w = 0, (u1+w1, u2+w2,u3+w3...un+wn) = (0,0,0,...0n)

i.e., u1+w1 = 0, then u1=-w1
if u1 = 1,then w1=-1 (is this valid?)

If your definition of V being a 'subspace' includes that V contains the zero vector, then there's not much to show. Is there?

 

[negation]

If your definition of V being a 'subspace' includes that V contains the zero vector, then there's not much to show. Is there?

Well, yea. But I thought it was a good way for the practice question to brush up my conceptual understanding of subspace. It's pretty nebulous at the moment.

 

[Dick]

Well, yea. But I thought it was a good way for the practice question to brush up my conceptual understanding of subspace. It's pretty nebulous at the moment.

Ok. But you might want to pick a little more substantial practice question.

 

[negation]

Ok. But you might want to pick a little more substantial practice question.

Looking forward to doing so.