Question

It is known that (V, +,.) is a vector space where V={(a,b,c) : a,b,c E R} and + and . are the "usual" operations on V.
Prove that H= {3t,0, 5t/2: t E R} is a subspace of V.

Answer 1

To show that a vector space is a subspace of another vector space, you have to show that \(H\) is closed under both operations (vector addition and scalar multiplication) defined for \(V\) and that both contain the zero vector.

It's clear that \(H\) contains the zero vector; this happens when \(t=0\).

To show closure under (vector) addition, take two arbitrary vectors from \(H\), add them together, and if their sum is of the form given for \(H\), then \(H\) must be closed under addition.

Similarly, to establish closure under (scalar) multiplication, take some arbitrary constant \(k\in\mathbb{R}\) and multiply it by a vector in \(H\). If the resultant vector also has the same form as any vector in \(H\), then the set is closed.

Satisfying all these conditions is required to be able to name a set of vectors a subspace of a vector space.

Answer 2

This helps a lot! thank you!!

Answer 3

You're welcome