In ℝ3 consider the vectors 𝑢 = (1, −1,3) and 𝑣 = (2,1,0) Determine the space generated by the vectors 𝑢 and 𝑣, that is: 𝑊 =<{𝑢, 𝑣} >

To determine the space generated by the vectors \mathbf{u} = (1, -1, 3) and \mathbf{v} = (2, 1, 0) in \mathbb{R}^3 , we need to find all linear combinations of these vectors.

Let \mathbf{w} = a \mathbf{u} + b \mathbf{v} , where a and b are scalars.

Then, \mathbf{w} = a(1, -1, 3) + b(2, 1, 0) = (a + 2b, -a + b, 3a) .

\mathbf{w}=(a,-a,3a)+(2b,b,0)

\mathbf{w}=(a+2b,-a+b,3a+0)

\mathbf{w}=(a+2b,-a+b,3a)

The space generated by \mathbf{u} and \mathbf{v} is the set of all vectors \mathbf{w} = (x, y, z) such that there exist scalars a and b satisfying (x, y, z) = (a + 2b, -a + b, 3a) .

Therefore, the space W = \text{span} \{ \mathbf{u}, \mathbf{v} \} is the set of all linear combinations of \mathbf{u} and \mathbf{v} in \mathbb{R}^3 .

\boxed{W = \left\{ (x, y, z) \in \mathbb{R}^3 \mid \text{ there exist } a, b \text{ such that } (x, y, z) = (a + 2b, -a + b, 3a) \right\}}