Calculate Subspace Spanned by A in R^3 using Linear Combinations | Plane x = z

[Fanta]

so, if I want to calculate the subspace spanned by A in:

A = {(1,0,1) , (0,1,0)} in R^{3}

c_{1}(1,0,1)+c_{2}(0,1,0) = (x,y,z)

i can make a system:

c_{1} = x

c_{2} = y

c_{1} = z

from which I can conclude that x = z, and so, the subspace spanned will be the plane given by x = z.

Is this right?

 

[ystael]

This is half a proof. You have shown that the subspace S spanned by \{(1,0,1), (0,1,0)\} is contained in the plane P = \{(x, y, z) \in \mathbb{R}^3 : x = z \}. You also need to show that S contains P. You could do this by direct calculation, or by an argument based on dimension.

 

[Fanta]

i see. So how would you calculate it then?

 

[ystael]

You showed that S \subset P by showing that if v \in S, that is, v is a linear combination of (1,0,1) and (0,1,0), then v \in P, that is, the first and third coordinates of v are equal.

Show that S \supset P by proving the reverse implication: if v \in P, that is, if the first and third coordinates of v are equal, then v \in S, that is, v is a linear combination of (1,0,1) and (0,1,0). You should be able to exhibit explicitly the coefficients in this linear combination, using the components of v.

 

[Fanta]

got it, thanks!