# Subspace spanned(Is this right?)

1. Jan 25, 2010

### Fanta

so, if I wanna 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?

2. Jan 25, 2010

### 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.

3. Jan 25, 2010

### Fanta

i see. So how would you calculate it then?

4. Jan 25, 2010

### 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 $$v$$.

5. Jan 25, 2010

### Fanta

got it, thanks!