Vector spaces Problem

**quah13579** · August 7th 2010, 08:47 PM

Let V1 = span {(1 0 2)}, and V2 = span{(0 0 1)} (they are colomn, not row)

Observe that V1 and V2 are both vector spaces. A new set of vectors, S, is constructed by taking any vector from V1 and any vector from V2 and adding these 2 vectors together.

why S will also be a vector space?
How to find a basis for S and the dimension of S?

**HallsofIvy** · August 8th 2010, 02:43 AM

In other words, $S= \{w |w= u+ v, u\in V_1, u\in V_2\}$ .

To show that a subset of a vector space is a subspace, you only need to show that the set is closed under addition and scalar multiplication of vectors. That is that if $w_1$ and $w_2$ are in S, so are $w_1+ w_2$ and $\alpha w_1$ where $\alpha$ is any scalar.

That can be done most conveniently, in one step, by showing that $\alpha w_1+ w_2$ is in the same set.

Well, if $w_1$ is in that set then $w_1= u_1+ v_1$ for some $v_1\in V_1$ and some $v_1\in V_2$ . If $w_2$ is in that set, similarly, $w_2= u_2+ v_2$ .

Now, $\alpha w_1+ w_2= \alpha(u_1+ v_1)+ (u_2+ w_2)= (\alpha u_1+ u_2)+ (\alpha v_1+ v2)$ . Do you see why that is again in S?

Any vector in S is of the form $w= u+ v$ . Since u is in the span of (1, 0, 2), $u= \alpha(1, 0, 2)$ and since v is in the span of (0, 0, 1), $v= \beta(0, 0, 1)$ . That means that any vector in S can be written $\alpha(1, 0, 2)+ \beta(0, 0, 1)$ . You should be able to see from that what the dimension is and a basis.

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