# Thread: Vectors x,y in V (this seems to easy)

1. ## Vectors x,y in V (this seems to easy)

V is a vector space over a field F and A is a subspace. $\displaystyle x,y\in V$ have two lin. ind. combinations which are in A. Show $\displaystyle x,y\in A$

Let $\displaystyle \lambda_i\in F$

$\displaystyle x=\lambda_1v_1+\cdots +\lambda_nv_n\in A$

$\displaystyle y=\lambda_1v_1+\cdots +\lambda_mvm\in A$

Since x = the linear combinations in A, x is in A. Is it really this simple?

2. ## Re: Vectors x,y in V (this seems to easy)

Originally Posted by dwsmith
V is a vector space over a field F and A is a subspace. $\displaystyle x,y\in V$ have two lin. ind. combinations which are in A. Show $\displaystyle x,y\in A$

Let $\displaystyle \lambda_i\in F$

$\displaystyle x=\lambda_1v_1+\cdots +\lambda_nv_n\in A$

$\displaystyle y=\lambda_1v_1+\cdots +\lambda_mvm\in A$

Since x = the linear combinations in A, x is in A. Is it really this simple?
Hi dwsmith,

I think you have misunderstood the question. $\displaystyle x,y\in V$ have two lin. ind. combinations which are in A, means that,

$\displaystyle \lambda_1 x+\lambda_2 y\in A\mbox{ and }\lambda_1 x+\lambda_2 y\in A\mbox{ for some }\lambda_1,\lambda_2,\lambda_3,\lambda_4\in F$

3. ## Re: Vectors x,y in V (this seems to easy)

Originally Posted by Sudharaka
Hi dwsmith,

I think you have misunderstood the question. $\displaystyle x,y\in V$ have two lin. ind. combinations which are in A, means that,

$\displaystyle \lambda_1 x+\lambda_2 y\in A\mbox{ and }\lambda_1 x+\lambda_2 y\in A\mbox{ for some }\lambda_1,\lambda_2,\lambda_3,\lambda_4\in F$
Sudharaka clearly meant $\displaystyle \lambda_3x+ \lambda_4y \in A$ for the second.

However, you still have to assume that "linear combinations" means "none of the coefficients are 0" which is not standard. Otherwise you would have $\displaystyle 0x+ 1y\in A$ and $\displaystyle 0x+ 2y\in A$ which are true for any y in A whether x is in A or not.