## anonymous 5 years ago Help with proving W = (1,2,4x+3Y) not a vector

1. anonymous

are there any more details to this question?

2. anonymous

x and y are real numbers and V = R3

3. anonymous

Help with proving W = (1x,2y,4x+3Y) not a vector , x and y are real numbers and V = R3

4. anonymous

not a vector or not a vector space?

5. anonymous

not a vector sub space

6. anonymous

okay, makes more sense now

7. anonymous

it appears to pass the conditions to be a subspace of R^3

8. anonymous

9. anonymous

Let $\vec{a}=(x_1,2y_2,4x_1+3y_1)$ and $\vec{b}=(x_2,y_2,4x_2+3y_2)$ be in W

10. anonymous

Now $\vec{a}+\vec{b}=(x_1+x_2,y_1+y_2,4x_1+3y_1+4x_2+3y_2)$ group and factor the third component to get $(x_1+x_2,y_1+y_2,4(x_1+x_2)+3(y_1+y_2))$

11. anonymous

since $x_1,x_2,y_1,y_2\in \mathbb{R}$ clearly this vector sum of arbitrary vectors in W is in W

12. anonymous

I should have made the statement about $x_1,x_2,y_1,y_2\in \mathbb{R}$ first of course

13. anonymous

Thank you I am reviewing to make sure I understand.

14. anonymous

would it work the same if W = x,2y,4x - 3Y

15. anonymous

it should, but I have not tried it, Proving a subspace is one of those "turn the crank" type proofs in linear algebra that may seem a little abstract at first but once yo get the process down is straight forward but sometimes tedious...

16. anonymous

Thank You

17. anonymous

IS W subspace of V, W is the set of all functions F(0)=1 , V=C($-\infty$,$\infty$)