anonymous 3 years ago How do you show that vector, x= (2,3,4)+ t1 (1,1,1) + t2 (1,2,3) is a subspace of R^n?

1. anonymous

Well for starters in order for a vector to be a subspace it must be linearly dependant

2. anonymous

Since ( 2,3,4)=(1,1,1)+(1,2,3), it follows that (2,3,4)+t1(1,1,1)+t2(2,3,4)=(t1+1)(1,1,1)+(t2+1)(2,3,4)

3. anonymous

hmmmm I would go about it just a drop differently

4. anonymous

$\left( \begin{array}{ccc} 2 & 1 & 1 \\ 3 & 1 & 2 \\ 4 & 1 & 3 \end{array} \right)$ I forgot the terms but basically x=(2+1t_1+1t_2)+(3+1t_1+2t_2)+(4+1t_1+3t_2) Now if you would row reduce this then you will see that this matrix is linearly dependant

5. anonymous

$\left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 3 & 1 & 2 \\ 4 & 1 & 3 \end{array} \right)$ 1/2R_1=R_1

6. anonymous

$\left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &- {1 \over 2} & - {1 \over 2} \\ 4 & 1 & 3 \end{array} \right)$ 3R_1+R_2=R_2

7. anonymous

$\left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 4 & 1 & 3 \end{array} \right)$ -2R_2=R_2

8. anonymous

you are going to get all 0s in the first column, with one 1 in each of the second and third columns. Thats because (2,3,4)=1(1,1,1)+1(1,2,3).

9. anonymous

$\left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 0 & -2 & -2 \end{array} \right)$ -4R_1+R_3=R_3

10. anonymous

$\left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 0 & 1 & 1 \end{array} \right)$ -1/2R_3=R_3

11. anonymous

$\left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 0 &0 & 0 \end{array} \right)$ -1R_2+R_3=R_3 As you see the last row is just zeroes this means that this matrix is linearly dependant meaning that it is a subspace

12. anonymous

thanks :)