Question

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

Answer 1

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

Answer 2

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)

Answer 3

hmmmm I would go about it just a drop differently

Answer 4

\[ \left( \begin{array}{ccc}
2 & 1 & 1 \\
3 & 1 & 2 \\
4 & 1 & 3 \end{array} \right)\]
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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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).

Answer 9

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

Answer 10

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

Answer 11

\[ \left( \begin{array}{ccc}
1 & {1 \over 2} & {1 \over 2} \\
0 &1 & 1 \\
0 &0 & 0 \end{array} \right)\]
-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

Answer 12

thanks :)