A community for students.
Here's the question you clicked on:
 0 viewing
derrick902
 2 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?
derrick902
 2 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?

This Question is Closed

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.1Well for starters in order for a vector to be a subspace it must be linearly dependant

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.1Since ( 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)

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.1hmmmm I would go about it just a drop differently

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.1\[ \left( \begin{array}{ccc} 2 & 1 & 1 \\ 3 & 1 & 2 \\ 4 & 1 & 3 \end{array} \right)\] 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

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.1\[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 3 & 1 & 2 \\ 4 & 1 & 3 \end{array} \right)\] 1/2R_1=R_1

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.1\[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 & {1 \over 2} &  {1 \over 2} \\ 4 & 1 & 3 \end{array} \right)\] 3R_1+R_2=R_2

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.1\[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 4 & 1 & 3 \end{array} \right)\] 2R_2=R_2

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.1you 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).

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.1\[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 0 & 2 & 2 \end{array} \right)\] 4R_1+R_3=R_3

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.1\[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 0 & 1 & 1 \end{array} \right)\] 1/2R_3=R_3

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.1\[ \left( \begin{array}{ccc} 1 & {1 \over 2} & {1 \over 2} \\ 0 &1 & 1 \\ 0 &0 & 0 \end{array} \right)\] 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.