Hi!
I don't understand how we can answer the following question.
How do we: extend the linearly independent set {(1,1,1,1,1),(1,2,3,4,5)} to form a basis for $\displaystyle R^5$ ?
Any help is appreciated.
Well the most simple basis to R^5
is the following 5 vectors because they are all linearly independant.
(1,0,0,0,0)
(0,1,0,0,0)
(0,0,1,0,0)
(0,0,0,1,0)
(0,0,0,0,1)
You have two vectors, you need 3 more that are linearly independant.
You can just easily use 3 of those or if you want to find more vectors you can u can just do trial and error.
The vectors you listed are probably linear combinations of the other vectors. You can check that by doing
v5= c1v1+c2v2+c3v3+c4v4
Usually if the vector contains a 0 component it is independant to a vector with numbers bc the c value would have to be 0 to make the vector 0 and if that happens it is unlikely to be a linear combination of the other vectors unless the other vectors have 0 in the same place.
Okay, so can I just add the vectors {(0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1)} to the set in order to extend the linearly independent set {(1,1,1,1,1),(1,2,3,4,5)} to form a basis for $\displaystyle R^5$?
To find out whether these vectors are linearly independent together I must test them. So, for scalars $\displaystyle c_{1},c_{2},c_{3},c_{4},c_{5} \in R $, in the linear combination:
$\displaystyle c_{1} (1,1,1,1,1) + c_{2}(1,2,3,4,5) + c_{3} (0,0,1,0,0) + c_{4}(0,0,0,1,0) + c_{5}(0,0,0,0,1)$
There can only be one solution to the scalars, $\displaystyle c_{1},c_{2},c_{3},c_{4},c_{5} = 0$
I know the answer to this question isn't unique but is my answer correct? Is this a right set (and a basis for $\displaystyle R^5$)?