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 ?

Any help is appreciated.

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- May 21st 2009, 11:29 PMRoamBasis
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 ?

Any help is appreciated. - May 22nd 2009, 12:28 AMscorpion007
Find 3 more linearly independent vectors to form a basis for . (You are given 2).

Recall that it is sufficient to find linearly independent vectors to form a basis for . - May 22nd 2009, 12:08 PMRoam
- May 22nd 2009, 12:23 PMIsomorphism
- May 22nd 2009, 12:38 PMSydneyBristow
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. - May 22nd 2009, 08:41 PMRoam
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 ?

To find out whether these vectors are linearly independent together I must test them. So, for scalars , in the linear combination:

There can only be one solution to the scalars,

I know the answer to this question isn't unique but is my answer correct? Is this a right set (and a basis for )? - May 22nd 2009, 09:56 PMscorpion007
Yes. (You can verify it by solving the system you mentioned)